FTC 2
If F(x)=∫_a^xf(t) dt, then F´(x)=f(x).

An alphabet lesson
What is ∫₀¹x²dx? In the last lecture, we saw with several different techniques that this is 1/3. Then let's play a little logical game.

• What is the value of ∫₀¹x²dx? The value is 1/3.
• What is the value of ∫₀¹w²dw? The value is 1/3.
• What is the value of ∫₀¹t²dt? The value is 1/3.
• What is the value of ∫₀¹θ²dθ? The value is 1/3.
• What is the value of ∫₀¹λ²dλ? The value is 1/3.

Simple example
Describe a function whose derivative is cos(x+e^x).
This is either very very easy, or, for reasons that will be discussed more in Math 152, very very hard. I will give the very very easy answer here.

One function is F(x)=∫_–3^xcos(t+e^t) dt. This is a function whose derivative is cos(sqrt(x)+e^x) by using FTC 2.

Another function is G(x)=∫₅^xcos(w+e^w) dw.

Let me compare the two answers. One has inside variable t and the other has inside variable w. This doesn't matter. They both are "satisfactory" answers to the question asked because FTC 2 implies that the derivative of a definite integral with a variable upper parameter is the integrand's value at that upper parameter (hey, I'm using all of the high-priced words!) One answer is a definite integral from –3 to x and the other is a definite integral from 5 to x. Using property (*) from last time, the two answers are different by this: ∫_–3⁵cos(s+e^s) ds. What is this? It is a constant (I'm just using another variable, s, to again bring up the idea of a "dummy variable"). Two functions which differ by a constant have the same derivative. So that's o.k.

By the way, it can be proved that no one can do much better in this problem than write the answer as a definite integral. There isn't any much simpler answer. Software can then plot this function, because people have spent a great deal of time and effort learning how to compute good numerical approximations to definite integrals very rapidly. The picture of F(x), the antiderivative of cos(x+e^x) which is defined above, on the interval [0,2] was produced by Maple in less than three-tenths of a second. I am sure other software would do the job as well.

More examples (a textbook problem) This is problem 23 in section 5.4. The graph of f is given to the right. The problem asks us to sketch a graph of A(x)=∫0xf(t) dt. I presume, since f's graph is given on the interval [0,4], that we are supposed to sketch a graph of A(x) on the same interval. There are various strategies for analyzing this sort of problem. Since the information is furnished graphically, I'd probably try to do the problem graphically, or at least begin it that way. I would think of the limits, 0 and x, and try to see how the definite integral (which I'd think of as signed area) varied as x varied.

Let's consider first x's between 0 and 1. Here I think of a vertical line somewhere over the [0,1] interval, moving to the right. The area between that line and the y-axis, under y=2, is ∫0xf(t) dt. Since the line has height 2, the area is 2x, and the graph of A is a straight line segment starting from (0,0) with slope 2.

When the vertical line passes x=1, we have accumulated 2 units of area, and A(1)=2. But now the "profile curve", f, changes height. It has height 1. We must add to the 2 units of accumulated area the new area we are getting between x and 1. Since the height of f is 1, and the base is x–1, we add on 1(x–1) units of area. The graph of A in this interval will be linear with a slope of 1, and it will start from (2,2).

At x=2, we have accumulated 3 units of area, 2 over the interval [0,1], and 1 more over the interval [1,2]. Now x moves to the right in the interval [2,3]. The height of f is –1, so the definite integral will decrease: the geometric area is below the x-axis. We know A(2)=3 becuase of the accumulated area. But if x is between 2 and 3, the change in A is (–1)(x–2). So we need to show the graph of A decreasing, and it will be decreasing linearly in x, with a slope of –1.

By the time the vertical line has gotten to x=3, it has accumulated 3–1 units of area. Yes, I know that 3–1=2, so actually A(3)=2, but somehow writing and thinking 3–1 helps me recall that + is assigned to areas above the x-axis and – is assigned to areas below the x-axis. Between 3 and 4, the height of f is 0 (zero: nothing). There is therefore no change in the area as we move the upper limit on the definite intergral to the right. And therefore there is no change in A, and the graph of A is a horizontal line segment beginning at (3,2).

From this we get a very good idea of the graph of A(x). It is 4 line segments with slopes of 2, 1, –1, and 0. It is a continuous function. The slopes of A correspond to the heights of f, and actually A is an antiderivative of f (the antiderivative with the initial condition A(0)=0), so f is the derivative of A. I think the graph looks like what's shown to the right.

Part B Well, there's actually a part B) to this problem. The graph to the right is given, and, again, we are asked to graph A(x)=∫0xf(t) dt. Here the graph is a bit more complicated (well, it is more complicated to me: the other function had exactly 4 values, and this function has lots and lots of values!). I'll use a strategy different from what we did in A) to understand and solve this problem.

Probably I would look at the graph and learn that A(0)=0 (there is no area from 0 to 0!). I would look a bit more and find that A(2)=1, because ∫02f(t) dt is the area of a triangle with base 2 and height 1, and (1/2)(2)(1)=1. Finally, I would see that A(4)=2, because there are two of those triangles. So I have the three dots shown to the right on the graph of A. Notice also that the graph of A is increasing because as we move the right limit to the right we are getting more and more area, and this is positive area since it is above the x-axis.

What can we say about the derivative of A in the interval [0,2]? By FTC 2, this derivative is the function f, and the function f is just (1/2)x. That means the second derivative of A is 1/2, and this is positive. So the graph of A between 0 and 2 is concave up, and connects (0,0) and (2,1). I actually know a formula for A, since A(0)=0 and A´(x)=(1/2)x. It must be (I antidifferentiate and get the correct constant!) A(x)=(1/4)x2. So what I see between 0 and 2 is a piece of a parabola.

Between 2 and 4, the function keeps increasing. But the slope of the function, A´, is positive: it seems to be a line segment from (2,1) to (4,0). But the derivative of that is –1/2, so the second derivative of A is negative. A is increasing and concave down. I think the graph is must look like what is displayed to the right. In fact, between 2 and 4, A´(x)=–(1/2)x+2. I also know that I have accumulated 1 unit of area by the time we get to 2, so that A(2)=1. I can solve this initial value problem: antidifferentiate to get A(x)=–(1/4)x2+2x+C. Use A(2)=1 to get C: –(1/4)22+2(2)+C=1, so C=–2, and the formula for A(x) in this interval is –(1/4)x2+2x–2. We can check this by plugging in x=4, so the result is –(1/4)42+2(4)–2=–4+2(4)–2=2 which is correct. Another thing which is not obvious is that the function A is differentiable. Since A is defined by two nice formulas, it isn't very surprising that A´ exists inside the two halves of the domain. In fact, A is differentiable at x=2, also. This can be checked with effort by looking at the difference quotient (there is a similar check in the workshop problem solution inserted in the Rutgers edition of the textbook). But I am willing to believe that A´(2) exists, because: If A(x)=(1/4)x2 then A´(x)=(1/2)x so A´(2) should be 1. If A(x)=–(1/4)x2+2x–2 then A´(x)=–(1/2)x+2 so A´(2) should be 1. I bet that the slope of the tangent line at 2 is 1.

Movement of a particle
Let's assume that the velocity of a particle traveling on the x-axis is given as a function of time: v(t)=4t²–t⁴ for t between 0 and 5. (As far as I know, this is a rather non-physical example, but I hope that the questions and ideas contributing to the solutions are correct and useful.)

1. For which values of t is the particle traveling to the right? To the left?
   "Travel to the right" in terms of velocity means "v(t)>0" and of course "to the left" is "v(t)<0". Since v(t)=4t²–t⁴=t²(4–t²). Notice that t² is non-negative, so that 4–t² has the sign information. We should consider 4–t² on the time interval [0,5]. Of course, 4–t²=0 when t is +/–2. So (considering everything) v(t)>0 when t is between 0 and 2 and v(t)<0 when t is between 2 and 5. Notice that velocity is 0 when t=0 and t=+/–2.

Last time we saw that if we had a function f defined on an interval [a,b], then we could introduce a rather elaborate collection of objects which are used to analyze f's behavior on all of [a,b].

If [a,b] has a partition P (which break up [a,b] into a number of subintervals) and a collection of sample points (the text's intermediate points) C, then we defined the Riemann sum of f using P and C:
R(f,P,C)=∑_j=1ⁿf(c_j)Δx_j.
If the length of the longest subinterval→0, then (for the functions we will consider in this course!) the Riemann sums approach a limit which is called the definite integral of f from a to b and written ∫_a^bf(x)dx.

Definite integrals have some very useful properties which people use frequently.

Property (*) and consequences
Suppose a<b<c. Then ∫_a^bf(x)dx+∫_b^cf(x)dx=∫_a^cf(x)dx and I will call this statement (*). I am happy if you believe this statement. I hope that the picture to the right is enough verification. The picture has red for area below the horizontal axis and green for area above the axis because I wanted to remind you how area is counted for the definite integral. It is possible to prove this statement from the definition with Riemann sums.

Unexpected consequences of (*)
While most people are willing to believe (*), there are some results which may take you some time to learn to use. For example, in the equation
    ∫_a^bf(x)dx+∫_b^cf(x)dx=∫_a^cf(x)dx
what if we change b to a? Then we seem to get
    ∫_a^af(x)dx+∫_a^cf(x)dx=∫_a^cf(x)dx
and the only way this could be true is if
    ∫_a^af(x)dx=0.
Most people find this easy enough to believe: this quantity is the area of a "region" with width equal to zero, and some height, and such an area should be 0. So any definite integral whose upper and lower limits are the same must be 0.

O.k., if you believe that one, then I will make another change in the equation:
    ∫_a^bf(x)dx+∫_b^cf(x)dx=∫_a^cf(x)dx.
Change c to a in this equation, and use the fact that the right-hand side will become ∫_a^af(x)dx which we already "know" is 0. So we get:
    ∫_a^bf(x)dx+∫_b^af(x)dx=0
and this certainly means that ∫_b^af(x)dx=–∫_a^bf(x)dx
so that when the limits of a definite integral are interchanged, then the sign of the integral is changed! (What we are in fact computed is called an oriented area, and more complicated than what I've told you so far.)

Example(s) Suppose f is the function whose graph is shown to the right. It is piecewise linear, and made of four line segments. Then all of the values of these integrals:     ∫02f(x)dx    ∫–11f(x)dx    ∫3–1f(x)dx    ∫44f(x)dx     ∫13f(x)dx    ∫–12f(x)dx    ∫30f(x)dx    ∫0–1f(x)dx are contained in this collection of numbers:     –1    0    1/2    1    3/2. Can you match each definite integral with one of these values?

Property (**) This is extremely useful when you are confused and need to make some sort of estimate. Suppose, somehow, you know that all of the values of f(x) on the interval [a,b] are between m and M. That is, you know m≤f(x)≤M for all of those x's. Then (look at the graph!) the "area" (actually the definite integral) will be trapped inside a box. So the result is:     m(b–a)≤∫abf(x)dx≤M(b–a). This is because the wiggly region contains the smaller box and is contained by the larger box. It looks silly, but the estimates can really be useful in checking computations.

I mentioned that definite integrals are used to compute many quantities of interest to engineers, includiong such things as work, power, energy, pressure, volume, length, etc. This may not be too clear right now, but a number of these are discussed in the next course (Math 152). The key is that for all of these quantities, (*) and (**) are true in some way.

Yet another example ...
Here is a final example of how to compute an area (or a definite integral) in the most direct fashion: by getting the area of a collection of approximating rectangles, and then analyzing what happens as the partition gets "finer".

The example is the area under y=x² from 0 to 1. This means the area enclosed by the x-axis, y=1, and y=x². It is an area in the first quadrant. We will approximate the area, which is the definite integral ∫₀¹x²dx by a collection of Riemann sums. Here f(x)=x².

1. Partition [0,1] into N equal subintervals, where you should think that N is some large positive integer. Since the original interval has length 1, each subinterval will have length 1/N. The actual partition points will begin 0/N, 1/N, 2/N, 3/N, 4/N ... and will end with (N–1)/N and N/N.
2. Get sample or intermediate points. In this computation, I will choose the right-hand endpoints of each subinterval as the sample points. So in the first subinterval [0/N,1/N], the sample point will be 1/N. In the second subinterval [1/N,2/N], the sample point will be 2/N. In the third subinterval [2/N,3/N], the sample point will be 3/N. ... In the N^th subinterval [(N–1)/N.N/N], the sample point will be N/N.
3. Compute the function values at the selected points. These numbers are the heights of the approximating rectangles, and they are f(1/N)=1²/N² and f(2/N)=2²/N² and f(3/N)=3²/N² and ... f((N–1)/N)=(N–1)²/N² and f(N/N)=N²/N². Wow!
4. Multiply each of the heights by the widths of the rectangles. Each of the widths is identical, however, so things aren't so bad. The results for the areas of the rectangles are:
   1²/N²·(1/N) and 2²/N²·(1/N) and 3²/N²·(1/N) and ... (N–1)²/N²·(1/N) and N²/N²·(1/N). More wow!
5. Add 'em up. So this is:
   1²/N²·(1/N)+2²/N²·(1/N)+3²/N²·(1/N)+...(N–1)²/N²·(1/N)+N²/N²·(1/N).
   More compactly written, the Riemann sum is ∑_j=1^Nj²/N³. I combined the powers of N "downstairs". You could also pull out the 1/N³ because it multiplies every term in the sum, and then the result would be (1/N³)∑_j=1^Nj².

N=5
If we had N=5, the approximation would be (1/5³)(1+4+9+16+25) which is (1/125)(55). The mysterious part is the 55, of course.

A magic formula
Well, everyone knows (it is in the book on page 317, so ... it is possible to know it!) that ∑_j=1^Nj² is (1/3)N³+(1/2)N²+(1/6)N.
We actually checked that if N=5 is plugged into this formula, the result is 55. That's nice. So maybe the formula is correct.

Getting the exact value of the definite integral
Since we know that ∑_j=1^Nj² is (1/3)N³+(1/2)N²+(1/6)N and that the Riemann sum is this divided by N³, then the value of the Riemann sum is (1/N³)((1/3)N³+(1/2)N²+(1/6)N) which is (1/3)+(1/2N)+(1/6N²) and certainly as N→∞, this must approach 1/3. So ∫₀¹x²dx=1/3.

Change the problem: make it harder!!!
These techniques seem to be (they are!) very intricate. Let me show you how to solve this problem by first making it more difficult. The people who discovered that the (seemingly) more difficult problem can be solved very neatly really had a lot of intellectual courage.

So instead of studying ∫₀¹x²dx, we will study the function
    A(b)=∫₀^bx²dx.
We really want to know A(1), but we'll consider this more general problem anyway.

What do we know?
Well, there is one value of the function that is extremely easy to compute: A(0)=∫₀⁰x²dx=0.

That seems to be the only simple thing. But this is a CALCULUS course, and maybe we should try to differentiate A(b). This, which certainly should seem almost ludicrous, turns out to be the successful approach.

In order to find A´(b) we will need to consider A(b+h)–A(b) and then divide the result by h.
Certainly (*) implies that
    ∫₀^bx²dx+∫_b^b+hx²dx=∫₀^b+hx²dx
so that A(b+h)–A(b) is ∫_b^b+hx²dx.

But we can use (**) now. f(x)=x² is increasing on [b,b+h], so for all x in the interval, b²≤f(x)≤(b+h)². So, with m=b² and M=(b+h)², we see that
    b²·h≤A(b+h)–A(b)≤(b+h)²·h
since h is the width of the interval. Now divide by h:
    b²≤[A(b+h)–A(b)]/h≤(b+h)².
Notice that both estimates →b² as h→0. This means that A(b) is differentiable, and that A´(b)=b².

An initial value problem
We have a function, A(b), with the following properties:
    A(0)=0 (initial condition);
    A´(b)=b² (differential equation).
This is an initial value problem. All antiderivatives of b² have the formula (1/3)b³+C. If A(b)=(1/3)b³+C then A(0) must be C, but the initial condition states that A(0)=0, so C=0. Therefore, A(b)=(1/3)b³.

Now solve the original problem
Since A(b)=(1/3)b³, then A(1)=(1/3). No problem!

Hey, things are easier than they look
In fact it turns out that we didn't need to figure out the C, because the value we need to compute the integral from 0 to 1 is A(1)–A(0), and any antiderivative will give the same result (the "+C" adds and subtracts, and doesn't change the answer).

The Fundamental Theorem of Calculus, version 1
If F´=f, then ∫_a^bf(x)dx=F(b)–F(a).
In fact, the notation usually used "abbreviates" F(b)–F(a) by F(x)|_x=a^x=b. And "Fundamental Theorem of Calculus" is also usually abbreviated as FTC.

Here is how the original area desired would be computed.

1. Let's find the area under y=x² from 0 to 1.
2. Recognize that this is the same as computing ∫₀¹x²dx.
3. We know an antiderivative of x²: it is (1/3)x³.
4. FTC then declares that the definite integral (1/3)x³|_x=0^x=1 and this is (1/3), the value of the area desired.

The area under a bump of sine
What is the area undo a bump of sine? Well, first think a bit: the idea of bump maybe means the area between two places where sine is 0 and where the curve is above the x-axis. So the area would be computed by the definite integral ∫₀^Pisin(x) dx. Usually having some estimate of the size of a quantity, even before computing, is a good idea. Otherwise you may make some simple error, and get some result which is a huge mistake.

One overestimate of this definite integral is obtained by looking at a rectangular box which contains the area. The width of the box is Π and its height is 1, so the integral should be less than Π·1=Π. An underestimate of the area could be gotten by looking at two triangles under the graph. Each has base Π/2 and height 1, so the total area of the two triangles is 2·(1/2)·(Π/2)·1=Π/2 (the 1/2 comes from the formula for the area of a triangle). One student suggested 0 as an underestimate. While that is certainly valid, I like under- and overestimates which are closer to the exact answer (as long as I don't have to work very hard!). Now we know
Π/2≤∫₀^Πsin(x) dx≤Π.

To compute this area exactly, we use FTC. We need an antiderivative of sine, and that happens to be –cos(x). So:
∫₀^Πsin(x) dx=–cos(x)|₀^Π=–cos(Π)–(–cos(0))=–(–1)–(–1)=2. The (potential!) confusion of minus signs occurs often when applying FTC, so some care is needed. The exact area is 2, and this is certainly between Π/2 and Π.

A fake computation: the "area" under a bump of cosine
I then tried to rush ahead and compute the "area" under a bump of cosine. Or, rather, I compute this definite integral: ∫₀^Πcos(x) dx. I asserted that the result should be ... well, should be ... something.

Here is the computation using FTC, since an antiderivative of cosine is sine.
∫₀^Πcos(x) dx=sin(x)|₀^Π=sin(Π)–(sin(0))=0–0=0.
What happened to the "area" under the bump? We computed a definite integral, and that's the limit of Riemann sums, and they count geometric area under the x-axis negatively. The graph of cosine on the interval [0,Π] exactly balances the positive over-the-axis region with a negative under-the-axis region. So it is 0 both computationally and geometrically!

Another bump but negative
I asked students what the curve y=x²(x–1) from 0 to 1 looked like. After a while we got a picture similar to what is shown to the right. I asked what the area was inside the bump. Please observe that the function is negative, so the definite integral would be negative, but the area would be positive.

A solution might go like this:
∫₀¹x²(x–1)dx=∫₀¹x³–x²dx=(using FTC)=(1/4)x⁴=(1/3)x³|₀¹=(1/4)–(1/3)=–1/12.
Therefore the area is 1/12.

Return of the second exam
I returned the second exam. Information about grading and statistics concerning the results are here. If you think there's been a mistake made with the grading of your exam (an almost certain event, considering the number of problems, the size of the class, and the too fallible human who did the grading), please check the grading guide, and then bring the exam to me.

I note that the median grade, which I regard as probably the most important number statistically for such happenings, is up from the first exam median.

Course grades will be strongly influenced by both your own efforts and also how the class as a whole does on the Math 151 final. Therefore the final exam results will be used to calibrate or measure or compare performance of the 153 students with the overall 151 group's work, and I will adjust course grades appropriately. I mean that the group achievement will be considered, so it is in the interest of individual members of the class that the whole class do as well as possible!

                      3          2
               (n + 1)    (n + 1)
12+22+...+n2 = -------- - -------- + n/6 + 1/6
                  3          2

                  8          7            6            4          2
           (n + 1)    (n + 1)    7 (n + 1)    7 (n + 1)    (n + 1)
           -------- - -------- + ---------- - ---------- + --------
              8          2           12           24          12

A question of blood... I asked how we could estimate how much blood the heart pumps out to the body, drawing an especially silly picture on the board (look left). Knowing the quantity of (hopefully!) oxygenated blood which is pushed around the body can be a useful part of health assessment. How can we measure the blood flow through the aorta, the major artery emerging from the heart? A better diagram of a heart is to the left. The heart is a very complicated and wonderful pump.

The Aztec solution
One method might be to take a knife, preferably an obsidian knife, and rapidly cut through the chest wall, sever the aorta, and cause the blood that would flow through it to fill up a bucket. In about five minutes we'd probably have a good idea of how much blood flows through the aorta. There are several criticisms possible of this "protocol". One criticism might be that, well, maybe the person involved might not be good for much afterwards. Another criticism could be that this would not show the actual pumping capacity of the aorta, because maybe within a minute or two some ... difficulties would occur. In some sense, this Aztec protocol is the ultimate in what might be called destructive testing. By the way, obsidian is "a dark glassy volcanic rock formed from hardened lava" and was actually used for certain similar purposes in that culture.

Another solution: cardiac catheterization
The American Heart Association and the NIH (National Institutes of Health) both have lots of information about this. Here are some quotes from these sources:

• The AHA begins:
  "This is a procedure done on the heart. In it, a doctor inserts a thin plastic tube (catheter) (KATH'eh-ter) into an artery or vein in the arm or leg. From there it can be advanced into the chambers of the heart or into the coronary arteries."
• Or from the NIH:
  "Cardiac catheterization is used to study the various functions of the heart or to obtain diagnostic information about the heart or its vessels. A small incision is made in an artery or vein in the arm, neck, or groin. The catheter is threaded through the artery or vein into the heart. X-ray images called fluoroscopy are used to guide the insertion."

Elapsed time	Fluid flow rate
1 min	30 cm3/min
2 min	20 cm3/min
4 min	40 cm3/min
5 min	25 cm3/min

The blood flow will be given in cm³/min (cubic centimeters per minute) and the time will be measured in minutes since the beginning of the procedure. The information learned is displayed to the right.

This isn't much information. But how can we use what we've learned? A standard way to use these numbers is the following. It is certainly very approximate in this case.

At the end of the first minute, we're told that the flow is 30 cm³/min. Hey, we could estimate how much blood is pumped by making the assumption that the flow rate is constant during the time interval involved. Then we see that 30 cm³/min·1 min=30 cm³ (flow rate multiplied by duration) would be pumped. I am not asserting that this is necessarily correct. It is an estimation. We are using the information, and the method is very simple. Similarly, during the second minute, we can estimate that 20 cm³/min·1 min=20 cm³ is pumped. The flow rate changed during the second minute. We're faced with a slightly different "challenge" in using the third line of the table. It has elapsed time (since the start of the procedure) as 4 minutes. So the duration has now changed from 1 minute to 2 minutes. (Why? Well, y'know, in the real world things maybe don't work as you'd like. Maybe the machine broke, or the technician didn't pay attention, or ... in any case, we need to work with the data we have!) So during those two minutes, we have (flow rate·duration) an estimate of 40 cm³/min·2 min=80 cm³ blood pumped. Finally, in the last minute of this procedure, we estimate that 25 cm³/min·1 min=25 cm³.

Let me abbreviate just a bit. The units are repetitive and I will omit them (this is a math course!). So here are the essential details of the computation:     30·1+20·1+40·2+25·1=155. That's what we're computing. If we wanted a picture of the computation, maybe what is shown to the right gives you some idea. The dots represent the actual known data. The area of the rectangles shown is given by the product of the appropriate flow rate and duration. Yes, there are other possible approximations using the data given. We will study some of them in Math 152, but here only a rather simple-minded approach will be used.

More information Elapsed time Fluid flow rate 30 sec 32 cm3/min 1 min 30 cm3/min 1 min 30 sec 28 cm3/min 2 min 20 cm3/min 2 min 30 sec –6 cm3/min 3 min –8 cm3/min 3 min 30 sec 34 cm3/min 4 min 40 cm3/min 4 min 30 sec 43 cm3/min 5 min 25 cm3/min Now let's refine the data. Let me suppose that we have no "technical" problems, and that we are recording the flow rate every 30 seconds. We could imagine the numbers given in the table to the right. Please notice that the data we've already obtained is part of this table -- I am imagining that we're measuring the same process as before, only we are sampling things more often. Here we have readings every thirty seconds with no data points are omitted. So we actually have 10 pairs of numbers. There is another complication which has occurred, and I hope you notice it. Some of the flow rates (2 min 30 sec and 3 min)are negative. Could this happen? Well, actually maybe yes. The heart is a terrific pump but maybe sometimes things don't always synchronize well or work perfectly: maybe the heart valves don't close all the way or don't close at precisely the correct time in the pumping cycle, etc. This complicates our task. We could ask what the heck we want to measure in this situation, and several answers are possible. We might want to measure the total pumping effort of the heart. Or we might want to know the net amount of oxygenated blood which is pushed out into the body. Let me look at the second quantity. In that case, the negative flow rates affect the total blood pumped because, indeed, that amount in that time should be subtracted, just like the sign of the flow rate indicates, from the total.

What should be computed here? If we do flow rate multiplied by duration, we should be consistent (the units!) and realize that, for example, 3 min 30 sec is 3.5 minutes. So the numerical computation here is     32·.5+30·.5+28·.5+20·.5+(–6)·.5+(–8)·.5+34·.5+34·.5+40·.5+43·.5+25·.5=119. And this estimate, using this data, is 119 cm3. (Ms. O'Sullivan asserts that the total is 109. Hah!) This is much less than the earlier figure, I guess because the data at 4 minutes over a duration of 2 minutes led to a big overestimate. A picture of what we computed is to the right, which I sort of laid over the old bar chart whose outlines where visible are indicated with dashed lines. I put in the new data points, and then the computation above is indicated by the areas of these rectangles. But the total area of the rectangles which are below the horizontal axis is subtracted from the total. Can you "see" the change from 155 to 119?

Ideally ... You might imagine that ideally we make lots and lots of measurements (in fact, this is more like what is done, because I think there is a computer recording the flow rates very frequently and then making the computations that I'm discussing in such a silly way). What should we think about? We could have lots and lots and lots of data points. And then the rectangles get very thin. And then the data points look very much like a curve. The number that's wanted is the total blood pumped out to the body, but the picture which might result could look a lot like what is shown here. I have again overlaid it with the rectangles from the previous discussion. So it seems there might be some kind of flow rate curve, and that the blood pumped out to the body, the net blood out, is the area of the colored regions above the curve minus the area of the colored region below the curve. I hope that you can see some resemblance between this and the discussion of the area of the hand which we did previously. Here we are approximating quantities of liquid, and there we were trying to approximate areas. It turns out that this computational idea can be applied in many situations, and it is used in essentially all fields of engineering and applied science. I now need to tell you the official vocabulary.

Mass from density measurements
I very briefly mentioned another almost realistic setup. We might have a bar with varying densities whose total mass we want to estimate. A sort of picture is to the right, with the different colors used to imply different densities. Direct measurement (lifting the bar, for example) might not be feasible. We could imagine taking small samples of the bar's materail at several points, and then measuring the densities of these samples. We could then multiply the density by the appropriate length of the bar and take the sum of these quantities as an estimate of the bar's mass. This would be a situation whose mathematical outline would be very similar to the blood flow analysis we've just done.

Vocabulary
Start with the following: a function f defined on an interval [a,b].

• Partition A partition of [a,b] is a collection of points in the interval always starting with a and ending with b. It looks like this:
  a=x₀<x₁<x₂<...x_n–1<x_n=b
• Subintervals Each partition defined a collection of subintervals. Here they are:
  [x₀,x₁] and [x₁,x₂] and [x₂,x₃] and ... [x_n–1,x_n].
  If there are n+1 points in the partition (counting from 0 to n) then there are n subintervals. The length of each subinterval is important, and the length of the first subinterval is x₁–x₀. This is frequently abbreviated Δx₁. So the length of the j^th subinterval, x_j–x_j–1, would be abbreviated Δx_j. The Greek letter Δ, Delta, is used as an abbreviation of Difference.
• Sample points or intermediate points You select an intermediate point in each subinterval. The text uses the phrase "intermediate point". The phrase "sample point" is what I am used to, and I think it is more widely used. It also makes me think more of physical measurements. I will try to use both phrases. The text uses the letter c_j for the j^th sample point. So c_j would be inside [x_j–1,x_j], and therefore x_j–1≤c_j≤x_j.
• Computing the sum The textbook uses the notation R(f,P,C) to mean Riemann sum depending on the function, f, the partition, P, and the choice of sample points, C. I don't know any generally accepted notation so let me stick with what is in the book. Anyway, here is the formula for the Riemann sum:
  f(c₁)Δx₁+f(c₂)Δx₂+...+f(c_n)Δx_n.
  This is too long to write and usually if people have to work with such things, they will write
  ∑_j=1ⁿf(c_j)Δx_j

What this means geometrically ... To the right is my attempt to show you what a Riemann sum might look like. But I've made it quite simple. There are only seven subintervals, and the curve which is the graph of f(x) is also not too complicated. The dots on the x-axis denote the sample or intermediate points, and the vertical lines above them are the lengths f(cj). The boxes have areas which are equal to the pieces of the Riemann sum, f(cj)δxj. The f(cj) is the height and the Δxj is the width of a rectangle.
An example To the left is a graph of a function, y=f(x), where f is defined on the interval [0,5]. f's graph is two line segments, the first horizontal from (0,1) to (2,1) and the second tilted from (2,1) to (3,0), followed by the lower half of a circle of radius 1 centered at (4,0), a circular arc going from (3,0) to (4,–1) to (5,0).
Here are some partitions, choices of sample points, and the resulting Riemann sums.
Partition 0, 2, 3, 5. Sample points 1, 3, 4. The function values at the sample points are: f(1)=1, f(3)=0, f(4)=–1. Riemann sum 1(2–0)+0(3–2)+(–1)(5–3)=0.
Choosing the sample point at 3 gives a zero contribution to the Riemann sum.
Partition 0, 2, 3.5, 5. Sample points 1, 3, 4. The function values at the sample points are: f(1)=1, f(3)=.5, f(4)=–1. Riemann sum 1(2–0)+.5(3.5–2)+(–1)(5–3.5)=1.25.
This was the QotD! Partition 0, 2, 3.5, 4.5, 5. Sample points 2, 2, 4, 5. The function values at the sample points are: f(2)=1, f(2)=1, f(4)=–1, f(5)=0. Riemann sum 1(2–0)+1(3.5–2)+(–1)(4.5–3.5)+0(5–4.5)=2.5.

The definition allows the sample point to be on the boundary of the subinterval, so two adjoining subintervals could possibly share their sample points. This is a bit wasteful in terms of information, though. Also notice that the rectangle on the rightmost interval has height 0.
Partition 0, 1, 2, 3, 4, 5. Sample points .7, 1.44, 2.5, 3.5, 4.75. The function values at the sample points are: f(.7)=1, f(1.44)=1, f(2.5)=.5, f(3.5)=–sqrt(3/4)≈–.866, f(4.75)=–sqrt(7/16)≈–.661. (The numbers for f's values at 3.5 and 4.75 come from the equation of the lower unit semicircle, –sqrt(1–x2), suitably translated to the right.) Riemann sum 1(1–0)+1(2–1)+.5(3–2)+(–.866)(4–3)+(–.661)(5–4)=.973.
Notice that putting another sample inside the interval from 0 to 2 didn't improve the approximation very much. If taking samples is difficult or expensive (and it might be in real life!) then probably increasing the sampling where functions wiggle a lot rather than where the values are (relatively) constant is a good idea. This is, in fact, what the approximation strategies inside most calculators try to do. Part of the impression I would like to leave with you, even with just a few examples, is that Riemann sums can be quite complicated objects. They can be irritating to compute, and there are so many of them that understanding them may be difficult.

Hand area
What is the area of my hand? There was, of course, some laughter. But the question and other similar inquiries, while superficially silly, is actually quite important. Appropriate drug doses are sometimes calculated in terms of BSA, body surface area. This is especially important when prescribing drugs for infants or drugs with strong side-effects (for example, some drugs used in cancer treatment). And knowledge of body surface area is important in trying to understand heat balance when people exercise. There are many formulas for BSA in addition to the Dubois and Dubois formula which was mentioned in an earlier workshop problem.

So my question is much more modest: how can we compute the area of my hand? We discussed this. There are many methods, but if you think about area, there are really not very many shapes whose area we can compute exactly. The most important and simplest shape is a rectangle, where we know the area is the product of the length and width. We should try to take advantage of this. So look at an image of my hand, and ... well, after discussion we came up with this:

1. Superimpose a square grid over an image of my hand.
2. Put boxes inside.
3. Put boxes outside.
4. Count the boxes totally inside. Count the boxes which contain any (even little!) piece of the hand image (these are the "outside" boxes).
5. Compute the area of the inside boxes and the area of the "outside" boxes.
6. The hand area will be between the inside box area and the outside box area.

I'll suppose first that I put the image of the hand in a square grid which is, say, 1 mm (millimeter) on a side. I don't know if this is physically too reasonable, but, what the heck, I could get something that looks like the picture to the right. Then I could count the grid boxes which are entirely within the hand image. There was some objection to this, because, you see, I could always be a little clever (?) and maybe sort of count some boxes which are almost entirely in and then compensate by dropping the count by 1 if I count a bunch of those.
I commented that I wanted a process (an algorithm) which could be done totally "mechanically", leaving nothing to choice, and which was extremely simple-minded.
I'll suppose first that I put the image of the hand in a square grid which is, say, 1 mm (millimeter) on a side. I don't know if this is physically too reasonable, but, what the heck, I could get something that looks like the picture to the right. I think if I count correctly there are a total of 18 boxes entirely inside the hand image.
Now for an overestimate. Let me start with the boxes which are entirely inside, and then add on any boxes that have even the slightest "tinge" (?) or hint of the hand image inside. Again, maybe there are objections: we could be cleverer (?) and sort of estimate how much of a box is inside, etc. But then we would have to make decisions, and some of those decisions maybe could be difficult. Here, if we have a simple-minded scheme, I'll bet that any two people will come up with the same answer. At least we get a result that's stable, and probably doesn't depend on the observer too much. There are 18 all-inside boxes, and (if I count correctly!) 60 more boxes are needed to cover all parts of the hand image. The total of the outside boxes is therefore 78.
The result? If I believe that area of bigger shapes is bigger, then I see that the area of the hand is between 18 mm2 and 78 mm2. And there is a huge gap between these over- and underestimates of 60 mm2. More information? How can we improve this? That is, how can we get over- and underestimates which are closer to the actual area of the hand, so the gap will be smaller? We discussed this. The idea of improving the estimates is not totally obvious, and the whole procedure that's being discussed in this lecture evolved over many centuries in at least 4 different cultures (in China, Greece, India, and perhaps also in Egypt and associated cultures). Lots of smart people thought about it.
We get try a grid with more subdivisions: this is called refining the grid. We could just halve the size of the square sides. Maybe the result is something like what is shown to the right. Now the idea is maybe not so easy to implement, at least for human beings whose eyesight and counting ability is not so perfect. But we can try anyway.
Well, if I have counted correctly, there are 63 new squares which are completely inside the hand image. This is in addition to the 18 old squares. But, wait: the new squares have sides which are half the length of the old squares. So each of the new squares has one quarter the area of the old squares. So actually we have now succeeding in underestimating the area by 18+(63/4)=25 and 7/8.
Now let's get an overestimate. We need to add on all of the little squares which contain any part of the hand image. If I have counted correctly (!!) there are 118 of them. That means we need to add one quarter multiplied to 118 to the previous underestimate in order to get the corresponding overestimate. So 118/4 is 29 and a half. Add this on to the previous 25 and 7/8, and the overestimate is 45 and 3/8.

Underestimate, overestimate, and gap
Let me summarize what we have done:

	Underestimate	Overestimate	Gap
1-by-1 squares	18	78	60
1/4-by-1/4 squares	25.875	45.375	29.5
And so on ...

The secret is that this sort of approach is just about all we can do with real physical quantities. You can really only make better (or worse!) estimates: over- or underestimates. The ideas outlined here, taking finer and finer grids, will lead to overestimates which shrink and underestimates which increase, so that (we hope!) the gap between them →0: refine until you get as close as you want. And this is about all we can do. I don't think the situation is hopeless, but I do think it is compicated.

Attribution
The neat idea of asking for the area of "my hand" is something I borrowed from Professor J. Rosenstein of our math department. I think it is clever. It is a question which brings up important ideas immediately and clearly.

An exact computation
We discussed if we could do an exact computation, not just an estimate, and this I was happy to try, since many people seemed rather distressed about the statements made previously. So I will present a simple example of a general idea. The general idea is very successful, although the simple example may seem to be overelaborate. I will try to compute the exact area "under" y=x for the interval [0,1]. (Yes, I am aware that we know this area -- the ideas are what is important -- as soon as we get the ideas we'll be able to do much more complicated computations.)

How to begin
We will try to get over- and underestimates of the area We could divide the interval [0,1] into 4 equal parts using 0 and 1/4 and 2/4 and 3/4 and 1, which I'll choose to call 4/4.

Then put rectangles underneath y=x in each subinterval. We will be slightly economical and put the largest rectangles we can under the y=x, but we will be slightly simple-minded and only look at rectangles which have sides parallel to the coordinate axes. Then we compute the area of these rectangles: 0 1 1 1 2 1 3 1 --- · --- + --- · --- + --- · --- + --- · --- 4 4 4 4 4 4 4 4 The 1/4's come from the width of the rectangles, and the varying heights (even including the somewhat absurd 0/4) from the heights of the rectangles.
Now we need an overestimate, and the true area will be "captured" between these two estimates. Again, we will use the interval divided into 4 equal pieces, and rectangles with sides parallel to the coordinate axes which just contain the region of interest to us. So the area of these rectanges is: 1 1 2 1 3 1 4 1 --- · --- + --- · --- + --- · --- + --- · --- 4 4 4 4 4 4 4 4
Finally, if we want to see how much gap or error there is between these over- and underestimates, we can look at the area that is inside the big rectangles and outside the small rectangles. The result (because y=x is so simple!) is 4 squares, each with side length 1/4. So the gap is 4(1/4)(1/4)=1/4.

The whole setup
We could do something similar with other numbers of subdivisions of [0,1]. Here are some numerical results:

# of subdivisions	Underestimate	Overestimate	Gap
4	.37500	.62500	.25000
10	.45000	.55000	.10000
25	.48000	.52000	.04000
100	.49500	.50500	.01000

What really happens, in general
Here it helps to get a little bit abstract. Suppose we divide the interval [0,1] into n equal pieces, where you should think that n is some large integer. Then the underestimate will have total area (using heights over the left-hand endpoint of each subinterval shown in the diagram to the right)

  0     1      1     1      2     1      3     1          n-1    1
 --- · --- +  --- · --- +  --- · --- +  --- · --- + ··· + --- · --- 
  n     n      n     n      n     n      n     n           n     n

So what happens as n→∞? Since the gap is 1/n, this will →0. So let me look at the overestimate. This is (1+2+3+...+n)·(1/n²). Let's get a formula for 1+2+3+...+n.

A formula
You may remember the trick (maybe done in high school when you studied arithmetic progressions). The trick is writing two versions of the sum in oppositite order. Look:

 1 +   2   +   3   +   4   + ... + (n-1) + n 
 n + (n-1) + (n-2) + (n-3) + ... +   2   + 1

I think we have verified that 1+2+3+...+(n–1)+n is [n(n+1]/2. And the overestimate of the area, is this divided by n². So a formula for the overestimate is [n(n+1]/[2n²]. What happens to this as n→∞? Well, we have studied such limits, and we could either divide top and bottom by n² or we could use L'H twice. In any case, the limit will be 1/2.

A spectacular result?
I have now proved (done an exact computation!) that the area of the triangle is 1/2. And I guess I knew that. But this whole approach is a template for more complicated reasoning.

Notation
We will need some notation. Lots of sums occur when we do area computations, and also many other computations. So a special way of writing these sums which takes less space has been developed. The Greek capital letter sigma, ∑, is used. Here are some examples:

Examples of summation notation
Written out	Abbreviated
1+2+3+4+5	∑j=15j
32+42+52+62	∑j=36j2
{1/2}+{1/4}+{1/6}+{1/8}	∑j=14{1/[2j]}
1–2+3–4+5	∑j=15(–1)j+1j

The notation, once you get used to it, is wonderfully economical. I should remark that the summation index, which is j in all of the examples above, doesn't really matter. That is ∑_j=1³j⁴ and ∑_k=1³k⁴ both mean the same thing, which is 1⁴+2⁴+3⁴.

How to approximate area
Suppose that we have a function, f, defined on an interval [a,b] and we want to compute the area under y=f(x). That's the area bounded by the x-axis (y=0) and the lines x=a and x=b on each side, and by the curve y=f(x) on top.

We can get approximations to the area of by dividing the interval into n equal parts (people don't always use equal parts, but we're just starting out!). Then we can look at "sample points" in each subinterval, compute f's value at those sample points, and multiply those heights by the widths of the subinterval to get areas of rectangles. Then (sigh!) finally take the sum of these rectangles, and that will be an approximation to the area we want to compute.

The textbook explains in detail how to write the approximating sums. In particular, it identifies three sums corresponding to three choices of sample points: left and right endpoints of subintervals, and the midpoints of subintervals. Below are the formulas, which you should not memorize, and what are supposed to be pictures of a piece of the graph of y=f(x) over (relatively) tall and (relatively) narrow subintervals with the areas of the relevant boxes displayed. Since the interval has length b–a, the width of each subinterval is {b–a}/n. Do not memorize but do try to understand.

Left-hand endpoints of subintervals	(∑j=1nf(a+[{j–1}{b-a}/n]))({b–a}/n)
Right-hand endpoints of subintervals	(∑j=1nf(a+[{j}{b-a}/n]))({b–a}/n)
Midpoints of subintervals	(∑j=1nf(a+[{j–1}{b-a}/n+(1/2){b-a}/n]))({b–a}/n)

QotD
I don't remember! I think it was something like "Compute ∑_j=1³(j+{2/j})".

Antiderivatives
A function F is an antiderivative of f if F´=f. I think the name itself is almost the definition. This turns out to be a very useful concept. It has accumulated other names. For example, F is also called an indefinite integral of f. The word "primitive" is also used, as in, "F is a primitive of f".

Example I think we should begin with a very simple example. Look at f(x)=x–3, quite simple. Then F(x)=(1/2)x2–3x is an antiderivative of f. Why it this true? Well, just compute the derivative of F. The 2 in the exponent moves down, and cancels the 2 in the 1/2. And –3x differentiates to –3. So the claim is verified. Actually (and this is useful to note!), you can always verify a claim that F is an antiderivative of f by just differentiating. Some qualitative aspects of even this "simple" example are worth noticing. The line is negative to the left of x=3, so the parabola (which happens to have roots at 0 and 6) must be decreasing to the left of x=3. The line is positive to the right of x=3, and there the parabola is increasing. Certainly, when x=3, the parabola must have a minimum, because to the left its derivative is negative and to the right its derivative is positive. The transition in first derivative sign from – to + indicates a local minimum (First Derivative Test).

More example and discussion Well, as several (more than several, actually) tried to tell me, there are other antidervatives. For example, (1/2)x2–3x+2 is also an antiderivative of f(x)=x–3. The graph now to the right I hope shows y=(1/2)x2–3x+2. It is the original parabola shoved up. I hope that you notice it has the same qualitative ({in|de}creasing) properties in the same intervals as the original parabola. And we could push the parabola down (I think y=(1/2)x2–3x–1.75 is displayed). Etc. What does this "Etc." mean here? +C If F(x) is an antiderivative of f(x), then F(x)+C is also an antiderivative of f(x) for any constant C. This is because "+C" differentiates to 0. On the other hand, if G(x) is any antiderivative of f(x), then the function F(x)–G(x) has derivative equal to f(x)–f(x), which is 0. The MVT asserts that such a function is a constant. So G(x) must be F(x)+C. So all antiderivatives of a function can be described in this way: (Any specific antiderivative)+C.

Here is a first very brief list of antiderivatives. This information is obtained just be reversing some derivative formulas. Please notice that if you don't believe a given formula, you can always differentiate and check. Also, the "+C" is generally omitted from tables of this type because it takes room and "one" should know it is there.

Function	An antiderivative
xn, n not equal to –1	(1/[n+1])x
1/x	ln(x)
sin(x)	–cos(x)
cos(x)	sin(x)
ex	ex
There will be many more entries!

General comment You already should feel that taking derivatives of functions given by formulas is straightforward. This is true. The reverse is emphatically not true. It is very easy to write a formula whose antiderivative is difficult to compute (even with the help of computers!). It is even easy to write a formula which has no nice antiderivative described by formulas (although it is not easy to confirm that fact). A big chunk of Math 152 is spent on showing you certain ways of guessing (?) antiderivatives. But there are lots of tricks, and the subject is difficult.

Notation
There is special notation for antiderivatives. For reasons which are not clear now but which will be clear (I hope!) in about two weeks, and which come from hundreds of years ago (really, this notation is an abbreviation of a Latin word) we would write the phrase, "F(x) is an antiderivative of f(x)" as ∫f(x) dx=F(x)+C. So, for example, ∫3e^x+x⁵dx=3e^x+(1/6)x⁶+C.

Example and vocabulary
A function can certainly have many antiderivatives (hey, if it has one, it must have infinitely many!). So how can we separate the various antiderivatives? The idea and the vocabulary come from simple physical considerations, where the derivative is velocity and the function is position (and x represents time). If the velocity is prescribed, then we should be able to tell where the object is if the starting position is given.

So this is the idea of an initial value problem. A differential equation is given for an "unknown" function, and then some value of the unknown function is given at some value of its argument (the initial condition). Let's look at a simple example.

Initial Value Problem
    Differential equation y´=2x⁷+8x.
    Initial condition y(1)=3.
Since y´=2x⁷+8x I know (consult the table above) that y must be (2/7)x⁸+4x²+C. I put in the "+C" so that all possibilities for y are covered. Now plug in x=1 and set this equal to the given value of y. We get: (2/7)+4+C=3, so C=3–4–(2/7)=–(9/7). The particular solution, the one specific solution for this initial value problem is therefore y=(2/7)x⁸+4x²–(9/7).

The idea here is that the differential equation tells how a quantity changes or evolves (?) with respect to its independent variable, and that giving a starting value together with a description of the rate of change should be enough to deduce the value of the quantity at any other time. Of course, whether or not this is practical or gives any useful information is something we need to investigate. Here is a very direct version of this idea: if you tell me that my weight last January 1^st was 180 lbs, and then you keep track of, say, my monthly gains and losses, you certainly should be able to tell my weight at the end of the year.

Superman
Suppose that Superman leaps straight up at 40 ft/sec from the top of a 20 ft tall hill. What is his maximum height, and when does he get highest? In this case, the presumed gravity of earth pulls down and we will say that g is –32 ft/sec².

It was my feeling that most students know how to do a problem like this using formulas, and I wanted to plod through it principally as a way of checking our vocabulary.

Translation
Let's measure height from ground level in feet, with + meaning above ground level. And we will measure time in seconds from when Superman jumped. What do we know? There are facts about three functions, and confusion is easy.

1. Gravity causes changes in velocity. So gravity refers to acceleration, a(t), which is the derivative of velocity, v(t). Velocity is in turn the derivative of position, s(t). In this problem, a(t)=–32 (I will start omitting units here). Or we could write v´(t)=–32. Or we could write s´´(t)=–32. Wow.
2. What do we know about velocity? Our main character "leaps straight up at 40 ft/sec" and this is at the initial time. So v(0)=40.
3. And how about distance? Well, s(0)=20 since the leaping is done "from the top of a 20 ft tall hill".

Solution

• From acceleration to velocity
  Since s´´(t)=–32, we antidifferentiate and get s´(t)=–32t+C. But s´(0)=40, and by plugging in t=40 in the general formula for s´ we see that our specific solution is s´(t)=–32t+40.
  
• From velocity to position
  Since s´(t)=–32t+40, we antidifferentiate again. The result is s(t)=–16t^t+40t+C (the 16 results from 32({1/2}t²). Plug in t=0. We know s(0)=20 but the formula states that s(0)=C (this C -- sorry, but people do tend to reuse the letter C in these problems; this can be confusing and lead to errors). So our specific solution is s(t)=–16t²+40t+20.
  
• Analysis of the solution to solve the problem
  I asked people to find the time and position when Superman was hightest, and this was the QotD.
  When Superman is on "top" (highest position) I bet that the velocity is 0. But the velocity at time t is –32t+40, so that time is t=40/32. The height at that time is s(40/32)=–16(40/32)²+40(40/32)+20. We've found the maximum height and the time at which the height is attained. (I was informed that the max height is also equal to 45 ft. Is that better?

Comments on this problem and its solution
This is not a difficult problem, and I am sure that lots of formulas memorized for physics can be used to get the answers. I want to be sure that you understand the language and the process, so that more interesting problems can be analyzed successfully.

The graph to the right has time in seconds as the horizontal axis and has height in feet as the vertical axis. The units differ. I plotted the height functon for the time interval [0,3]. My Superman has somehow landed below ground level on his flight. Oh well. In fact, I think my Superman would hit the darn hill he leaped from (unless someone else has moved it during his flight) at time 2.5. I've never found this sort of problem too believable.

Car acceleration So I need to buy a car. One car I am thinking about goes from 0 to 60 mph in 9.5 seconds. How many g's (g is acceleration of gravity) is this? How can we set this up as a differential equation problem? Again, right now, I am more interested in the process than the answer. Well, 60 mph is 88 feet per second. Assume we begin at t=0 seconds, with both position and velocity equal to 0: s(0)=0 and v(0)=0. We know that v(9.5)=88. Also assume acceleration is constant, so a(t)=K. Now v(t)=Kt+C, and (using v(0)=0) we get v(t)=Kt. Since v(9.5)=88, I see that K(9.5)=88, and K=88/9.5, approximately 9.263. Now g (in these units) is 32, so this car accelerates (for this task) at 9.263/32, approximately .289 g. And then I wondered what distance the car takes to get to 60 MPH, surely a relevant safety question. For this we need to know s(t). But s(t)=(1/2)Kt2+C, and again C–0 (since s(0)=0). So s(9.5)=(1/2)(9.263)(9.5)2, about 417.993 ft, or a bit less than a tenth of a mile (hey, more than a football field). This car's list price is $16,500. Now another car surely which should be bought by a faculty member will go from 0 to 60 in 3.4 seconds. Let's go through the numbers: K(3.4)=88, so this K is 25.882, and g is 25.882, about .81 g. Huh: about three times the acceleration. And the distance needed to get to 60 mph is s(3.4)=(1/2)(25.882)(3.4)2, about 150 feet, half the length of a football field. This list price of this reasonable vehicle is $123,600. Indeed. Which should I buy?

Maintained by greenfie@math.rutgers.edu and last modified 11/17/2009.