Mr. Nakamura discussed a version of L'H with you:

L'Hôpital's Rule (version 1, for 0/0)
Suppose that f and g have continuous derivatives, and that f(a)=g(a)=0 (the eligibility criterion). If lim_x→af´(x)/g´(x) exists, then lim_x→af(x)/g(x) exists and is the same value.

Comment 1 It is very important to check the eligibility criteria. For example, if you try to compute lim_x→2x²/(5x+7) by instead computing lim_x→22x/5 you will report the false answer 4/5 instead of the correct answer 4/17.

Comment 2 L'H is so appealing maybe because it looks like what we all believe the Quotient Rule should be. It is a version of paradise, maybe. (More theology is below, so please just wait.) That's why people want to use it, even when they shouldn't.

The L'H above is called the 0/0 ("zero over zero") form. There are other forms. For example, one version is very useful in certain practical applications.

L'Hôpital's Rule (version 2, for ∞/∞)
Suppose that f and g have continuous derivatives, and that f(a)=g(a)=∞ (the eligibility criterion). If lim_x→af´(x)/g´(x) exists, then lim_x→af(x)/g(x) exists and is the same value.

scale, then one might consider that x could be in nanoseconds, so large values of x could be more relevant than one might initially suspect.

One way to compare these functions is to consider their ratio and see what happens when x gets large. So we looked at lim_x→∞x^10,000/e^.001x. Of course people said that the top→∞and the bottom→∞ separately. So this is eligible for L'H. The result, after separate differentiation of the top and bottom is
lim_x→∞[10,000x^9,999]/[.001e^.001x]. Is this any improvement? A completely naive person might not see much improvement, but again this is eligible for L'H since the top and bottom both→∞. So the separate differentiations, done carefully, give
lim_x→∞[10,000·9,999x^9,998]/[(.001)²e^.001x]. Now I hoped that people would see a pattern.

(L'H)^10,000
We can compute lim_x→∞x^10,000/e^.001x by applying L'H ten thousand times in succession. I hope that you can see the result, almost. After 10,000 separate differentiations of the top and bottom we need to evaluate
lim_x→∞(10,000!)/[(.001)^10,000e^.001x]
Here by 10,000! I mean the product of the integers 10,000 and 9,999 and 9,998 and 9,997 and ... all the way down to 1. This is called a factorial and you will see many factorials in Math 152. Actually, for the purposes of this computation I don't care much about specific constants, and really I think of the limit as
lim_x→∞CONSTANT₁/[CONSTANT₂e^.001x]
because then it is clear that the exponential, which is really growing and growing and growing will make the limit equal to 0.

The final result
So, in fact, eventually e^.001x grows faster than x^10,000.

Angels and humans
We could think of functions with exponential growth, so e^Ax with A>0, as angels, and functions with power of x growth, x^A with A>0, as humans. (If A happens to be an integer we get a monomial, and sums [families?] would be polynomials.) So if you think about it, eventually any angel is stronger than even the most powerful human.

Another not so obvious use of L'H
Here was another pair of functions we considered.

Formulas for the functions	Value when x=5	If 1 unit is an inch, then ...
(ln(x))300	1.005·1062	This is only about 1.7·1044 times the theoretical diameter of the universe!
x1/3	1.7099...	Almost two inches!

We use L'H to compare the rates of growth. So look at lim_x→∞[(ln(x))³⁰⁰]/x^1/3. You need to really believe in (?) logs and powers of x, but, if you do, you should see that the top and bottom both →∞separately. So apply L'H by computing derivatives, and this limit results:
lim_x→∞[300(ln(x))²⁹⁹(1/x)]/[(1/3)x^–2/3]. This computation seems more intricate to me than the earlier one. The Chain Rule is needed. The result is a compound fraction, and here it is worthwhile to simplify the compound fraction. The (1/x) makes an additional power of x appear "downstairs" and we get (after we recognize that 1+(–2/3)=1/3 !)
lim_x→∞[300(ln(x))²⁹⁹]/[(1/3)x^1/3]. Again L'H applies (yes, we check the eligibility!) and the result is
lim_x→∞[300·299(ln(x))²⁹⁸](1/x)/[(1/3)²x^–2/3]. Look! 1/x is on top, which changes the power on the bottom from x^–2/3 to x^1/3. The structure of the computation should be observed.

(L'H)³⁰⁰
We can compute lim_x→∞[(ln(x))³⁰⁰]/x^1/3 by using L'H three hundred times in a row, each time carefully (!) checking (!!) the resulting quotient for the eligibility (!!!) of the top and bottom. Ha. The result will be lim_x→∞(300!)/[(1/3)³⁰⁰x^1/3] and this easily has limit equal to 0. Maybe it is amusing to notice we can get the exact values of certain constants, but for the final result of this computation, the exact values of the constants don't matter.

The final result So, in fact, eventually x1/3 grows faster than (ln(x))300. And now demons ... Well, if humans are postive powers of x, then maybe demons could be any powers, no matter how large, of logs. So one feeble human might be x.000005 and one ferocious demon might be (ln(x))600,000. Eventually even the most feeble human will overpower the most terribly ferocious demon. Moral (?) lessons concerning this hierarchy A former student, Mr. Harrison, gave me, almost casually, the lovely definition of hierarchy as "stratified layers of power". It is certainly appropriate. Anyway, below is an effort by another former student, Mr. Park to note my comments, and to the right is a picture. I like pictures.

This ends discussion of material eligible for the second
exam which will be given one week from this lecture.

Another story

A line segment joins the points (3,0) and (0,2) in the first quadrant. A rectangle in the first quadrant with sides on the x- and y-axes sits inside the resulting triangle, and has a vertex on the line segment. What is the largest area that such a rectangle can have?

Comment
Wel, darn it, I tried to write the specifications very carefully. After class, Mr. Orrico and Ms. O'Sullivan convinced me that my paragraph was not carefully written enough. I think they were correct. Instead of the phrase "sits inside the resulting triangle, and has a vertex on the line segment", I had written "[has] a vertex on the line segment". In that case (future lawyers should follow this carefully!) a rectangle with vertices at (0,0), (3,0), (3,2), and (0,2), that is a rectangle with the line segment as a diagonal will have all the other eligible rectangles inside it. So it would definitely be the largest. This isn't what I wanted. So my specification was not written precisely enough. Let me go on with the changed specification written above. Probably the first thing I would do when meeting a problem of this type is to draw a picture. I think my sketch would look something like what's to the right. Certainly, I would need the line segment joining the points (3,0) and (0,2) in the first quadrant which is drawn. Then I likely would try to draw a typical rectangle. Although I tried to write the specification of the rectangle in a direct way, my desire to be brief may have made this difficult to understand. The rectangle has sides on the x- and y-axes as shown. It also has a vertex on the line segment. Legitimate questions include, "What's a vertex?" and "What line segment?" I hope, even if you've not heard the word before or this use of the word, that you would know "vertex" means a corner of the rectangle. And the first sentence of the problem defines "the line segment".

Almost surely in my mind I would play around with the problem as pictured, and try to see some of the eligible rectangles. The pictures I would look at would include the two below. If the vertex were very near (0,2), as in the left picture, I see that one side of the rectangle would be almost 2 units long, while the other is very small. The area of that rectangle would be small. On the other hand (right picture), if the vertex were very near (3,0), one side of the rectangle would be almost 3 units long, while the other is very small. Again the area would be small. This encourages me to think that, yes, there is a maximum and almost surely the max area will occur for some rectangle with vertex "inside" the segment (at a critical point of the algebraic model we will build).

Two "extreme" rectangles
Small area	Also small area

What to do ...
I mentioned in class that there is sort of a translation process, going from the problem statement and creating a calculus problem. I find this, even though I have a great deal of experience, to be difficult. I generally need to read the problem statement very carefully, and usually I need to read the statement several times. Then there is an analysis step, where the calculus problem is solved. This is usually more straightforward. It isn't simple, because of course the opportunity for error is always present: arithmetic errors, algebra errors, calculus errors. Certainly I have made all of these, but somehow those can be fixed, at least for me, more easily while the translation process definitely stubbornly proves to be more difficult, at least for me.

Building the algebraic model
We will need the equation for the line segment. In class I tried to describe how I would "guess" this equation. Since this is top secret and also silly, I will just write it here: (x/3)+(y/2)=1. Well, o.k., my guessing method goes like this: the equation will involve both variables since the line is neither vertical (x=const) nor horizontal (y=const). Therefore it could be written (x/something)+(y/something else)=1. Then plugging in both (0,2) and (3,0) give me the values of something and something else. What about the area of the rectangle? If the coordinates of the point p are (x,y), then the area is xy.
I'm almost "there". The objective function is xy. The constraint situation certainly involves (x/3)+(y/2)=1 but there is just a bit more. Notice we are in the first quadrant, so that both x and y are non-negative. Now my model follows:
    Constraint(x/3)+(y/2)=1 and x≥0 and y≥0.
    Objective Maximize xy.
As before I'll use the constraint to reduce the number of variables in the objective. So (x/3)+(y/2)=1 becomes y=2[1–(x/3)] and xy becomes 2x(1–(x/3)). But don't forget the remainder of the constraint, which becomes a domain restriction: x is in the interval [0,3].

The resulting calculus problem
Find the max value of f(x)=2x(1–(x/3)) when x is in the interval [0,3]. We know that the maximum of f on [0,3] occurs either at endpoints or critical points.
• Endpoint evaluation f(0)=2·0(1–0)=0; f(3)=2·3²(1–(3/3))=0. I bet the max is inside, at a c.p.
• Critical point analysis Since f(x)=2x(1–(x/3)) I know that f´=2(1–(x/3))+2x(–1/3)=2–(4/3)x. Therefore the only c.p. is where 2–(4/3)x=0 so x must be 3/2. f(3/2)=2(3/2)(1–(1/2)) is positive, so this is the maximum value. Yes, it "simplifies" to 3/2. The answer to the problem is 3/2. Notice that the problem requests the largest area. We should attempt to answer the question that is asked (so remarking that the dimensions of the rectangle are ... and ... is not responsive!).

You should ...
Do 8 or 10 of these problems yourself, possibly together with other students. Certainly what I wrote about is much more than I expect you to write, but I would hope that, after sufficient experience, you would think through such problems in much the same way as what's above. I wanted to help you learn the process.
Practice is essential to your own ability to answer questions of this type. The "translation" skill, that is, constructing the correct mathematical model, is very important in applications.

A story about smallest

A line segment joins a point on the positive x-axis with a point on the positive y-axis. The line segment goes through the point (3,2) and creates, with the appropriate parts of the x- and y-axes, a triangle. Find the smallest area of such a triangle.

We discussed this a bit. We sketched the geometric situation relatively quickly, but then students seem to have found making the transition from the geometry to an algebraic description not so easy. There's a diagram to the right. It shows a "typical" line segment which joins a point on the positive x-axis with a point on the positive y-axis and which also goes through the point (3,2).

As I wrote in a discussion of a previous geometric "story", I would play around with the problem. When the point on the x-axis is very close to (3,0), then the line would tilt "way up", very steeply. The length of the line segment would be large and the area would be large also (the length of the base →3 while the height →∞. The area, which is (1/2)(base)(height) would therefore be large. Similarly, if we were to pull the point on the x-axis far away from (3,0), the slope would be close to 0, and the length of the line segment would be large. The length of the base →∞ while the height →∞. Again, the area would be large.

This seems to suggest that the area varies, and that "somewhere in the middle" there is a minimum, and the minimum will occur at a critical point. To the right is a very vague picture of x and A(x), the area of the triangle when (x,0) is one end-point of the segment. Here we note that the x's are in the interval (3,∞) because otherwise we won't get a triangle -- the line segment won't hit the positive y-axis. Also, I drew the graph so that as x→3⁺ the area function gets very large, and a similar behavior occurs as x→∞.

How about the transition from picture to an algebraic description? The area is (1/2)xy, the Objective Function. How can we relate x and y (the Constraint)?

If a diagram of the situation is labeled, then some restrictions almost yell at the reader. Look to the right. Certainly x>3 and y>2. And there are bunch (well, three) similar triangles which give some further relationships between the variables. For example, the big triangle and the lower right small triangle tell me that y/x=2/(x-3), so that y=(2x)/(x-3). And therefore the function we need to minimize is f(x)=(1/2)(x)(2x)/(x-3). What is the domain of this function when used to describe this problem? Certainly, x>3. There is no other restriction. Now we have turned this into a calculus problem.

A calculus problem
What is the minimum of f(x)=x²/(x-3) for 3<x<∞? (I canceled the 2's and combined the x's.)
• Endpoint evaluation There are no endpoints! But I would still begin with some sort of analysis or understanding of what happens out towards the edges. So maybe I should call this
    Edge analysis What happens as x→∞? Well, lim_x→∞f(x)=lim_x→∞sqrt(x²/(x-3)]. When x gets very large, we have a degree 2 polynomial on top and just a degree 1 polynomial on the bottom. We analyzed such situations before, and the result is that x→∞.
Now what happens as x→3⁺? Well, x²→3². But 1/(x-3) is like 1/(something small and positive), which is large and positive. So I bet that the product of these two →∞, just as we should expect from our previous thoughts.

Comment I like the idea of wearing both belt and suspenders to keep my pants from falling down. Well, maybe not exactly that, but I certainly do like the idea of reinforcing my geometric understanding of the problem with another, algebraic approach. Both methods should be used! Also, this way, I can be sure that if we identify critical points, the resulting value(s) of the function will include the minimum, and won't be a maximum. As I mentioned in class, there have been some interesting real world failures of engineering projects where this identification (max vs min) was neglected.

• Critical point analysis If f(x)=x²/(x-3) then f´(x)=[(2x)(x-3)-(1)x²]/(x-3)² (Quotient Rule). Then f´(x)=0 when the top is 0, so we need to solve 2x(x-3)-(1)x²=0. We can factor this (it is a toy problem!) and get x[2(x-3)-x]=0 which has roots x=0 (first factor) and x=6 (second factor). We can discard 0 since that is not in the domain for this problem. Therefore the minimum area occurs when x=6 (we only know this is a minimum because we did the edge analysis, otherwise we'd need more work), and f(6)=36/3=12.

Resembles...?
The last two stories had pictures which resembled each other quite a bit. We minimized an area, and we maximized an area. The two problems are actually related to each other. They are examples of what are called dual problems. In economics, such problems occur when we want to maximize profit. This might be the same as minimizing cost. In physics, the ideas are minimizing work and maximizing potential energy.

My darling, struggling in the ocean!
The last story for class.

I am standing on a straight beach and my darling is swimming in the ocean, a quarter mile from shore. The closest point to my darling is two miles down the beach. The sharks attack, and I must get to my darling as soon as possible. I can run a mile in 10 minutes and swim a mile in 40 minutes. How can I get to my darling in the least time?

As I mentioned in class, this is actually not such a toy problem. Similar problems arise in optics frequently: minimizing travel time when the speed of light in different materials varies. I've attempted to sketch the situation, as seen from "above". How can I get from my initial position to my darling?
Pure strategy #1 Swim all the way! Swim directly. The distance is sqrt((1/4)2+22), about 2.01555 miles. At 40 minutes per mile, this takes about 80.6226 minutes.
Pure strategy #2 Run as much as possible. I run 2 miles down the beach, and then swim. So the 2 miles running take me 20 minutes, and the 1/4 mile swimming takes 10 minutes. The total time is 30 minutes.
A "mixed" strategy? Run for a while, then swim. It is not clear but maybe some blend of the two is faster. So I could run part of the way, and then swim directly to my darling.
Suppose the "breakpoint" between these two activities is x miles from the point on the beach which is closest to my darling. Then I'd run 2-x miles, which would take 10(2-x) minutes. I would need to swim the length of the hypotenuse of a triangle with legs x and 1/4 long: that's sqrt(x2+(1/4)2) miles, and that would take 40sqrt(x2+(1/4)2) minutes. The total time would be f(x)=10(2-x)+40sqrt(x2+(1/4)2). The domain of interest is [0,2].

Amazing!!! To the right is a fairly careful graph of f(x) (which I requested as the QotD). Notice that there is a critical point in the interval [0,2]. The critical point is close to 0. I used my "friend" Maple to find the critical point. Partly this is because I am lazy, but it is more because I am tired. This computation could be done by hand, because the most "difficult" part of it is just solving a quadratic question. The first instruction defines f as the algebraic mess we have above. Maple echos the definition so that I can check it is what I want (I make lots of typing errors!). The second instruction differentiates this formula. The third instruction, which uses the word solve, sets the previous expression (that is what % means) equal to 0. The answer is exact and comes from the quadratic formula. Then I substituted this into the expression for f. Since I didn't "understand" the expression, I used evalf to find a 10 digit approximation to the answer. > f:=10*(2-x)+40*sqrt(x^2+(1/4)^2); 2 1/2 f := 20 - 10 x + 10 (16 x + 1) > diff(f,x); 160 x -10 + -------------- 2 1/2 (16 x + 1) > solve(%); 1/2 15 ----- 60 > subs(x=%,f); 1/2 1/2 1/2 15 2 16 15 20 - ----- + ------------- 6 3 > evalf(%); 29.68245837 At the all swimming endpoint, the time needed was about 80.6226 minutes. At the most running endpoint, the time needed was 30 minutes. The ideal strategy (at least for minizing time!) gets me to my darling in less time than that: 29.6824 ... minutes. I will certainly happily admit that this is not a great difference in time. But you should see that there is a difference, and in other problems the difference might be significant. The speed of light in vacuum (air is about the same) is 299,792,458 meters per second (I copied this!). This speed is frequently called c. "Denser media, such as water and glass, can slow light much more, to fractions such as 3/4 and 2/3 of c." This difference is responsible for light "bending" or refracting, because light travels (!) to minimize time.

L'Hôp
This is the next topic in the syllabus. L'Hôpital's (also L'Hopital's or L'Hospital's) Rule is a method of evaluating certain very special limits. It is worth showing to you because it is another neat application of "local linearization", and some really important limits become easy to evaluate. However, it must be used with some care. One quote I found on the Internet declared, "Giving l'Hopital's Rule to a calculus student is like handing a chainsaw to a three year old."

The example I was going to begin with is lim_x→0[sin(9x)/(e^6x-1)]. "Plugging in" gets no information, since we have 0/0. By the way, 0/0 limits can have strange and varied behavior. For example, consider as x→0 the three expressions x²/x and x/x and x/x². They all result in 0/0 when we just plug in. But if you think about them a bit, the first has limit=0, the second has limit=1, and the limit for the third expression does not exist. So 0/0 can conceal lots of different behaviors.

You'll learn more about L'H on Monday.

Graphing a function
I gave data about a function in a rather strange way.

• f is a differentiable function whose domain is all real numbers except for –2 and 1.
• f´ is positive only in the intervals x>3 and –2<x<0.
• These limits are known:
  lim_x→1^–f(x)=–∞; lim_x→1⁺f(x)=+∞; lim_x→–2⁺f(x)=–∞;
  lim_x→–2^–f(x)=–∞; lim_x→+∞f(x)=–1; lim_x→–∞f(x)=0.

Thinking about the graph of y=f(x)
The process is important. I'll try to go slowly through every bit of information. I may make mistakes, and I may have to fix things up. I'll only be able to sketch somethng which is qualitatively correct.

Fact f is a differentiable function whose domain is all real numbers except for –2 and 1. Response I'd probably try to indicate somehow to myself that the graph I'll draw should not appear when x=–2 and on x=1.
Fact f´ is positive only in the intervals x>3 and –2<x<0. Response f is increasing in those intervals where the derivative is positive. An important word in the "Fact" sentence is only. To me this says that f is increasing only in the intervals from –2 to 0 and from 3 upwards. In interval notation, these are (–2,0) and (3,∞). Probably f will be decreasing in other intervals.
Fact limx→1–f(x)=–∞. Response The symbols x→1– mean x is getting close to 1 from the negative (left) side. And then –∞ means that f(x) is getting large, but large negative, so the graph is going down. What I would expect is something like what is shown here.
Fact limx→1+f(x)=+∞. Response Here we have x→1+ and we should consider x getting close to 1 from the right, positive side. The +∞ means that f(x) will be getting very large positive. The direction of approach and the largeness combine to get a sort of piece of a graph which looks like what is shown.
Fact limx→–2+f(x)=–∞; limx→–2–f(x)=–∞. Response I'll do these next two as a pair. The notation is somewhat intricate here. We've got x→–2+ and x→–2–. So x is getting close to –2 from the right (+) and from the left (–). The result in both cases is –∞. The f(x) is getting large negative as x gets close to –2.
Fact limx→+∞f(x)=–1. Response This is new (at least in the lectures). The x→+∞ indicates the idea that x is traveling far to the right. As it does, f(x), the y coordinate on the graph, is getting close to –1. I don't really know exactly what I should draw. I drew two possible candidate graphs in magenta. Both of the pieces of curves drawn satisfy the limit statement (there are, as I mentioned, even other possibilities to draw). BUT we can select exactly one as a valid candidate using other information. This piece of y=f(x) is drawn in a region where the derivative is positive, so the graph should be increasing. Therefore we can drop the top candidate and the other, wiggly (?) possibilities.
Fact limx→–∞f(x)=0. Response We've gotten to the final limit statement. Here x→–∞ means that x is moving "far" to the left. And the =0 tells me that the graph is getting close to the x-axis. Again I've drawn two possible candidates. Which one will be appropriate? Here we need knowledge once more about {in|de}creasing behavior to make a selection. We are outside of the region where we know f(x) is increasing, and since we had that word "only" we should expect in this region that f(x) is decreasing. The lower alternative, the bottom candidate, is the one we should select. Please note that in the behavior on the right, I have drawn only the surviving candidate as supported by the previous picture's discussion.
Further discussion We've got to complete the graph. Let me move from left to right. I know that the graph must be differentiable and therefore also continuous. In the region to the left of –2, I think the graph should be decreasing. I have indicated a tentative way of joining the two pieces we've already drawn, and you should check that what's suggested is indeed decreasing. Now consider the region between –2 and 1. This is a bit more complicated. We need to be increasing until 0, and then, probably, decreasing. And we should draw a continuous graph which connects what the limit statements gave us. So I have tried to suggest a good candidate. Notice, please, that although I am fairly sure f has a local maximum when x=0, I really have no idea where the local maximum is (I don't even know if the value is positive or negative). I've only been given qualitative information. So my suggestion is only one of many which would be consistent with what's required.
And more discussion What happens between 1 and 4 in the picture? To the right of 3, that is, for x>3, f should be increasing. So we guess something like the magenta piece shown. And to the left of 3, in the region with 1<x<3, we know that f is not increasing, so probably (!) it should be decreasing. The result could be what is shown in this magenta piece. but I've inserted large question marks because something strange has happened: if we accept both suggestions as drawn the result will certainly not be continuous. The graph should not have a jump in it. How can we fix this? You need to think a bit.
The last piece A curve which completes the graph validly, satisfying all of the specifications, is shown to the right. From 1 to 3, the graph decreases, dropping below the horizontal asymptote y=–1. Then, at x=3, the curve begins increasing, and blends into the required asymptotic behavior as x→+∞. This graph has two vertical asymptotes (x=–2 and x=1) and two horizontal asymptotes (y=0 and y=–1). To the right is sketched one correct answer to this problem. Again, the problem is mostly qualitative, and there could be many specific graphs which are correct answers to the problem. I do know that for any solution to this problem, there will be two critical points. There will be one at 0 which will be a local max, and there will be one at 3 which will be a local min. Additionally, there will be at least one inflection point, somewhere to the right of 3. (As I mentioned in class, we could imagine the curve wiggling a bit, not changing its {in|de}creasing behavior, so I can't declare that there will be only one inflection point -- there will be at least one.)

Comment
I began the "construction" of the previous problem by sketching a graph and working backwards. I "read off" the limit statements from the graph. If you just wrote limit statements at random without starting from information that you knew was consistent, the resulting specifications might be impossible to fulfill. Also, I really believe I could create a function defined by an algebraic formula which had a graph similar to what was drawn above.

Limits at +/–∞
I wanted to give examples of some of the algebraic manipulations resulting in horizontal asysmptotes which you should know about. So here they are.

1. What is limx→∞[x3+7x–17]/[x4+3x+9]? Here I would take the fraction [x3+7x–17]/[x4+3x+9] and multiply the top and bottom by 1/x4. The result is a fraction equal to the original. If the algebra is done carefully, this result is [{1/x}+{7/x3}–{17/x4]/[1+{3/x3}+{9/x4}]. Now consider carefully and separately the pieces of this fraction. These terms: {1/x} and {7/x3} and –{17/x4 and {3/x3} and {9/x4 are →0 as x→∞. The only term that "survives" is the 1 on the bottom. I think that the limit is [0+0+0]/[1+0+0], and this is 1.
I would like to illustrate what's happening with a graph. There are some difficulties in presenting the information graphically. I'll give two pictures. One of them is what we "think" the situation looks like. This graph is to the right. Please look carefully at the scales on the horizontal and vertical axes. They are not the same. x goes from 4 to 20 and y goes from 0 to 0.35. There is a disproportion -- a factor of 45 to 1. That's remarkable. Below is the graph shown with its true proportions. Look at how flat it is! Generally I will show you pictures similar to what is on the right. The flat picture below doesn't "show" me very much.
2. What is limx→∞[5x4+2x2+33]/[x4+3x+9]? Here divide again top and bottom by x4. The result is [5+{2/x2}+{33/x4}]/[1+{3/x3+{9/x4}]. What happens here is slightly different. A bunch of terms ({2/x2} and {33/x4} and {3/x3 and {9/x4}) still go individually to 0 as x→∞. But there are two non-zero terms, one each in the top and bottom. As x→∞, the ratio→[5+0+0]/[1+0+0]. The limit should be 5. To the right is a graph, about as good as I can have the machine draw for this example. It is a graph of the ratio between 4 and 20, with y between 4 and 5. The graph gets quite close to 5 rapidly. It is not clear to me how helpful this picture is.

3. What is limx→∞[8x5–5x3+88x]/[x4+3x+9]? In this case the degree of the top is one greater than the degree of the bottom. Dividing the top and bottom by x4 gives [8x–{5/x}+{88/x3}]/[1+{3/x3}+{9/x4}]. The bottom is 1+{3/x3}+{9/x4} and I think as x→∞ the bottom gets close to 1 rather rapidly. The top is [8x–{5/x}+{88/x3}. The second and third terms are negligible as x gets large, and the top seems to be nearly 8x. I think that the quotient will behave as 8x/1 for large x, and the limit will be ∞. A graph is shown to the right. Please notice that in this case, x goes from 0 to 10 and y, from 0 to 80. The graph appears to resemble closely a straight line of slope 8, which is what the algebraic manipulation suggests.

A weird one ...
I also consider the function f(x)=(5x+7)/sqrt(x²+3). Here let me begin by displaying a graph, since the algebraic analysis did not proceed very well in class. Below is a graph of y=f(x) for x between –30 and 30. Also on display (the dashed green lines) are the horizontal lines y=5 and y=–5. Look at the graph, and observe (I hope!) that the limit as x gets large positive of f(x) seems to be 5, and the limit as x gets large negative of f(x) seems to be –5. The situation is more complicated than the previous examples.

What's happening? Look at sqrt(x²+5) and "massage" it algebraically. Well, sqrt(x²+5)=sqrt(x²·(1+{5/x²})).

Square root facts
The function which is sqrt Sqrt is a function whose domain is [0,∞) and whose range is [0,∞). Example The value of sqrt(4) is 2. The equation x2=4 has two roots. We could indicate these two roots by writing +/–sqrt(4), but sqrt(4) without any "decoration" always means 2.
Multiplication works well sqrt(A·B)=sqrt(A)·sqrt(B) if both A and B are non-negative. Example sqrt(400)=sqrt(4·100)=sqrt(4)·sqrt(100)=2·10=20.
Addition does not work with square root There is no simple relationship between sqrt(A+B) and sqrt(A)+sqrt(B). Example sqrt(25)=5 and sqrt(16)=4 and sqrt(9)=3. Although 16+9=25, notice that 3+4=7 which is not equal to 5. That is: sqrt(16+9) and sqrt(16)+sqrt(9) are very different.

Therefore sqrt(x²·(1+{5/x²})) is the same as sqrt(x²)·sqrt(1+{5/x²}). Please notice that if x>0, then sqrt(x²) is the same as x. So:
If x>0 then f(x)=(5x+7)/sqrt(x²+3)=(5x+7)/[x·sqrt(1+{5/x²})]. Divide the top and bottom by x and the result is (5+{7/x})/[sqrt(1+{5/x²})] and I hope that now you can see the limit as x→∞ is 5.

If x<0 then sqrt(x²) is the same as –x. You can check this: try, say, x=–5 and see what happens. So look at f(x):
f(x)=(5x+7)/sqrt(x²+3)=(5x+7)/[–x·sqrt(1+{5/x²})]. Dividing top and bottom by x gets (5+{7/x})/[–sqrt(1+{5/x²})]. As x→–∞, this becomes –5.

Damped oscillation The function f(x)=sin(x)/x is actually something I'm more comfortable discussing with engineering students, since things that it "models" are easily observed. As x→∞, the bottom, x, grows, and the top, sin(x), is caught between –1 and 1. This function should approach 0. In fact, an appropriate version of the Sandwich Theorem can be used. By this I mean, please, just realize that:     –1/x≤sin(x)/x≤1/x. Since both –1/x and 1/x →0 as x→∞ f(x), which is caught in between, also approaches 0. A graph of y=f(x) is shown to the right. The "envelope curves", y=+/–1/x, are dashed green curves. You will study oscillations whose amplitude go to 0 (think of a spring vibrating in a viscous fluid).

A story about numbers

The sum of two non-negative numbers is 20. How should the numbers be selected so that the product of the square of one with the other is largest?

So the two non-negative numbers will be called x and y. We know that x≥0 and y≥0. We also know that their sum ... is 20 so x+y=20.
We want to know How should the numbers be selected to make something largest. What's the quantity we should try to maximize? It is the product of the square of one with the other and this is, I think, x²y.

The problem can be translated to the following algebraic statement:
Suppose x+y=20 and x≥0 and y≥0. Find the maximum value of x²y.
In economics and some other disciplines, the first sentence is called the constraint. It relates the variables of the problem and restricts which values should be considered. It usually is related to the appropriate domain of the problem. The second sentence is the objective function. It describes what should be "extremized" -- in this case, what should be maximized or made largest. Here the objective involves two variables, x and y. But the constraint tells us that y=20–x. And therefore the problem description changes.

Suppose f(x)=x²(20–x). The domain for this problem is [0,20]. How can we get the largest value of f?
Notice that the constraint, which shouldn't be forgotten, has resulted in the domain statement when we write this version of the problem.
Now we can use calculus. The maximum of f on [0,20] occurs either at endpoints or critical points.
• Endpoint evaluation f(0)=0²(20–0)=0; f(20)=(20)²(20–20)=0. I bet the max is inside, at a c.p.
• Critical point analysis If f(x)=x²(20–x), then f´(x)=2x(20–x)–x². This is 0 when =2x(20–x)–x²=0 or x(40–2x–x)=0 or x(40–3x)=0. So we get x=0 (already checked) and x=40/3. Since f(40/3)=(40/3)²(20–40/3) is positive (20 is 60/3), this is where the max value of f occurs.

The answer to the question is 40/3 and 20–40/3 (we were asked How should the numbers be selected).

Another story

A line segment joins the points (3,0) and (0,2) in the first quadrant. A rectangle in the first quadrant which has sides on the x- and y-axes is inside the resulting triangle, and has a vertex on the line segment. What is the largest area that such a rectangle can have?

Comment
I really tried to write the specifications in this problem carefully. After class, Mr. Orrico and Ms. O'Sullivan convinced me that my paragraph was not carefully written enough. I think they were correct. Instead of the phrase "which has sides on the x- and y-axes is inside the resulting triangle, and has a vertex on the line segment", I had written "has sides on the x- and y-axes and a vertex on the line segment". In that case (future lawyers should follow this carefully!) a rectangle with vertices at (0,0), (3,0), (3,2), and (0,2), that is a rectangle with the line segment as a diagonal, will definitely have (I think!) all the other "eligible" rectangles inside it. So it would definitely be the largest. This isn't what I wanted. So my specification was not written precisely enough. I am sorry. I believe (currently, about 5 hours later!) that what I've written above is what I wanted and really do want. Sigh.

I will continue the discussion with the changed specification written above. Probably the first thing I would do when meeting a problem of this type is to draw a picture. I think my sketch would look something like what's to the right. Certainly, I would need the line segment joins the points (3,0) and (0,2) in the first quadrant which is drawn. Then I likely would try to draw a typical rectangle. Although I tried to write the specification of the rectangle in a direct way, my desire to be brief may have made this difficult to understand. The rectangle has sides on the x- and y-axes as shown. It also has a vertex on the line segment. Legitimate questions include, "What's a vertex?" and "What line segment?" I hope, even if you've not heard the word before or this use of the word, that you would know "vertex" means a corner of the rectangle. And the first sentence of the problem defines "the line segment".

Almost surely in my mind I would play around with the problem as pictured, and try to see some of the eligible rectangles. The pictures I would look at would include the two below. If the vertex were very near (0,2), as in the left picture, I see that one side of the rectangle would be almost 2 units long, while the other is very small. The area of that rectangle would be small. On the other hand (right picture), if the vertex were very near (3,0), one side of the rectangle would be almost 3 units long, while the other is very small. Again the area would be small. This encourages me to think that, yes, there is a maximum and almost surely the max area will occur for some rectangle with vertex "inside" the segment (at a critical point of the algebraic model we will build).

Two "extreme" rectangles
Small area	Also small area

I will finish this problem on Thursday and discuss some additional problems.

• If f´>0 on an interval, f is increasing on that interval. Therefore if you "move" from left to right on the graph, you'll also move up.
• If f´<0 on an interval, f is decreasing on that interval. Now if you "move" from left to right on the graph, you'll move down.

What second derivatives can tell you
The sign of the second derivative on an interval also has some neat geometric information. Here is the logic:

If f´´>0 on an interval then f´ is increasing on that interval. So moving from left to right means that the slope of the tangent line increases. I don't know enough about animations to put my walking demonstration of this on the web (I wish I did, but I am lazy), but to the right is a static picture of what I am trying to describe. This curve is concave up.

Of course, if f´´<0 on an interval then f´ is decreasing on that interval. So moving from left to right means that the slope of the tangent line decreases. A typical geometric picture of this situation is to the right. (Since I am lazy, all I did was flip the previous picture -- drawing programs are wonderful.) This curve is concave down.

The Second Derivative Zoo
Before we went on, I thought giving several (relatively) simple examples would be useful. I learn far more from examples than from definitions, and even from theorems. So here is my "zoo" or collection of concavity examples.

Examples and discussion	Graphs
x3 Here f(x)=x3 so f´(x)=3x2. Away from x=0, this is certainly positive, so f is increasing in (–∞,0] and in [0,∞). Actually, if we "glue" the intervals together, we can see that f is increasing in all of the real numbers. Since f´´(x)=6x, I know that f is concave up where 6x>0: this is (0,∞). Similarly, f is concave down where 6x<0. This is (–∞,0). At 0, the x-axis (equation:y=0) is a horizontal tangent line. Also at x=0, the concavity changes from down to up. This is called an inflection point.
x1/3 Well, this function is a bit weirder. Here f´(x)=(1/3)x–2/3. You need to think about this for a moment. The derivative is a negative power of x. This means that the domain of the derivative does not include 0 (we don't divide by 0!). Geometrically, this function is the flip over the "main diagonal" y=x of y=x3: it is the inverse of x1/3. The previous horizontal tangent line becomes a vertical line, which has no slope. Since =(1/3)x–2/3=(1/3)(1/x2)1/3, I know that the derivative (where it exists!) is always positive. This function is increasing always. Since f´(x)=(1/3)x–2/3, we know that f´´(x)=–(4/9)x–5/3. Again, the domain is non-zero x's. But what is the sign of f´´(x) for those non-zero x's? Here please notice that 5 and 3 are both odd. This is important. A negative number raised to an odd integer power or an odd integer root is negative. But look closely: in front of the formula, before the (4/9) is –: a minus sign. So f´´ is positive when x is negative, and therefore the graph is concave up for negative x. For similar reasons, the graph is concave down when x is positive. And x=0 is a point on the graph where concavity changes, and therefore is an inflection point.
1/x There is a tricky bit of language here, which I will mention later. Since f(x)=1/x=x–1, I know the f´(x)=–1/x2. Squares are always positive, so this derivative is always negative, and therefore, inside each interval in which this function is defined, the function is decreasing. This means f is decreasing inside (–∞,0) and f is decreasing inside (0,∞). Notice, though, that f(1)=1 and f(01)=–1. Even though –1<1, we have f(–1)<f(1): f is not decreasing "across" the intervals. If we compute correctly, f´´(x)=2/x3. 3 is an odd power, so the graph is concave down in (–∞,0) and concave up in (0,∞). The domain of f does not include 0, so there is no point of inflection!
|x| I put this in to stimulate some conversation. If you are clever and just look at the graph, you can "read" the derivative. The function f(x)=|x| is differentiable except for x=0. When x<0, f´(x)=–1, and when x>0, f´(x)=1. Actually, the second derivative also "exists" when x is not 0: f´´(x)=0 for all x not equal to 0. The graph has no concavity as it is usually understood.
x4 If f(x)=x4, then f´(x)=4x3 and f´´(x)=12x2. If x is not equal to 0, then the second derivative is positive. The graph is concave up. Notice that although f´´(0)=0, this graph has no point of inflection! The concavity on both sides of x=0 is "up": the graph is concave up in both (–∞,0) and (0,∞). Since concavity is the same on both sides of 0, the origin is not a point of inflection. Comment People will say that y=x4 "is" or "looks like" a parabola. Well, right here (to the right, here!) are graphs of y=x2 and y=x4. I will agree that the graphs resemble each other in many ways. They are both concave up always, go through (0,0), and are symmetric with respect to the y-axis. But please look a bit more closely. y=x4 is flatter near the origin (higher powers of small numbers are smaller!). As soon as the x's get out of [–1,1], that curve grows much larger than y=x2 (higher powers of large numbers are larger). The parabola is y=x2 and it has some interesting geometric and physical properties (for example, mirrors are made to have parabolic cross-section to help aim light beams and telescopes) which are not shared by y=x4.

Inflection point
x is an inflection point of f if

• x is in the domain of f.
• The concavity of the graph of f on the two sides of x (less than and greater than) is different.

The Gaussian (bell-shaped) curve
I wanted to graph a "random" (not!) curve defined by a formula. What we looked at was f(x)= e^(–x2). This turns out to be a rather important function in analyzing experimental results. Any "real" repeatable experiment with a measurable outcome is likely to have the numbers scattered so they resemble this bell-shaped curve. This phenomenon is not obvious at all!!

The exponential function's outputs are never negative and never 0. Therefore no part of the graph will be on or below the x-axis. Since –x² is always less than or equal to 0, and these numbers serve as inputs to exp, I bet that the outputs from exp will be less than or equal to 1, and will be equal to 1 only at x=0. Also this graph is even, symmetric with respect to the x-axis, since (–x)²=x². I hope this explains to you why the graph looks the way it does. But let me try analyzing the graph using the first and second derivative.

If f(x)=e^–x2 then f´(x)=(e^–x2)(–2x). The exponential function is very nice. It is never 0 and always positive. Therefore the only x for which f´(x)=0 is when –2x=0. So x=0 is the only critical number. Now reasoning using the Intermediate Value Theorem says that f (which is certainly continuous!) can have only one sign for x<0 and one sign for x>0 (or else f(x) would have to have to be 0 again). We can check signs at, say, x=1 (where the derivative is negative) and x=–1 (where the derivative is positive). f is increasing in (–∞,0) and f is decreasing in (0,∞). Naturally 0 represents a local (and indeed, absolute!) maximum.

What information can we get from the second derivative? If we use the product and the chain rule correctly, then f´´(x)=(e^–x2)(4x²–2). Logic similar to the preceding asserts that this is 0 exactly when the non-exponential factor is 0. But 4x²–2=0 when x={+|–}1/sqrt(2). Again, we can check signs in between the 0's of f'', and f will be concave up for x<–1/sqrt(2) and for x>1/sqrt(2). For x between –1/sqrt(2) and +1/sqrt(2), the graph will be concave down. The points where x={+|–}1/sqrt(2) are where the concavity of f changes: these are called inflection points. These particular inflection points are related to the standard deviation, which represents dispersal from an average when this function is used in statistics. The center of the curve, at x=0, is related to the mean of the data observed.

The diagram/picture above is an attempt to indicate how a person could assemble the information about signs of the first and second derivative, and patch together template pieces. The patched-together curve should be continuous (no breaks) and differentiable (smooth). This all takes some practice!

Graphing x³(x–1)⁴
Here I invited students again to use a graphing calculator and try to see what y=x³(x–1)⁴ "looked like". I did remark that the "action" took place somewhere between –1 and 2. The result of this was something like the graph shown to the right. I believe that calculators and graphing devices are wonderful, but sometimes they almost conceal what's going on.

Here f(x)=x³(x–1)⁴. If we want to find out where f is increasing and decreasing, we really should look at f´(x). For this we need the product rule and the chain rule. So:
f´(x)=3x²(x–1)⁴+x³4(x–1)³.
Generally I am against "simplifying" because I view it as mostly a chance to make lots of mistakes. But here some simplifying will reveal structure in the derivative. So please notice the common factors, and what you get is as follows:
f´(x)=x²(x–1)³(3(x–1)+4x)=x²(x–1)³(7x–3).

What can we tell about where the derivative is 0 and where it is positive and where it is negative? Well, the different factors allow us to deduce that the derivative is 0 at x=0 and x=1 and x=3/7.

If x is very large positive, say, then f(x) is a product of three factors, all of which are positive. And if x is very large negative, then the x² is positive and the (x–1)³ is negative and the 7x–3 is negative. Therefore f(x) in that range is positive also. So we have learned (using logic from the Intermediate Value Theorem as before) that the derivative is positive on at least the intervals (–∞,0) and (1,∞). There is a chance for the derivative to change signs at x=0, but the factor which controls sign change there is x²: since 2 is even, there is no sign change at x=0. But there is a sign change at x=3/7 and at x=1. So now we have broken up the real line into pieces based on the signs of the derivative of f:

    Deriv is +     Still +          Now it is –        Back to + here 
<---------------0-----------3/7--------------------1------------------->
 Func increases   increases     The func decreases   Here it decreases

So from this I learn that f has critical points at 0 and 3/7 and 1. We can also learn that f has a local max at 3/7 and a local min at 1. This is not entirely clear from the initial graph. Actually, if we just look at the graph from –.1 to 1.1, you can see some of the structure. This is shown to the right. Please notice that the vertical scale of this graph is very small. This might all be difficult to see without looking at the calculus first.

Let's consider the concavity of this function, and make some guesses about the number and location of inflection points. We can be sure about this if we find the second derivative. I will use the product rule, and make the second factor, a product itself. So: f´(x)=x2(x–1)3(7x–3)= x2((x–1)3(7x–3)) f´´(x)=2x((x–1)3(7x–3))+x2(3(x–1)2(7x–3)+(x–1)3(7)) So let me try to "simplify" f´´(x). We will get: x(x–1)2(2(x–1)(7x–3)+3x(7x–3)+x(x–1)7)= x(x–1)2(2(7x2–10x+3)+21x2–9x+7x2–7x)=x(x–1)2(42x2–36x+6) So the second derivative is 0 at x=0 and at x=1 and at the roots of 42x2–36x+6=0: those are x=(3/7)+/–sqrt(2)/7 (approximately .227 and .631). The second derivative does not change sign at 1 because the factor is (x–1)2, an even power. It does change sign at 0 and the two other numbers which are on either side of the local max. There are three inflection points. I needed to work fairly hard to get everything correct in the preceding example. Second Derivative Test There is a result which allows you to tell when a critical point is a local max or local min if you have information about the second derivative at that point. It goes like this:     If f´(x)=0, and if f´´(x)>0, then f has a local min at x. Example –x2 at 0.     If f´(x)=0, and if f´´(x)>0, then f has a local min at x. Example x2 at 0.     If f´(x)=0, and if f´´(x)=0, then nothing can be deduced. Example –x4, x4, and x3 at 0. I don't often use this result, because I am lazy and computing the second derivative (and evaluating it!) may be work. Usually I can examine the first derivative, if needed. Inflection points as you drive! Suppose you are driving along a road, which happens to be the graph of a function. Please refer to the picture to the right. This discussion is an effort to persuade you that inflection "occurs" in the "real world". My car is the weird object shown at the lower left, and it is moving in the direction of the red arrow. This is a one-way road and I won't hit anything (I hope). As I drive along the road, I steer to the left (that's the A region of the road). Eventually I come to a place where the road begins to bend the other way, at B. After that point, in the C region, I steer right. The bending place, where the switch from left steering to right steering occurs, is an inflection point. Please notice that from the point of view of the car, it actually doesn't matter (no gravity, we are looking from above) whether the curve is decreasing or increasing. The steering occurs as a response to how the curve is bending. When the bending changes (inflection!) we must change steering direction -- if we want to stay on the road, anyway. What could have been the QotD To the right is shown a graph of the derivative of f(x). The function is intended to have domain all of R, the real numbers, and the arrows at the end of the graph are intended to indicate that the graph goes on forever behaving nicely. Please answer the questions in A first, and then draw the graph requested in B. A. Use this graph to answer the following questions as well as you can. What are the critical points of f? In what intervals is f decreasing? In what intervals is f increasing? For each critical point: is it a local max or a local min or neither? Where is the graph of f concave up? Where is the graph of f concave down? Where are the inflection points of f? B. Sketch a graph of y=f(x) using the information just described. There will be many different valid graphs, but they will all share the same qualitative features. Please try to write a solution on your own. Then you may wish to look at a solution I wrote.. Do this yourself first, please. This is a valid exam question, and you should check your own ability with such material.

• Problem 53 of section 4.2 Find the max and min values of f(x)=(ln(x))/x on the interval [1,3].
  Endpoints f(1)=ln(1)/1=0 and f(3)=ln(3)/3.
  The derivative f´(x)=[(1/x)(x)–1·ln(x)]/x². The top is 1–ln(x), and this is 0 when ln(x)=1, so x=e.
  Plugging in f(e)=ln(e)/e=1/e.
  Biggest/smallest of the available candidates Max must be 1/e and min must be 0.
  
  Comment We needed to compute 1/e: its decimal expansion begins .3678, and ln(3)/3: its decimal expansion begins .3662. I certainly can't "see" which number is larger. And, really the graph hardly helps. (Later in this lecture we'll see another way to detect the highest point without extensive computation.)
  
• Problem 54 of section 4.2 Find the max and min values of f(x)=3e^x–e^2x on the interval [–1/2,1].
  Endpoints f(–1/2)=3e^–1/2–e^–1 and f(1)=3e–e². The derivative f´(x)=3e^x–2e^2x. When is this 0? Well, we can factor: e^x(3–2e^x)=0. Since e^x is never 0, this is 0 when 3–2e^x=0 so e^x=3/2 and x=ln(3/2). Plugging in f(ln(3/2))=3e^ln(3/2)–e^2ln(3/2)=3(3/2)–(9/4)=9/4.
  Biggest/smallest of the available candidates Max must be 9/4 and min must be 3e–e².
  Comment Again, you can compute to compare the three values, or, in this case, maybe the comparison can be done by "pure thought". So, for example, f(1)=3e–e²=(3–e)e is about .3(2.7), approximately .8 And f(1/2)=3e^–1/2–e^–1 is about 3/2–1/3, which is 7/6. And the other value is 9/4. So the 9/4 is largest, and the (approximately) .8 is smallest.

Rolle's Theorem
If f is continuous on[a,b], and if f(a)=0 and f(b)=0, and if f is differentiable inside the interval, then there is at least one number c inside the interval where f´(c)=0.

To the right is a typical picture of the situation described in Rolle's Theorem. Since the function is glued down (?) on the x-axis at a and b, either the function is always 0 (so there are lots of c's) or the absolute max and absolute min occur inside the interval. There may be more than one of each. Since the function is differentiable, these max's and min's occur at critical points where the derivative is 0, as shown.

Below is a gallery of possible "Rolle" situations. The first, to the left, is the simplest and silliest. The function is 0 in the whole interval. Then any c is a valid candidate for the consequence of the theorem. The second picture shows what happens if f is positive somewhere. Then the absolute max must occur inside, and it must occur at a local max inside the interval, and since f is differentiable, f´(c)=0 there. The next is the negative situation, and the last picture is what might happen if there f had a mixed sign behavior.

Rolle's Theorem tilted
Rolle's Theorem seems very specialized. What would happen if we tilted it? That is, we took the coordinate axes and rotated the whole picture? Then the graph is no longer glued down at the endpoints. This is the important generalization that people use constantly. Of course, the picture is relabeled and everything is more conventionally drawn.

Here is one of the two results in this course that anyone who uses calculus will think about quite frequently.
The Mean Value Theorem (MVT)
Suppose f is differentiable in [a,b]. Then there is at least one c inside the interval so that f´(c)=(f(b)–f(a))/(b–a).

Discussion To the right is a tilted and relabeled picture. The line segment joining (a,f(a)) and (b,f(b)) has slope equal to (f(b)–f(a))/(b–a). The other indicated line segments are pieces of tangent lines which are parallel to the segment joining (a,f(a)) and (b,f(b)). Since they are parallel, their slopes must be equal. But the slope of a tangent line at c is f´(c), and that's the equation which appears above.
If you want an algebraic verification of MVT, then look at the end of section 4.2. This picture is sufficient for my purposes. I want to show you some of the ways people use this result.

Simple (?) observation #1
Suppose f is differentiable, and in some interval I we know for some reason that f´ is always 0. What happens? Well, take two points, x₁ and x₂ in I with x₁<x₂. MVT says that
    (f(x₂)–f(x₁))/(x₂–x₁)=f´(c)
But the right-hand side of the equation is 0. So the quotient on the left-hand side must then have 0 on top: f(x₂)–f(x₁)=0, which means f(x₂)=f(x₁) for any choices of x₁ and x₂. This means that f is constant.
    If f´=0 always, then f is constant.

Physical meaning
Well, one way to get a physical (?) interpretation of this statement is to imagine that f(x) is the position depending on time, x. (Sorry: you could rotate the x and get a t, if that makes you happier.) So f(x) reports the position on a line. Then the statement above translates to:
    If velocity is always zero, then position is constant.
Well, it sure doesn't look profound, but maybe it does make sense. By the way, this "simple" deduction which is so clear physically wasn't verified mathematically for about 150 years. Stupid human beings. (!) (?) Or maybe this stuff is somewhat subtle?

Simple (?) observation #2
Suppose f is differentiable, and in some interval I we know for some reason that f´ is always positive. What happens? Well, take two points, x₁ and x₂ in I with x₁<x₂. MVT says that
    (f(x₂)–f(x₁))/(x₂–x₁)=f´(c)
Now let's see: the right-hand side is now supposed to be positive. The bottom of the left-hand side is positive (since x₁<x₂ means x₂–x₁>0). So we have a fraction with a positive bottom equal to a positive number. So the top of the fraction should be positive. That means f(x₂)–f(x₁)>0 so that f(x₁)<f(x₂). So what do we know:
    IF f´ is positive, THEN x₁<x₂ implies f(x₁)<f(x₂).

Physical meaning
Well, if derivative corresponds to velocity, and position corresponds to the original function (with higher corresponding to more right position on the line), then:
    If velocity is always positive, then position is moving steadily right.
Another way of thinking about this is graphical. If you wish, you could imagine position as a function of time. Then the "progress" of time might be the horizontal axis, and position in this diagram would be the vertical axis. Position getting bigger would mean that the graph of position versus time would be getting higher. So a qualitative picture of this sort of graph is shown to the right.

Simple (?) observation #3
I will reverse the sign of the derivative. So here suppose f is differentiable, and in some interval I we know for some reason that f´ is always negative. What happens here? Well, take two points, x₁ and x₂ in I with x₁<x₂. MVT says that
    (f(x₂)–f(x₁))/(x₂–x₁)=f´(c)
Now let's see: the right-hand side is now supposed to be negative. The bottom of the left-hand side is positive (since x₁<x₂ means x₂–x₁>0). So we have a fraction with a positive bottom equal to a negative number. So the top of the fraction should be negative. That means f(x₂)–f(x₁)<0 so that f(x₁)>f(x₂). So what do we know:
    IF f´ is negative, THEN x₁<x₂ implies f(x₁)>f(x₂).

Physical meaning
Again, derivative corresponds to velocity, and position corresponds to the original function (and higher still means position is more to the right, and lower position means more to the left on the line), then:
    If velocity is always negative, then position is moving steadily leftt.
Again, we could think graphically. Position is a function of time, the "progress" of time is the horizontal axis, and position is the vertical axis. Position getting smaller would mean that the graph of position versus time would be getting lower. A qualitative picture of this sort of grpah is shown to the right.

Further definitions
A function f is increasing on an interval I if whenever we take x₁ and x₂ in I with x₁<x₂, then f(x₁)<f(x₂).
A function f is increasing on an interval I if whenever we take x₁ and x₂ in I with x₁<x₂, then f(x₁)<f(x₂).

Facts
MVT implies:
    If f´ is positive on an interval, then f is increasing on that interval.
    If f´ is negative on an interval, then f is decreasing on that interval.

What we can say with graphical (qualitative) evidence
This is basically a discussion of problem 24 in section 4.3. The problem gives a graph of the derivative f´(x) of a function f(x). Please let me repeat: the graph is the derivative of the function. I've attempted to copy the graph in the picture to the right.

First, we are asked to "find the critical points of f". Since f is differentiable (otherwise we wouldn't be looked at a graph of f´, darn it!) the critical points here are where f´(x)=0. Look at the graph. The curve crosses the x-axis at three values of x: –1, 0.5, and 2. These are the critical points of f.

Then we are asked to determine whether these critical points are local max or local min or neither. So let me think through this with you.

Local analysis near x=0.5 Look closely at the graph of f´ in a small interval to the left of x=0.5. The graph (inside the very light blue region) is above the x-axis. So in this interval, the derivative is positive and therefore the function is increasing. Look now at the graph of f´: in a small interval to the right of x=0.5. The graph (inside the very light green region) is below the x-axis. So in this interval, the derivative is negative and the function is decreasing.
What should the graph of f (not the derivative!) look like? To the left of 0.5, the function is increasing. The derivative at 0.5 is 0, so the tangent line is horizontal. To the right of 0.5, the function is decreasing. Surely f has a local maximum at 0.5. I can't tell the value of f(0.5), but I can tell you that value is larger than f(x) for nearby x's, on either side of 0.5 f has a local maximum at 0.5.

Local analysis near x=2 The situation is similar but reversed near x=2. A "blow up" of the graph of f´ near x=2 is shown to the right. In a small interval to the left of x=2, the graph of the derivative (inside the very light blue region) is below the x-axis. So in this interval, the derivative is negative and therefore the function is decreasing. The graph of f´: in a small interval to the right of x=2 is inside the very light green region, above the x-axis. So in this interval, the derivative is positive and the function is increasing.
What should the graph of f (again, the graph of f, not of its derivative!) look like? To the left of 2, the function is decreasing. The derivative at 2 is 0, so the tangent line is horizontal. To the right of 2, the function is increasing. Surely f has a local maximum at 2. Again, we don't know the actual value of f(2), but we do know its relative value. f(2) is smaller than f(x) for nearby x's, on either side of 2. f has a local minimum at 2.
Local analysis near x=–1 I saved the most interesting (or maybe most irritating!) for last. What do we see when we look very closely at the derivative near x=–1? Well, to the left of –1 the derivative is positive, above the x-axis, so the function itself should be increasing in that interval. And to the right of –1, the derivative is also positive, so the function is also increasing there. That's what the graph of the derivative of f declares.
The graph of f will show some delicate (?) behavior near –1. In a small interval to the left of –1, we know f should be increasing (as we walk from left to right, the graph will go up). And the same sort of qualitative behavior will be true for the graph of f in a small interval to the right of –1: up again. In fact, the graph is just going up, increasing in the whole interval. The most fascinating (?) aspect is that since f´(–1)=0, we'd better draw the graph of f so that it has a horizontal tangent line at –1. And that's what I've tried to show to the right. If you are not used to this sort of kink in a graph, take a look. f has neither a local minimum nor a local maximum at x=–1.
An office hour question Mr. Orrico asked me in my office before class about problem 28 of section 4.3. Here y=x(x+1)3 and you are asked to find the critical points and find intervals in which the function is increasing and decreasing. He found y&3180; and found that it was equal to 0 at exactly two values of x, –1 and –1/4. So the derivative is not zero in the interval (–∞,–1) and in the interval (–1,–1/4) and in the interval (–1/4,∞). How many SIGNS can the derivative have in the interval (–1/4,∞)? Well, I think the derivative, which is a polynomial, is continuous. If it is both positive and negative in that interval then the Intermediate Value Theorem tells me that the derivative must be 0 inside the interval. But it is not zero. Therefore the derivative can have only one sign inside (–1/4,∞). Similar logic tells me the derivative can have only one sign in the the interval (–1,–1/4) and only one sign in the interval (–∞,–1). I don't know what the signs are, but I do know that the sign doesn't change. How can I discover what the sign is in the interval (–1/4,∞)? I can check one value of f´ inside that interval. Any value will work, and I should just choose something that I find simple to compute with. I could choose 37 or 1,028,405.008. Or just 0. In fact, I don't even need to compute too well -- I just need enough information to check what the sign is. So that logic makes the problem easier.

Maybe you'll believe this one ...
Here is a machine-generated graph of f(x)=x³ for x "near" 0 (well, for x between –1.1 and 1.1). Look closely at it. The local behavior is very much like the local behavior of the text example near –1.

The derivative is 3x². This is positive if x is not 0, so certainly the function is increasing for x<0 and is increasing for x>0. We paste these pieces together, and I think that f is increasing for all numbers. There are several things to notice "at 0" (actually, both at 0 and near 0). The tangent line is horizontal. The curvature (!!) changes. The curve to the left of 0 is concave down and, to the right of 0, it is concave up. I will spend more time on such phenomena during the next lecture.

The First Derivative Test
Here is how, maybe, you could detect whether a critical point is a local minimum or a local maximum or neither.

Suppose f has one critical point in an interval.
IF		THEN
f´>0 to the left of the c.p.	f´<0 to the right of the c.p.	The c.p. is a local max.
f´<0 to the left of the c.p.	f´>0 to the right of the c.p.	The c.p. is a local min.
f´ has any other sign assortment (positive/positive or negative/negative)		The c.p. is neither. It is not a local max and not a local min.

Actually I think about the local pictures and don't memorize the algebra (with only 4 brain cells, for every fact I learn, it seems that I must forget two). I hope that you can see that appropriate local pictures for the last line are (positive/positive) and (negative/negative . You may prefer to think about the algebraic versions, but the key idea is that the sign of the first derivative on both sides of the critical point provides enough information to decide local max/min questions.

A simple algebraic example Here is problem 33 of section 4.3. The general instructions are to find critical points and the intervals where the function is increasing and decreasing, and then apply the First Derivative Test to each critical point. Here f(x)=x4+x3. Certainly f´(x)=4x3+3x2 and we set this equal to 0, factoring: x2(4x+3)=0. The roots are 0 and –3/4. These are the only places where f´ is 0. Notice that f´ is continuous. This means if we test the sign of f´ "between" the critical points, then we know the sign in each of the intervals whose boundary is a critical point. This is because if the sign changes, then (continuity, Intermediate Value Theorem) there would have be a place where f´ is 0. This is a very useful observation if you want to be lazy and efficient. Testing the signs We need to look "between" (on both sides of!) 0 and –3/4. Well, I bet that f´(10)=4·103+3·102 is positive. I bet also that f´(–10)=–4·103+3·102 is negative (thousands are way bigger than hundreds, and the parity [even/oddness] of the powers makes this easy. What about really between 0 and –3/4? I think in class we looked at x=–1/2. Then f´(–1/2)=4(–1/2)3+3(–1/2)2=(–1/2)+(3/4) is positive. So now in my head I have a picture of the real line: What's important here is not the specific example but the process. With a very small amount of work (compute derivative, find critical points, take a few carefully chosen values of the derivative) I can conclude: f is decreasing in the interval (–∞,–3/4]. f is increasing in the interval [–3/4,0]. f is increasing in the interval [0,∞). f has a local minimum at x=–3/4. f has neither a local minimum nor a local maximum at x=0. Actually, f is increasing in the whole interval [–3/4,∞). The type of each critical point is clear by using the First Derivative Test. If you don't believe me, then take a look at a computer-generated graph of y=x4+x3 to the right. Here, I will make a bold statement: if I've done my critical point analysis correctly, I think I would believe more in what it told me than in a computer-generated graph. Formulas that make deceptive graphs are known.

My friend Francine drives again ...
I don't want to get into legal trouble. The "logo" of the (beloved!) New Jersey Turnpike is shown to the right. I am using it here totally for educational purposes. Sigh. Only education. Really. No DVD's, no music videos, no Latin American film rights, no web things. Really, really.

After this assurance from the cowardly and worried writer, I remind you that a previous drive of Francine on the Garden State Parkway had been analyzed earlier using the Intermediate Value Theorem.

The turnpike has a restricted number of entrances and exits. Suppose Francine

• enters the turnpike at Interchange 4 ("NJ73, Camden, Philadelphia"), Milepost 34.5, driving north at 8 AM.

• exits the turnpike at Interchange 9 ("NJ-18, New Brunswick"), Milepost 83.3, at 8:40 AM.

Let's model the situation the following way. Suppose f(t) describes Francine's position in miles on the turnpike, and t is the time, in hours, after 8 AM. Then what we know is the following:
     f(0)=34.5 and f(2/3)=83.3 (40 minutes is two-thirds of an hour).
Then MVT asserts there is some time c between 8 AM and 8:40 AM so that f´(c)=(83.3–34.5)/(2/3–0)=48.8/(2/3)≈73.207. Since the speed limit on that portion of the turnpike is 65 MPH, MVT shows that Francine was speeding.

The Intermediate Value Theorem provides some information about Francine's driving. It tells us that she passes through every place (position) on the highway between Milepost 34.5 and Milepost 83.3 at some time between 8 AM and 8:40 AM. This is somewhat indefinite.
The Mean Value Theorem gives more quantitative information, also in a rather indefinite style (heck, we're given only outline information!). It declares that the velocity (the derivative) must be exactly 73.207 at some time between 8 AM and 8:40 AM.
The chunks of information have different flavors but mostly I think that MVT information has more uses.

Semester project, honors course Why doesn't the state actually use the information it has for such things? It does have entrance/exit times and specifications of which interchanges occur, etc. Explain in detail, with citations of specific laws. Sigh.

Quantitative information from the derivative
Here is a Math Problem. Suppose you know the following about a differentiable function:
     f(5)=7 and, for all x, f´'s values are between 40 and 60
     (that is, 40≤f´(x)≤60).
     What can you say about f(10)?
This seems to be very abstract and very silly. Let me translate it into physical language, however.

A Physical Representation of the Math Problem:
Suppose f(t) represents the miles that have been traveled down a road at time t (in hours, I guess AM!).
     At 5 AM, you are 7 miles does the road.
     Your speed is always between 40 and 60 mph.
     What can be said about your position at 10 AM?
Somehow, phrased this way, the problem seems much more easily handled, even to me, with a brain accustomed to math stuff. "Golly," I might say, "In 5 hours you drive at least 200 [that's 5·40] miles and at most 300 [that's 5·60] miles, so your position is between 207 and 307 miles down the road."

We are actually using MVT reasoning on this. Look:
    By MVT, [f(10)–f(5)]/[10–5]=f´(c), for some c between 5 and 10.
But the derivative is between 40 and 60, and f(5)=7.
    So 40≤[f(10)–7]/[5]≤60
Multiply by 5 and then add 7.
    And 207≤f(10)<307.
The advantage of the mathematical approach is that it will apply to all situations where we have a model and some estimate of the derivative and knowledge of the function at one value. We then can "predict" some estimate of the function at another value.

And even more (a possible QotD!)
Here is what would have been the QotD, with my apology because I ran out of time.

Suppose you know that f(0)=7 and f´(x)=sqrt(1+x³). What can you tell me about f(2)?

MVT says that f´(c)=[f(2)–f(0)]/[2–0]=[f(2)–7]/2. Since f´(c)=sqrt(1+c³) and c is between 0 and 2, I bet that
1≤f´(c)≤3. Why is this? Well, f´ is increasing (you can take the derivative and the derivative is always positive in that interval so the function is increasing) and thus 1=sqrt(1+0³)=f´(0)≤f´(c)<f´(2)=sqrt(1+2³)=3.
Therefore 1≤[f(2)–7]/2≤3. Multiply by 2 and add 7. The result is 9≤f(2)≤13.

---

Thursday, October 22 (Lecture #15)

We begin today the last half of the course.

Special added attraction (I hope)

I have been getting "customers" at my Tuesday office hour and I would like to encourage this. So I will enlarge the office hour (?): it will run from 11 AM to 1 PM. Part of my job, which I do happily, is to help you learn and office hours can be useful. I do remark, though, if you haven't visited, that my policy is to sit in my chair, sluglike, and have students write things on the board. You do the work. And as I remarked in class, if I'm asked a question, I will frequently respond by requesting that people look in the textbook -- the text should be used. Otherwise it is a rather expensive decorative item.

Today we discussed the material in section 4.2, which is very important, perhaps the core of most applications of mathematics in engineering and applied science. Section 4.2 covers the ideas well, and the class discussion (it was as close as we've gotten to a discussion, at times) was an effort to familiarize people with the ideas and methods which are quite important. I just wrote the boxed items from section 4.2 on the board, and we went through them one by one.

The difficulty now (from the instructor's point of view!) is controlling the number and complexity of examples. There are many, many, many interesting examples coming from real applications.

The class interaction was rapid, and maybe I spoke too fast at times (I'm sorry). The examples below are generally similar to the ones discussed in class.

Extreme values on an interval
I wrote the textbook's definitions of Absolute minimum and Absolute maximum on the board. I remarked that I never understood definitions until I had examples that did and did not satisfy the definitions. So I will assume that you have a textbook next to you (I don't want to just copy what's there). You should know the definitions. The situation the definition deals with is an interval, I, which is the domain of the function, and a function, f, defined on that interval.

YES!
Here is an example of a function and an interval, where there is both an absolute maximum and an absolute minimum.
     Here I=[0,1] and f(x)=x. Then f has an absolute maximum at x=1, where f's value is 1, and f has an absolute minimum at x=0, where f's value is 0.

another YES!
Another example, where the interval is an open interval and the function is still fairly familiar.
     Suppose I=(Π/4,7Π/4) (this does not include endpoints!) and f(x)=sin(x). This f has an absolute maximum at x=Π/2 where f's value is 1. It has an absolute minimum at x=3Π/2, where f's value is –1.

NO!
The first NO! example we had was rather subtle. I think it was something like this.
    The interval was (–1,1) (an open interval, which does not contain its endpoints. The function was something like f(x)=x². There is an absolute minimum at (0,0), but ... what would be "your" candidate for an absolute max? Any number you choose which is close to 1 will have a number even closer to 1 and bigger. For example, if you were to assert that f has an absolute max at .99998, I would tell you that f's value at .9999999998 was bigger. If you said to me that .9999... (infinitely repeating) was where an abs. max was located, I would reply that if you are sophisticated enough with decimal notation to know about such things, this indicates a certain geometric series whose sum is 1: the notation .9999999.... is just another "address" for the number 1. And in this case, the domain does not include 1 and does not include –1. So this function in this domain does not have an absolute max.

NO!
Another no example, which I think is fairly distressing, is gotten by taking I=(0,1) (the open interval which does not have endpoints) and the function f(x)=x. This is certainly a simple function and a simple interval. This function on this interval has no absolute max and no absolute min. I think a person who is understanding this example for the first time is really augmenting their intuition.

Why doesn't it have abs max or min? Well, certainly x=1 and x=–1 are not eligible locations for them, since both of these x's are not in I. Now let's consider any other x in I. If we microscopically (?) examine the local appearance of the graph of f(x) near that x, it will always be a tilted line segment. To the left, there are lower points on the graph, and to the right there are higher points on the graph. That certainly implies our candidate is not valid.

Another NO!
Here is maybe a simpler example. Look at f(x)=tan(x) with the interval (–Π/2,Π/2). I've only supplied a little doodle (?) of the graph to the right.

Certainly there will be no absolute maximum and no absolute minimum, since tan(x)→+∞ as x→(Π/2)^– and tan(x)→–∞ as x→(–Π/2)⁺.

A more artificial NO!
Suppose the domain is [–1,1], which is a closed interval (includes its end points) and which is bounded. Those two attributes (closed and bounded) turn out to be important in the theory, as is described just below. Now define f(x) piecewise: it is 0 when x=0 and it is 1/x when x is not 0. Part of the graph of this function is shown to the right. And if you think this is a silly example, I will totally agree. But why is it silly, and what is there about it which is not "nice"? Understanding one's feelings about such things is sometimes very helpful. I don't like this example because the function is quite clearly not continuous at 0 (the limit as x→0 of f(x) does not exist). I also don't like this function because it certainly does not have an absolute maximum and does not have an absolute minimum.

Existence of extreme values on a closed and bounded interval
In many situations mathematical models are created which need maximum and minimums. In business, we might be interested in lowest cost (min) or highest profit (max). In more physical situations, many people want lowest energy or least work. Or greatest volume or ... the applications are numerous. Before searching for max or min, we might want to know that such things exist (apparently they don't always -- look at the examples we have for NO!).

The following result can be proved, that is, deduced from generally accepted and more obvious statements. The process of deducing it is rather lengthy, and takes up a substantial amount of Math 311. I request politely that you, as student engineers and applied science types, be willing to accept the statement as true. Most of the proofs of this statement do not supply any method for actually getting the predicted max and min. So:

A theoretical result
Suppose I is a closed and bounded interval (so it is [a,b]), and suppose that f is continuous. Then f has an absolute maximum and an absolute minimum on I. So there are x_M and x_m in I so that f(x_m)≤f(x)≤f(x_M) for all x's in I.

This result does not tell how to find x_M and x_m. It does not tell how many x_M's and x_m's there may be (there may be many, but at least one of each is guaranteed). For functions that are likely to be used in practice, we will outline a useful strategy. First some more definitions.

Local extreme values
f has a local maximum at x if f's value at x is the largest in some open interval containing x. The most important and maybe most confusing part of that sentence is "open interval". The dual definition is: f has a local minimum at x if f's value at x is the smallest in some open interval containing x.

YES!
Here is a very simple example which is the same function and domain as before. The absolute min is a local min. The absolute max is a local max. I have put magenta in the domain and on the graph to indicate the relevant open intervals.

A shocking NO!
Again look at (0,1) and f(x)=x, choices which are very simple.

This function has no local min and no local max. Just as before, the only points you might want to consider are the endpoints. But notice, please, that the endpoints are no inside an open interval of the domain. The right endpoint has no open side to the right, and the left endpoint has no open side to the right. This example is really quite annoying and unexpected if you're new to this game.

A zoo of YES and NO
Below is a collection of pictures. This probably is "Too much information" but I tried to write all of the possibilities. It wouldn't be difficult to write some fairly simple formulas whose graphs qualitatively have the same behavior. I wanted to show that "local" and "absolute" have no necessary logical relationships.

Critical points
These turn out to be very important. f has a critical point at x if either f´(x)=0 or f´(x) does not exist.

The picture below is a zoo of critical points. Again, maybe this is too much information. I do hope you get the feeling that local maxes and local mins occur at critical points. This is true.

Geometrically, the derivative doesn't exist at the "point" of |x| because the graph is not locally linear: no matter how much the graph is magnified near (0,0), the result does not begin to resemble a straight line. As for x^1/3, this graph is the flip over the main diagonal of x³. the x³ graph has a horizontal tangent line at (0,0), and the geometrically valid tangent line for x^1/3 at (0,0) is certainly vertical. But the derivative computes the slope of the tangent line, and vertical lines have no slope. So there is no value for the derivative at x=0 (the derivative formula is of course (1/3)x^–2/3 and because of the negative exponent 0 is not in the domain of f´(x) since we are not allowed to divide by 0. The function x^2/3 has derivative (2/3)x^–1/3 and this is certainly o.k. away from 0. But, again, the behavior near (0,0) isn't locally linear. The "point" on this graph is sharper (!) than on the graph of |x|. It has a name: it is called a cusp.

Fermat's theorem on local extreme values
If there is one single result which has kept math people prosperous over the last three centuries, this is it.

If f is differentiable at x and if f has a local max or a local min at x, then f´(x)=0.

In this case I think it is useful for a practicing engineer to have some feeling for why this result is correct. Look at the picture to the right.

I drew what happens if you've got a local max. To get the local min picture, just flip things over. What should you notice about this picture? Well, the secant line from the left has a positive slope and the secant line from the right has a negative slope. IF the derivative at the local max exists, it will be the limit of things with different signs. The only way this can occur is if the slope "at" the local max is 0.
Notice, though, that we need information from both sides. This is why we are considering local maxes and mins. If we don't have the two-sided information, we won't get the disagreement of signs which compromises (?) at a zero derivative.

Extreme Values on a closed and bounded interval
Here I describe a procedure which provides a method for computing the absolute max and absolute min of a function in an interval in many cases, including almost all situations which arise in practice.

1. Look at a continuous function on [a,b]. Theory says there's an absolute max and an absolute min.
2. Check the endpoints, a and b. The abs max and abs min could be there.
3. If the abs max and abs min are inside then: either the function is differentiable (and by Fermat the derivative is 0) or the function is not differentiable. So find all the critical points inside the interval.
4. Compare the values of f at the candidates you found: the endpoints a and b and the critical points inside the interval.

Theory and practice
Mr. Theory asserts that there must be absolute max and absolute min for functions and intervals which satisfy the hypotheses of the theorem. But the theorem doesn't give any advice about the location of these values. Indeed, if you are naive, you might waste a huge amount of time just trying random values of the function and trying find big and small values.
Ms. Practice says to follow the process above, and you've got to find the max and min. The advantage is that in most "real" cases there are only a few critical points, and the procedure allows you to narrow the search for abs max and min to only a few numbers.

I will admit that in practice, though, it can be confusing. You look at f(a) and f(b). You compute f´(x). Where is it 0? Where does it not exist? Take those numbers and plug them back into f. So there are two functions, f and f´, wandering around. You need to understand what's necessary.

I concluded the lecture by solving a few of the many problems in section 4.2. It can be argued that (as I said) these are toy problems but you should get familiar with the method using these toys, and then you can play (?) with more realistic examples.

Section 4.2, exercise #16
Find the extreme values of 2x³–9x²+12x on [0,3] and [0,2].

I think I discussed [0,2] in class. Let me analyze [0,3] here. I will try to be strict and not distract myself with pictures. I see my initial task as creating a list of the possible locations of the extreme values. Well, we will include the endpoints 0 and 3. Now we need any critical points inside [0,3]. The derivative is 6x²–18x+12. The derivative formula for this nice function, a polynomial, is valid everywhere. Let's see where it is equal to 0. So:
    6x²–18x+12=0 means 6(x²–3x+2)=0 means 6(x–2)(x–1)=0.
Please note that I sincerely doubt you will encounter very many polynomials which can be factored in real applications. So the roots are at 2 and 1.

The list of candidates is: 0 and 3 (endpoints) and 1 and 2 (critical points). The values of the original function 2x³–9x²+12x at these points are: 0 (at x=0), 9 (at x=3), 5 (at x=1), and 4 (at x=2). The absolute max is 9 and the absolute min is 0 on the interval [0,3].

I don't need to examine any other values. We're done.

Section 4.2, exercise #38
We should look at y=(1–x)/(x²+3x) on the interval [1,4]. The only initial worry is whether this formula defines a continuous function on the interval, but all of the input x's are positive, and I am sure that the bottom, x²+3x, won't be 0 for positive x's.

My candidate list includes the endpoints 1 and 4. We need dy/dx. This is (Quotient Rule)

(–1)(x2+3x)–(1–x)(2x+3)
------------------------
     (x2+3x)2

Where is this equal to 0? The only way a fraction can be 0 is if the TOP is 0 so forget about the BOTTOM! We solve (–1)(x²+3x)–(1–x)(2x+3)=0.
    (–1)(x²+3x)–(1–x)(2x+3)=0 means –x²–3x+(x–1)(2x+3)=0 means –x²–3x+2x²+x+3=0 means x²–2x+3=0 means (x–3)(x+1)=0.
Again, this is not "real". Real polynomials rarely have nice roots, and almost always the roots can only be found approximately. Well, the TOP=0 when x=3 or x=–1. We discard x=–1 because it is not in the interval of interest, [1,4]. We only need critical points inside the interval.

Our list of candidates in this case is composed of the endpoints 1 and 4 and the critical point 3. The values of the original function (1–x)/(x²+3x) are: 0 (at x=1), –3/28 (at x=4), and –2/18 (at x=3). The absolute max is at 0 (that's easy). Then (if I'm doing this "by hand") I need to worry a bit about which of the other two values is more negative. In fact, –2/18=–1/9 is less than –3/28 (because that is larger than –3/27 -- or just compute them). So the absolute min is –1/9.

Again, the whole idea is that we really don't need to check many values of the function. This turns out to be terrific. BUT in order to do this successfully, even with toy problems, and to do things which occur later, every day in this course, you will need to be able to compute the derivatives of functions defined by formulas. Please know how to do this.

---

Tuesday, October 20	(Lecture #14)

Two very useful tricks
The lecture today will discuss two extremely useful "tricks" which are constantly used, both computationally and theoretically. The relevant textbook sections are 4.1 and 4.8. Please do not get distressed at the numbering of the sections. The two techniques are quite related -- they use the basic idea that a tangent line is close to the curve.

Some numerical evidence
I began by presenting some numerical "evidence". Here is a table of numbers and their square roots which was examined by students. The table has 20 digit accuracy, far more than any real world application I know of. But the point of what's given here is to help people think about things. The table has a few more entries than I wrote on the board, and since I got it straight from the computer, it has all 20 digits.

Number	20 digit square root of the number
4.	2.00000 00000 00000 0000
4.1	2.02484 56731 31658 6933
4.01	2.00249 84394 50078 5728
4.001	2.00024 99843 76952 8199
4.0001	2.00002 49998 43751 9531
4.00001	2.00000 24999 98437 5020
4.000001	2.00000 02499 99984 3750
4.0000001	2.00000 00249 99999 8438

I wanted people to examine the table for patterns. The first column was, of course, 4{period}{some 0's}1. The second column, well, since the square root of 4 is 2, and the numbers in the first column were close to 4, then since square root is a continuous function, the numbers in the second column were close to 2. But what's the pattern of closeness? There are more 0's between the 2 and the first non-zero digit. But the non-zero digits seem to have some pattern. For example, consider the entry corresponding to 4.0001. That entry is 2.000024999 8437519531. There are 4 zeros between "." and "2". But the structure of the non-zero digit string following is 24999998 and it sure looks like it wants to be a bit simpler, just 25000000. What's going on?

The algebraic way
Well, we have written the following a large number of times.
    f(x+h)=f(x)+f´(x)h+Err(h)
and we know that the Err(h) term should →0 faster than a multiple of h. In the case we have above, f(x)=sqrt(x) and we're considering things that happen what x=4. Well, f´(x)=(1/2)1/sqrt(x). At x=4, we get f(4)=2 and f´(2)=1/4 and this is .25 decimally. So the equation becomes
    sqrt(4+h)=2+(.25)h+Err(h)
(Here the Err(h) term was also called H.O.T., for "higher order terms" earlier in the course. If we just want an approximation, hey, we could forget the Err(h) term and get
    sqrt(4+h)≈2+(.25)h
Here I am using the weird symbol ≈ for the phrase "approximately equal" and I am deliberately not being too precise, right now, about what this means. But let me rewrite the table above, now with more decorations (!).

Number	20 digit square root of the number
4.	2.00000 00000 00000 0000
4.1	2.02484 56731 31658 6933
4.01	2.00249 84394 50078 5728
4.001	2.00024 99843 76952 8199
4.0001	2.00002 49998 43751 9531
4.00001	2.00000 24999 98437 5020
4.000001	2.00000 02499 99984 3750
4.0000001	2.00000 00249 99999 8438

Here what is underlined is essentially the 2+(.25)h stuff. Look: the Err(h) term is shrinking twice as fast as the h is shrinking. And it is negative. So for 4.0000001, with 6 zeros, the 2+(.25)h with h=.0000001 appears in 16 places (except you need to think a bit since the correction is negative, and therefore it appears as a digit string of 24 followed by 9's). So there is evidence (yes, I admit it, this is a "toy" problem, and all the numbers are very easy!) that the Err(h) is sort of h².

Geometric interpretation
What does 2+(.25)h mean? Look at the picture. Everything is not meant to be totally, literally accurate, but I'd like the ideas to be correct. The circular area on the graph is magnified. We have a point (x,f(x)) on the graph. x is moved (I drew the picture so that h is a small positive number, and x+h is to the right of x). In the local, magnified picture, there's a right triangle. One leg, the "adjacent", has length h. The other is labeled ?. The line is tangent to the graph at (x,f(x)). But the slope of that tangent line is f´(x). The slope of the line is the tangent of the angle that line makes with respect to a horizontal line, such as the leg labeled h. So f´(x) is the quotient of ? and h: f´(x)=?/h. And ? must be f´(x)h, and this length is added on to the vertical height that (x,f(x)) has from the x-axis. The result is f(x)+f´(x)h. Hey! Using the tangent line to make that little approximation is exactly the same as starting from (x,f(x)) and making the adjustment from x to x+h by riding along the tangent line. This is why
    f(x+h)f(x)+f´(x)h
is frequently called the tangent line approximation. It also explains why, in our specific case, the approximation is always greater than the true value, because the parabola on its side is underneath the tangent line.

In class ...
I actually found the formula for the tangent line, y–2=(1/4)(x–4) and then substituted in 4+h for x to get y–2=(1/4)h so that y=2+(1/4)h on the tangent line. So the tangent line value is actually this approximation. That's why it is called the tangent line or linear approximation to f(x+h).

All together?
I would like you to see how all of these points of view actually reinforce each other. As I've remarked in class, I like the pictures, but I know that many people prefer a formula, and certainly still others like to see convincing numbers. That's why I've tried to show you these varied approaches.

What about the error?
What the heck is the error like? Is it big or small? What can we say? Well, I first remarked that we could see in our specific case that as we go from left to right on the graph, the tangent lines got flatter (that is, the slopes of the tangent lines themselves decreased). Each tangent line was on top of the curve. The linear approximation was an overestimate, and (as we will see later) since f´ is decreasing, the second derivative, f&180;´, is negative. The sign of the error is more or less determined by the sign of the second derivative. The linear approximation is larger than the true value when f´´(x)<0 and it is smaller when f´´(x)>0. I didn't say anything about the size (the magnitude) of the Error.

Irving
Irving is someone who remembers a formula and tries to fit every situation to a formula. So Irving remembers f(x+h)≈f(x)+f´(x)h and something about sqrt(4)=2. He decides to approximate sqrt(100) in the following manner:
     f(x+h)≈f(x)+f´(x)h
becomes
     sqrt(100)=sqrt(4+96)≈sqrt(4)+(1/4)(96)=26.
Here Irving uses f(4)=4 and f´(4)=1/4 and h=96. I don't think that 26 is a very good approximation to sqrt(100)=10. Irving may remember the formula, but I think the formula has been used inappropriately.

It turns out that the magnitude or size of the Error is roughly proportional to h². When h is teensy-weensy (maybe that word is too technical?) then h² is much teensier than h. This is what was shown above in the table. That we only get a discrepancy of 16 (from 10 to 26) in this case is not as bad as it could be. I am joking here, but the decision about when to use (and "trust"!) the linear approximation may not be too easy in practice (as Ms. O'Sullivan managed to get me to admit!).

Fifth roots
Here is another "classroom" example of this approximation.
I know that 32^1/5 is 2. What is a (useful) approximation to (32+h)^1/5? Here f(x)=x^1/5 and f´(x)=(1/5)x^–4/5. So the formula f(x+h)≈f(x)+f´(x)h with x=32 becomes (32+h)^1/5≈32^1/5+(1/5)(32)^–4/5h and (after arithmetic, since 32^–4/5 is 1/16) this is (32+h)^1/5≈2+(1/80)h.

For example, 31^1/5 is reported as 1.987340 by a computer, and the approximation (with h=–1) is 31^1/5=(32–1)^1/5≈2+(1/80)(–1)=1.9875000. I didn't expect that the approximation would be this good. Notice that f´´(x)=(1/5)(–4/5)x^–9/5 so f´´(32)=(1/5)(–4/5)32^–9/5 is negative. So just as in the square root example, the estimate is larger -- an overestimate.

A more real problem... One of the early experiments attributed to Galileo is observation of a pendulum. He asserted that the period of a pendulum depended on the length of the pendulum and was independent of the size of the angular oscillation. This is really remarkable and means that maybe you can use a pendulum clock to time things accurately --- only the length of the pendulum matters and not how far or wide it swings. It turns out this assertion is actually true only for small oscillations. Galileo supposedly only observed small oscillations. A mathematical model of pendulum motion (go to your physics instructor!) gives d2θ/dt2 =(Constant)sin(θ) (here θ is the angle between the pendulum and the vertical direction). This differential equation is too darn hard to "solve" (you will see why later) and so people want a simpler model. If θ is very small, then (since sin(0)=0 and sin´(0)=cos(0)=1) we can substitute θ for sin(θ) and the equation becomes d2θ/dt2 =(Constant)θ. This equation is much simpler (I'll actually be able to tell you all about it in a few weeks). And the predicted independence of period turns out to be true, with this small oscillation assumption.

What is the square root of 4? Or the square root of 2?
Sigh. I'm fairly sure that the square root of 4 is 2. I would like to know the square root of 2. Generally, there are many occasions when we would like to solve f(x)=0. This turns out to be impossible in terms of standard functions and formulas even for relatively simple f(x)'s. Really what I wanted to discuss was the single most widely used method of taking an approximation to the root of an equation and improving the approximation, that is, getting closer to the root. The improvement technique is called Newton's Method. I called the old guess, G, and the (hoped for) improved new guess, N.

Newton's Method
Newton's method is a way to (try to) improve a guess at a root of f(x)=0 when f is a differentiable function. The guess, G, is (hopefully!) improved with the following process (as you read this, please glance at the picture to the right). First, go up from G until you "hit" the graph of y=f(x). The point will be (G,f(G)). Then "slide down" the tangent line of the graph. The slope of that tangent line must be f´(G) because it is tangent at (G,f(G)). The point that this line hits the x-axis is the new guess, N. The picture is rather simple and shows the new guess closer to a root of the function. Again, this is a simple picture, and is the way we would like the method to work. I will discuss more horrible possibilities later, but right now I would like to get a formula for N in terms of G. Well, the slope is f´(G), but this slope is also equal to OPPOSITE over ADJACENT because the slope of a line is the tangent of the angle that the line makes with the x-axis (this is the same key idea we've already used once before in this lecture). Here OPPOSITE is f(G) and ADJACENT is G–N. Therefore f´(G)=f(G)/(G–N) which is the same as G–N=f(G)/f´(G), so that

For square roots
Suppose I want to compute or approximate sqrt(A). What I am describing is the method that is actually used in most calculators and computers. I need a function which has a root at sqrt(A). The simplest candidate seems to be f(x)=x²–A and then f´(x)=2x, so that G–{f(G)/f´(G)} becomes G–{[G²–A]/2G} which is {2G²–[G²–A]}/2G which is [G²+A]/2G which is (1/2)[G+{A/G}]. That is, "Improve the guess to a new guess, N, by taking the average of the old guess, G, and A/G: N=(1/2)(G+[A/G])."

Here is the complete "coding" (writing and running of the "program") for Newton's method to compute the square root of A=2 using Maple. Yes, this is silly, but you should see again what the numbers look like.

> N:=G->(1/2)*(G+(2/G));
                                  N := G -> 1/2 G + 1/G
> N(2.);
                                       1.500000000
> N(%);
                                       1.416666667
> N(%);
                                       1.414215686
> N(%);
                                       1.414213562
> N(%);
                                       1.414213562
> sqrt(2.);
                                       1.414213562

The first command, N := G -> 1/2 G + 1/G defines how to replace the old guess, G, by a new guess N. The symbol "%" means "Use the previous answer." If we initially "guess" 2 as sqrt(2), then only 5 uses of Newton's Method gets sqrt(2) to 10 decimal place accuracy.

So just a few iterations (repetitions of N) makes the estimate close enough to sqrt(2) that we can't tell the difference (to 10 decimal digits)! This is pretty darn fast, and this is actually how your calculator computes square roots. There is considerable evidence suggesting that this method of computing approximations to square roots has been known for 4,000 years. Try the key words Babylonian square roots in a search engine. Human beings can be quite clever.

The picture to the right is an attempt to show you how Newton's method works geometrically for the specific function f(x)=x²–4 with a specific initial guess of 3. So there's the graph, y=x²–4, the parent parabola pushed down 4 units. Take the initial guess at 3 (what's described now is all in magenta) and go up until you hit the parabola. Slide down the tangent line until you hit the x-axis. Then up to the parabola again, slide down the tangent line, etc. This is very silly, but look at the magenta line segments. It is rather difficult to draw them so that they can be seen, because they tend to pile up so quickly at the point where the parabola and the x-axis intersect. This is remarkably fast when it can be used, much faster than bisection. Newton's method generally doubles (!) the number of decimal digits of accuracy at each repetition. (This is really really really good.)

Dividing is difficult Of the arithmetic things we all learned early in grade school, certainly division is the one that I and many other people believe is the most difficult. To find, say, 1/7, you need to make some guesses at places in the computation, and sometimes the guesses are just wrong. Some "backtracking" is needed. What the heck can we do if we want to tell machines how to compute this? Here, very briefly, is what we can do. Suppose A is a positive number and we want to compute an approximation to 1/A. Well, 1/A is a root of f(x)=(1/x)–A. Then f´(x)=–1/x2. Notice in all this, please, that computing derivatives must be totally routine, or else everything becomes much too hard. Then the Newton's method equation, N=G–{f(G)/f´(G)} becomes     N=G–{[(1/G)–A]/[–1/G2] and this can be simplified (changing compound fractions to simple fractions):     N=G+[(G)–AG2]=2G–AG2 This means that to compute 1/7 we should do the following: replace an old guess, G, by a new guess, 2G–7G2. Please appreciate how easy this is compared to long division, with guessing and subtracting and then doing it again! Here is what happens when we try to use this: > INV:=x->2*x-7*x^2;   (Comment This defines how to replace the old guess.) > INV(.1); 0.13 > INV(%); 0.141 > INV(%); 0.1417 > INV(%); 0.14284777 > INV(%); 0.1428571422421897 > INV(%); 0.14285714285714285450 This last answer is accurate to 17 decimal digits. This is almost surely more than any moderately sane human being needs. QotD What is 101/3? More precisely, define a fairly simple function f which has 101/3 as a root. Then use the initial estimate of x=2 in Newton's method twice. Here the function that I thought of is f(x)=x3–10. And f´(x)=3x2. The Newton's method would take a guess, x, and replace it by x–(x3–10)/(3x2). If the initial guess is x=2, the replacement, which we hope is closer to the true value, would be 2–(23–10)/(3[22]). This is 2–(–2)/12=2+1/6=13/6. Then replace 13/6 by {13/6}–({13/6}3–10)/(3{13/6}2) The exact result is (not clearly!) 3277/1521 (yeah, I did not do this by hand!). How about decimal computation? If the initial guess is 2, then the replacement is 2.1666666667 and that, in turn, becomes 2.1545036160. Sigh. The computer reports that 101/3 is 2.1544346900 so that we seem to have gotten about three decimal places of accuracy. Three iterations gets 8 places of accuracy, and four iterations gets 15 places of accuracy. Is this exciting? Bad things can happen The Bisection Method is much slower than Newton's Method, but the Bisection Method (assuming that the hypotheses are satisfied!) always works. Newton's Method is notorious for fouling up, sometimes especially when you don't want problems. Let me show you what can happen. Look at the picture to the right. It is a graph of a function, y=f(x). The function has one root, at x=R. Let's examine the effect of various initial guesses. If the initial guess is x=A, then bouncing up and sliding down, etc., works really nicely. The repeats of Newton's Method converge nicely and rapidly to the root, R. Suppose the initial guess is at x=B. Then up to y=f(x), and the tangent line pushes the next guess way to the left. One should have (I certainly don't have!) any feeling about what would happen out there. Certainly things may not come back to the root. Suppose the initial guess is at x=C. Then Newton's Method pushes the following guesses out and out and out to the right (think of a curve that's asymptotic to the x-axis but has no root to the right). So again the situation is highly unsatisfactory. In practice people may check to see if things are working well. Usually (experience!) it is when you don't check that Newton's Method misbehaves.

---

Thursday, October 15	(Lecture #13)

How a rectangle changes
A rectangle is changing with time. At a certain time, its length is 7 inches, and is decreasing at .6 inches per second. At the same time, its width is 3 inches, and is increasing at .5 inches per second.
    Question 1 At that time, is the area of the rectangle increasing or decreasing? At what rate is the area changing?
    Question 2 At that time, is the length of the diagonal of the rectangle increasing or decreasing? At what rate is the diagonal changing?

What I might think about this
Here is how I might think about this problem. I don't know how many of these details I would write as I analyze the problem, but almost surely I would think what I write here.

1. The nouns of interest are rectangle, length, width, area, and diagonal.
2. I'd probably make a sketch of as much of the situation as possible. The sketch might look something like what it shown to the right.
3. I would label the sketch with as many of the "nouns of interest" (?) as possible. I usually try to choose labels that sort of match the nouns or ideas they designate (so I use L for length). I also try to make the sketch match as much additional information as possible in the problem. For example, I labeled the longer side of the rectangle L mostly because in the problem statement, L was 7 at the "certain time", while W, the width, was 3, and 3 is less than 7. There are two diagonals in the rectangle, and the diagonals have equal length. The diagonal I chose to draw and label with D is that one which completes a right triangle with legs L and W, and the reason I chose that diagonal is I was thinking ahead to the next step.
   I am not trying to be silly about this problem. But, as much as possible, I am trying to slow down and give a precise, step-by-step, discussion of my approach to this problem.
4. I would try to connect the variables algebraically, somehow making the picture into equations. For example, I'll call the area, A, and, although A is not shown in the picture, I do know that A=L·W. Also, since I "completed" a right triangle with D as hypotenuse, then the Theorem of Pythagoras tells me that D=sqrt(W²+L²).
5. Now what? Maybe I can
   Answer the first question Since A=L·W, and we need to know about dA/dt, I can try to d/dt the equation. I assume in everything that I'm doing that I can differentiate things. So the Product Rule gives us (since both L and W are functions of time) that dA/dt=(dW/dt)L+W(dL/dt). At the "certain time" we know the values of the four terms which appear on the right-hand side of the equation. dW/dt is .5 (I'm not worrying about the units right now) and W is 3. L is 7 and dL/dt is –.5 (here the interesting thing is the –, and we need the minus sign because of the word decreasing in the problem statement). Therefore dA/dt=(.5)(7)+(3)(–.5)=3.5–1.5=2. The area of the rectangle is increasing at the rate of 2 square inches per second.
6. And I can
   Answer the second question Here D=sqrt(W²+L²), and we must d/dt this equation also. Some care is needed. Again W and L are functions which vary with time, that D's relationship with them is complicated. Use the Chain Rule with some care. The result seems to be: dD/dt=(1/2)(W²+D²)^–1/2(2W(dW/dt)+2L(dL/dt)). Now we "instantitate" with the values we have at the "certain time" (but do it carefully!): dD/dt=(1/2)(3²+7²)^–½(2·3(.3)+2·7(–.6)). Again, watch the darn signs. The result, if I do the arithmetic correctly (!) will be dD/dt=(58)^–½(–3.3). So the length of the diagonal is decreasing (!) at the rate of (58)^–1/2(3.3) inches per second.
7. Again, I will agree that this is not a profound problem. But I'll also write what I stated in class: the {in|de}creasing nature of the results sure isn't clear from the problem statement (in fact, I had to arrange the numbers carefully to get the qualitative differences: that the area is increasing and the diagonal is decreasing: slightly weird).
   As I mentioned in class, I am experienced but I generally need to read this type of problem two or three times (at least) to make more likely that I understand what is varying, and what information is given and what information is requested. Even if the person who wrote the problem is trying to be helpful, there can still be real difficulty in comprehension.

How an annulus changes
An annulus or annular region is a region in the plane which is between two circles which have the same center (concentric circles). For example, you can see such a region if you drop a rock into a quiet pond. There are likely to be a series of circular ripples which spread radially and evenly from the point of impact of the rock. Suppose we have an annular region changing with time. At a certain time, the outer radius is 6 inches and decreasing at .3 inches per second, and the inner radius is 4 inches and decreasing at .2 inches per second. At that time, is the area of the annulus increasing or decreasing? What is the rate of change of the area?

This was the QotD.
Here's a discussion of a solution. The geometry of this annulus is determined by the radii (that is the plural of radius, I think). There is an inner, smaller radius which I'll call r, and an outer, larger radius which I'll call R. The area of the region is the difference of the area of the region bounded by the outer circle, Π R², and the area of the region bounded by the inner circle, Π r². The area, A, of the annulus, is given by A=Π R²–Π r². The extra minus sign, together with the two uses of the word "decreasing" in the problem description make the solution of the problem interesting to me. I can't easily guess whether the area is increasing or decreasing.

If we differentiate the equation carefully, the result is dA/dt=2R(dR/dt)–2r(dr/dt). At the "certain time", we get dA/dt=2·6(–.3)–2·4(–.2)=–3.6+1.6=–2 inches per second. So the area of the annulus is decreasing at a rate of 2 inches per second at that time.

An ant crawls on a parabola
An ant is crawling on the parabola y=x². (Fairly absurd, but maybe slightly more realistic and useful than the first two problems.) Suppose that the horizontal, x, coordinate of the plane is increasing at 6 units per second when the ant crawls through (?) the point (2,4). How fast is the y coordinate of the ant's position changing at that time?
So x and y are functions of time and y=x². If we d/dt the equation, the result is dy/dt=2x(dx/dt). When the ant is at (2,4), this becomes dy/dt=2·2·6=24.

A bit more ant crawling
Suppose that the ant crawls on y=x² so that the rate of change of its first, x, coordinate is always 6 units per second. What happens to the rate of change of the y coordinate as the x moves far to the right? I'll try to both compute this and explain it.

Let's "think" about the situation. The picture to the right is to help with thinking. If we take a tiny piece of the parabola, and blow it up, the piece becomes almost a straight line (the parabola is locally linear). For a piece near (2,4), say (similar to the lower piece shown), the line is not too steep. For a piece "far" to the right (as the upper piece shown), the line is rather steep. Imagine the ant traveling in such a manner that the x coordinate changes steadily with a rate of 6 units per second. There will be different y changes to allow the ant to stay on the curve. In the more right, more up box, the y change will be larger than the y change needed in the lower box. So I conclude that dy/dt will grow as the ant moves more to the right.

Now let's look at the algebra. y=x² again implies dy/dt=2x(dx/dt) and if dx/dt is always 6, then dy/dt=12x. Indeed, when x→∞, dy/dt→∞.

Change the crawling
What if the ant crawls so that the vertical, y, coordinate changes steadily at 6 units per second. Again the motion of the ant along y=x² moves to the right and up. What happens to the x-coordinate change as the ant moves steadily up?

It might be useful for you to again consider the local linearization pictures, which I hope will tell you in this case that if dy/dt is constant, dx/dt must get smaller as the ant moves to the right. And the algebra: since dy/dt=2x(dx/dt) and here dy/dt is supposed to be 6 always, we see that dx/dt=3/x. As x→∞, dx/dt→0.

Of course what I am doing is considering the velocity vector of the ant's motion under various hypotheses concerning components of the vector. This velocity vector is tangent to the ant's path, and this geometric fact forces the other conclusions

A conical reservoir is being filled
A right circular cone is an object which shows circular slices when cut by a plane perpendicular to its axis of symmetry. The sides of the cones are formed by straight lines through the vertex. The vertex of the cone is the pointy part.
Now suppose we have a reservoir which is a right circular cone with its vertex at the bottom. The height of the cone is 30 feet, and the base radius is 5 feet. The cone is being steadily filled by a fluid at the rate of .2 cubic feet per minute. How fast is the depth of the fluid changing when the fluid is 20 feet deep in the reservoir?

Oh mechanical engineers: "A fluid is defined as a substance that continually deforms (flows) under an applied shear stress regardless of the magnitude of the applied stress." I asked for some suggested fluids and got almost no answers. Sigh.

The pictures show a right circular cone with vertex at the bottom. The oblique view is prettier, I think, but a transverse "section" through the axis is probably better for analyzing this problem. A formula which certainly is needed follows: V=(Π/3)R²H is the volume of a right circular cone with height H and base radius R. If the fluid level is H and the radius of the fluid-filled volume is R, then the volume filled by the fluid is V=(Π/3)R²H. We know also that dV/dt=.2 (positive: the cone is being filled). Well, V seems to have too many letters for this course (in calc 3, functions involving several variables are dealt with). But there is a geometrically forced relationship between R and H which I hope you see. Look at the transverse picture and at several similar right triangles. Ratios of corresponding sides must be equal for the two right triangles shown, so 30/H=5/R and R=H/6 (I want V in terms of H because the problem statement asks for the rate of change of the depth of the fluid, which I believe is dH/dt). Therefore V=(Π/3)R²H=(Π/3)(H/6)²H=(Π/108)H³. If we d/dt the equation V=(Π/108)H³ being careful to use the Chain Rule appropriately, the result is dV/dt=(Π/108)(3H²)(dH/dt)=(Π/36)H²(dH/dt). Now plug in the known dV/dt and the depth, H=20, to get .2=(Π/36)(20)²(dH/dt). Therefore the rate of change of the depth when the depth is 20 feet is .2/[(Π/36)(20)²] feet per minute.

Please notice that since dV/dt is constant (.2) we know that dH/dt=.2/[(Π/36)H²]. So we see that as H increases, the rate of change of H decreases. This is because the cross-sections of the cone's volume are steadily increasing so the steady volume increase means that the height is not increasing as much.

Another way (as done in class)
This isn't what I did in class. There I took the whole equation V=(Π/3)R²H and d/dt'd it to get (Chain Rule and Product Rule) dV/dt=(Π/3)2R(dR/dt)H+(Π/3)R²(dH/dt) . If H=30 and dV/dt=.2 we have .2=(Π/3)2R(dR/dt)(30)+(Π/3)R²(dH/dt) but 6R=H (deduced above and in class) implies first that when H is 20, R is (20/6). Also it implies that 6(dR/dt)=dH/dt or dR/dt=(1/6)(dH/dt). Then .2=(Π/3)2R(dR/dt)(30)+(Π/3)R²(dH/dt) changes to .2=(Π/3)2(30/6)(21/6)(dH/dt)(30)+(Π/3)(20/6)²(dH/dt). We can then solve the equation for the desired dH/dt. I think this answer is the same as obtained by the earlier method.

---

Tuesday, October 13	(Lecture #12)

I think lecture #11 was the first exam. So here is (more or less) what we did today.

Implicit differentiation can help find the derivatives of inverse functions. The inverse function to sine was discussed yesterday by Mr. Nakamura, but I will include a diary entry from a previous 151 course about this material.

arcsine What's happening in the pictures below (left to right): The first picture is supposed to be a portion of the graph of sine. It is 2Π periodic, and its range is [–1,1]. The green line is the "main diagonal", y=x, which also happens to be tangent to y=sin(x) at (0,0). This is because the slope of the tangent line is the derivative of sine, which is cosine, and cos(0)=1. To get the inverse function, we interchange inputs and outputs. Geometrically we flip the graph over the mBBain diagonal, and get the second picture. The tangent line is still tangent, but now, look at the red line. This demonstrates that the flipped graph is not the graph of a function. It fails the vertical line test to be a graph of a function. Thus we need to cut away (!) part of the graph. The "clouds" in blue-green (?) demonstrate what will be cut away. And what's left is shown in the third picture. This is the official graph of y=arcsin(x): domain [–1,1] and range [–Π/2,Π/2]. It has arcsin(0)=0, and the tangent lines seem always to slope up, so the derivative should be positive. And if we are very careful, the lines tangent to sine at +/–Π/2 are horizontal, so the lines tangent to the flipped curve will be vertical and have no slope so there will be no derivative at +/–1. The derivative of arcsin should have domain (–1,1), the interval without endpoints. Consider this process: y=arcsin(x)   The inverse function sin(y)=sin(arcsin(x))   Using the inverseness sin(y)=x   Recognition of inverseness cos(y)(dy/dx)=1   Implicit differentiation dy/dx=1/cos(y)   Solving for dy/dx Since sin(y)=x, x2+(cos(y))2=1, and so cos(y)=+/–sqrt(1–x2). Which sign to take? In the range we are considering between –Π/2 and Π/2, cosine is positive. Also the slope of the tangent line is positive for arcsine. We will take the + sign. And cos(y)=sqrt(1–x2).   Solving for dy/dx stuff in x So arcsin´(x)=1/sqrt(1–x2)   Statement of formula We will go through this a few more times today. But in any case, the derivative is 1/sqrt(1–x2). Notice that for –1<x<1, this is positive, and that the derivative formula is not valid at +/–1 as the geometric evidence predicts. Example The derivative of arcsin(5x3–ex) is [1/sqrt(5x3–ex)](5·3x2–ex) using the chain rule.

arctan
What's the picture below supposed to show? The initial picture is y=tan(x). This function is periodic with period Π, and its domain does not include odd multiples of Π/2. The function is rather simple looking (!), always tilted up, and has vertical asymptotes at odd multiples of Π/2. Flipping to get an attempted inverse function reveals lots of problems (I omitted the red line here). The standard restriction is to throw out the "branches" that don't intersect the horizontal axis, and that's what I've attempted to suggest with the blue-green "clouds". Again, y=x is a tangent line to both arctan and tan at (0,0) because the derivative of tangent is (sec(x))², and sec(0)=1/cos(0)=1. Arctan is very useful. It "compresses" all of the real numbers into the interval from –Π/2 to Π/2, so if you have lots of data and you don't know ahead of time how big (+ or –) the data will be, composing it with arctan will at least control it a bit. Now for the derivative. Notice that the derivative of arctan should always be positive (the tangent lines all tilt up) and that the slopes→0 as x→+∞ or as x→–∞, since the arctan curve gets flatter there.

1. y=arctan(x)   The inverse function
2. tan(y)=tan(arctan(x))   Using the inverseness
3. tan(y)=x   Recognition of inverseness
4. (sec(y))²(dy/dx)=1   Implicit differentiation
5. dy/dx=1/(sec(y))²   Solving for dy/dx
6. Since tan(y)=x, x²+1=(sec(y))² (much easier than arcsine!).   Solving for dy/dx stuff in x
7. Thus arctan'(x)=1/(1+x²)   Statement of formula

The derivative function, 1/(1+x²), is always positive as we thought. And as x→+∞ or as x→–∞, the function 1/(1+x²) does →0, so this formula does have the desired behavior out towards the "edges".

Examples
I did several, maybe something like this.
The derivative of arctan(e^(5x4)) is 1/(1+(e^(5x4))²· e^(5x4)·5·4x³: somewhat of a mess, using the chain rule several times.
The derivative of arctan([x+1]/[x–1]) is 1/(1+{[x+1]/[x–1]}²)·{[1(x–1)–(x+1)1]/[(x–1)²]}

Looking at a constantly accelerating rocket ship
Here is almost a real computation. It is my effort to show how arctan can be used.
Let's consider a rocket ship launching vertically upward with constant accelaration a. We know from physics class that at time t, the rocket ship is {1/2}at² high. Suppose we are standing a distance D from the launch point of the rocket ship. The rocket ship starts from velocity 0 and seems to move slowly. Then it seems to move faster. Finally, when it is high up in the sky, its apparent motion is slow again, although it is actually moving very fast. Let's call θ the angle betwwen the ground and the rocket ship. How does θ change with time? Well, tan(θ)={1/2}at²/D={a/(2D)}t² so that θ=arctan({a/(2D)}t²).

The rate of change of θ with respect to time is dθ/dt, which we could call angular velocity. What is it? The Chain Rule tells us.
   dθ/dt={1/(1+[{a/(2D)}t²]²)}{a/D)}t I will act against many previous comments and simplify this, because I will want to do further computations. So, after simplifying, we have
   dθ/dt={[(a/D)t]/(1+[{a/(2D)}t²]²)}={[(a/D)t]/(1+[{a²/(4D²)}t⁴])}.
The angular velocity begins small because the rocket ship is moving very slowly. Later this gets larger as the rocket speeds up. But, eventually, as the rocket gets very high in the sky, even though its velocity might be quite large, the apparent position of the rocket seems to change only slightly, so the angular velocity decreases towards 0. The curve to the right describes this qualitative behavior. If you consider the algebraic formula we just got for dθ/dt you will see:

• If t is small, the bottom is 1+(Constant)t⁴, and t⁴ is very small (small numbers get smaller even when powers are done). So the bottom is ≈1. The top is (Another constant)t, so we have that dθ/dt is about (Another constant)t/1 so it must be small when t is small.
• If t is very large, then the top is t¹ but the bottom has t⁴ so certainly the bottom overpowers the top. (Note: The word "overpower" is supposed to be sort of a joke ["powers of t" -- get it?] and I will show you later other ways to decide that this formula for dθ/dt is small when t is large.)

So the formula and my understanding of the physical situation lead me to realize that maybe there is some time, t_MAX, where the angular velocity is largest. What is that time? Well, it will occur when the tangent line to the curve (the dθ/dt curve) is horizontal. That means we need to find t so that the derivative of dθ/dt is 0. So we need to consider what we could call angular accelaration.

Let's differentiate {[(a/D)t]/(1+[{a²/(4D²)}t⁴])}. we need the quotient rule. Please realize that a and D are constants, and t is the variable.

(a/D)(1+[{a2/(4D2)}t4])-[(a/D)t](0+{a2/(4D2)}(4t3) 
-------------------------------------------------
              (1+[{a2/(4D2)}t4])2 .

This is (wonderful but traditional notation!) d²θ/dt². The wonderfully unsymmetric placement of the 2's is done because ... because ... someone did it centuries ago. Well, for which t is this 0? So we have a fraction, TOP/BOTTOM. It will be 0 exactly when (and only when!) the TOP is 0. So we need to find t so that
   (a/D)(1+[{a²/(4D²)}t⁴])–[(a/D)t](0+{a²/(4D²)}(4t³)=0.
This was confusing enough in class, and I can only hope that my messy formulas can be understood here. Let's distribute. Some "miracles" occur.
   (a/D)+[{a³/(4D³)}t⁴]–{a³/(D³)}(t⁴)=0.
Now multiply by 4D³ (certainly this is not the only way of making sense of this mess, but I like to get rid of fractions when possible -- when you present this problem, go ahead and do the algebra the way you like!).
   4aD²+[a³t⁴]–{4a³(t⁴)=0.
The last two terms combine.
   4aD²–3a³(t⁴)=0.
And now we "solve" for t by pushing the minus term to the other side and dividing by some constants and taking the fourth roots (1/4):
   So t_MAX=((4/3)(D²/a²)^1/4.

What a lot of work. You can now find the maximum angular velocity by plugging in the expression for t_MAX into the formula for dθ/dt. The result is complicated and doesn't tell me much. There is one thing which is somewhat amazing to me. Let me find the angle, let me call it θ_MAX, at which the maximum angular velocity occurs. So we need to plug t_MAX into θ=arctan({a/(2D)}t²). Here we go:
θ_MAX=arctan({a/(2D)}(t_MAX)²)=arctan({a/(2D)}(((4/3)(D²/a²)^1/4) ²)=arctan({a/(2D)}((4/3)(D²/a²)^1/2).
Now look again at the input, the "argument", of arctan. The ((4/3)(D²/a²)^1/2 stuff is really [2/sqrt(3)](D/a) and this multiplies {a/(2D)} so lots of stuff cancels. Therefore θ_MAX is arctan(1/sqrt(3)) which is Π/6 (o.k., darn it, 30^o) always. There is no a and there is no D in this answer. I find this amazing and amusing (?). I wish that I understood why this is true physically -- that is, if there were some reasoning without all this algebra to convince me that the answer should not have a and D. I don't know such reasoning.

Drop a rock (on any planet!) Suppose you were at the top of a tall cliff on Planet X (for any value of X!). And a distance D away on the horizontal top of the cliff an alien dropped a rock. It would fall with some unknown acceleration due to local gravity. If the cliff were tall enough (!!), as you looked at it, the angle of inclination would increase. You would turn your head faster and faster, until, exactly at angle -Π/6, the angular velocity of the dropping rock would be a maximum. Then this angular velocity (how fast you would be rotating your vision) would begin to slow. WHY doesn't this depend on the distance between you and the alien, and why doesn't it depend on the local acceleration of gravity? I don't know (in some real "physical" way) and I wish I did. If you have some insight, please tell me. I thank you in advance.

arcexp (?)
Should this be called arcexp? Well, it isn't. The inverse function to an exponential function is a logarithmic function. This logarithm function is very important. The log functions you may deal with include log₁₀ (used in the definition of pH, and for hand calculation in "the old days") and log₂ (used in some computer science applications).

This picture shows exp, the exponential function, e^x. Since this function is one-to-one, its inverse will be a function: no more blue-green clouds! The green line is the tangent at (0,1) which has slope=1. Then we flip it and get the graph of ln. What about the derivative?

1. y=ln(x)   The inverse function
2. exp(y)=exp(ln(x))   Using the inverseness
3. exp(y)=x   Recognition of inverseness
4. exp(y)(dy/dx)=1   Implicit differentiation
5. dy/dx=1/exp(y)   Solving for dy/dx
6. Since exp(y)=x, we're done! This is even easier still   Solving for dy/dx stuff in x
7. Thus ln'(x)=1/x   Statement of formula

The reason this log function is so important is that its derivative is 1/x. You will see how this is used more and more as we go through calculus.

Examples If f(x)=ln(x³–5x+78.6), then f´(x)=(1/[x³–5x+78.6])(3x²–5).
If f(x)=sec(ln(2x+5)), then f´(x)=sec(ln(2x+5)tan(2x+5)(1/[2x+5])2.

ln is important because its derivative is 1/x. Other things about ln are amusing (such as ln(AB)=ln(A)+ln(B) and ln(A^B)=Bln(A)) and were useful for several centuries (tables of logs helped hand computation, and slide rules, analog devices acting like tables of logs, did a similar job). But now these properties are much less important, and it is the derivative of ln which remains central to computations in calculus.

a^x
What if you needed to find the derivative of 10^x? Well, if y=10^x, you might do the following:

1. y=10^x   The original equation
2. ln(y)=ln(10^x)=[ln(10)]x   Take log's (ln's?) and use log properties
3. (1/y)(dy/dx)=ln(10)   Differentiate using the Chain Rule but realize that ln(10) is a constant multiplier on the right
4. dy/dx=[ln(10)]y=[ln(10)]10^x   Solving for dy/dx stuff in x

In fact, the derivative of a^x, when a is a constant, is a^xln(a). If we wanted the nicest, simplest derivative, we would want ln(a) to be 1, and that happens when a is e.

The derivative of 10^x is (ln(10))10^x≈(2.303)10^x. So if you want lots and lots of 2.303's in your answers, please compute with 10^x in calculus. I won't. The entries of the derivative table below have been developed to be as simple as possible, really, really, really.

An absurdity: (sin x)^x
Suppose you wanted the derivative of (sin x)^x, a fairly silly function. Here is a way to compute it.

1. y=(sin x)^x   The original equation
2. ln(y)=ln((sin x)^x)=x·ln(sin x)   Take logs and use log properties
3. (1/y)(dy/dx)=1·ln(sin(x))+x·(1/sin x)(cos x)   Differentiate using the Chain Rule, realizing please that the outermost "structure" on the right is a multiplication
4. dy/dx=(1·ln(sin(x))+x·(1/sin x)(cos x))y=(1·ln(sin(x))+x·(1/sin x)(cos x))(sin x)^x   Solving for dy/dx stuff in x

As I remarked in class, I don't know any natural phenomena which are modeled by such functions. Their natural habitat is calculus exams. Below is about what we'll need for computing derivatives in this course.

Function	Derivative
f(x)	limh→0[f(x+h)–f(x)]/h
xn	nxn–1
CONSTANT	0
ex	ex
f(x)+g(x)	f´(x)+g´(x)
f(x)·g(x)	f´(x)·g(x)+f(x)·g´(x)
CONSTANT(f(x))	CONSTANT(f´(x))
1/f(x)	–f´(x)/[f(x)]2
f(x)/g(x)	[f´(x)g(x)–g´(x)f(x)]/[g(x)]2
sin(x)	cos(x)
cos(x)	–sin(x)
tan(x)	[sec(x)]2=1/[cos(x)]2
sec(x)	sec(x)tan(x)
f(g(x))	f´(g(x))g´(x)
arcsin(x)	1/sqrt(1–x2)
arctan(x)	1/(1+x2)
ln(x)	1/x
ax	axln(a)
In mathematics you don't understand things. You just get used to them.

---

Tuesday, October 6	(Lecture #10)

This is the part of lecture #10 which covers material not tested on the first exam.

A "simple" tangent line problem
Suppose we consider x²+y²=1, the unit circle (center (0,0) and radius 1). What is an equation for the line tangent to this circle when x=1/2 and the point of tangency is in the upper half of the circle.
I just looked up "tangency" and it is a word. It means "the state of touching". So I could have said, solving for y, that the line is tangent to the circle at y=sqrt(3)/2, but using an infrequent word is ... silly and therefore I did it.

More interestingly, I do know three or four ways of computing the answer to this problem. I will show you two of them. The problem itself is, again, not profound. But knowing a variety of ways to solve such problems is extremely useful. Sometimes one way or the other can't be used. Enough data for the solution is: a POINT, which is (1/2,sqrt(3)/2), and a SLOPE, which is what I will concentrate on. So the answer will be (y–sqrt(3)/2)=SLOPE(x–1/2), and we'll compute the SLOPE.

First solution
Since x²+y²=1, we solve for y as a function of x. Thus, y²=1–x² and y=+/–sqrt(1–x²). We need to figure what +/– means. Here since I specified "in the upper half of the circle" there is little to worry about. "upper" means +, so y=sqrt(1–x²). The Chain Rule now allows us to compute the derivative. So dy/dx=(1/2)(1–x²)^–1/2(–2x), and when x=1/2, this is (1/2)(1–(1/2)²)^–1/2(–2(1/2)). That number is the SLOPE we need. It is a negative number, which does correspond to the slope of the tangent line shown in the graph.

Second solution
Well, if x²+y²=1, I could try d/dx'ing the whole equation. In this I will be guided by the following idea: that somehow the equation defines y as a function of x. But, right now (forgetting the first solution above!), I don't know the function. So y is some unknown function of x. Let me d/dx the equation. The right-hand side is easy, since the derivative of a constant is 0. What about the left-hand side? It is the sum of two functions, x² and y². The derivative will be the sum of two derivatives. The derivative of x² is 2x. What about y²? y is some unknown function of x. So I will use the Chain Rule, and I can't assume anything about the function y:
    d/dx(y²)=2y(dy/dx).
I think of y² as a composition. The "outside" function is squaring, and the "inside" function is y, the unknown function. The result of differentiation is 2y multiplied by the derivative of the unknown function. So put all this together.
    The derivative of x²+y²=1 is 2x+2y(dy/dx)=0.
Now we can solve for dy/dx: 2y(dy/dx)=–2x and so dy/dx=–(2x)/(2y)=–x/y.
In our specific situation, x=1/2 and y=sqrt(3)/2, so dy/dx=–(1/2)/(sqrt(3)/2))=–1/sqrt(3). You can check that the previous answer is actually the same number as this.

Key assumption
The key assumption in the second method is that the equation defines, somehow, y as a function of x, and then that this function is differentiable. It's possible to prove this statement under circumstances which commonly arise in Math 151. The customary practice is just to assume without further mention that the assumption is true. Sometimes things can get weird if the assumption is not valid, but this rarely happens.

Language
In the first method above, we solved for y as a function of x. People say that y is an explicit function of x. The second method has y defined by the equation it sits in. People say that y is defined implictly. A dictionary definition for the word "implicit" is "implied though not directly expressed; inherent in the nature of something". The differentiation trick above is called implicit differentiation and people use it whenever it is difficult or impossible to solve for y as a function of x.

An implicit problem
The point (–2,1) is on the curve defined by the equation y²=x³–3xy+3. We can check this: plug in x=–2 and y=1 and get 1²=1 on the left, and (–2)³–3(–2)(1)+3=–8+6+3=1 on the right.

Let's find an equation for the line tangent to this curve at the point (–2,1). Again we need the SLOPE. It is certainly possible to solve for y as a function of x since the equation defining the curve is a quadratic equation in y and we could use the quadratic formula. But that would be messy and tedious, so I won't even try the first method.

Computing the derivative is "easy" if you realize it should be done implicitly. We need to d/dx the whole equation y²=x³–3xy+3, and be careful to use the Chain Rule and the Product Rule.
    2yy´=3x²–3y–3xy´
Let's insert x=–2 and y=1:
2(1)y´=3(–2)²–3(1)–3(–2)y´ this gets us 2y´=12–3+6y´ so that y´=–9/4. The picture shows, by the way, that the slope of the tangent line is negative at (–2,1), so we have a little bit more evidence that the answer is correct.
Therefore an equation for the tangent line is (y–2)=–9/4(x–(–2)).

Explicit versus implicit
It is frequently impractical or even impossible to solve an equation for one of its variables. We will see other examples.

---

Maintained by greenfie@math.rutgers.edu and last modified 10/13/2009.