Clairaut's Theorem (equality of "mixed" partial derivatives)
Suppose f(x,y) is a function of two variables, and the mixed partial derivatives f_xy and f_yx both exist and are both continuous. Then these mixed partial derivatives must be the same.

Certainly in Math 251, the hypotheses of the theorem will be satisfied. There are examples (similar in nature to the bizarre functions previously given) where things aren't the same. But in this course, the mixed partials will be continuous and therefore will agree. The verification of this result is in the textbook and uses the Mean Value Theorem of 1 variable calculus. As I said, this result will apply to almost all functions we will meet in 251. The result implies, for example, that if we look at the "crowd" of all the possible third partial derivatives of a function of two variables:
f_xxx   f_xxy   f_xyx   f_yxx   f_xyy   f_yxy   f_yyx   f_yyy
it may seem that there are eight possibilities. But due to Clairaut, there are only these four:
f_xxx   f_xxy   f_xyy   f_yyy
The effect of the "concentration" gets even stronger as the number of derivatives increases.

Return to the game
How does Clairaut influence my original question? Again, I was perhaps not the most helpful person in leading the discussion, but eventually the most relevant fact appeared. The function f(x,y)=(sin(y⁴)x-7)³ is a cubic (degree 3) polynomial in x. That is, it can be written as
(Stuff₀)x⁰+(Stuff₁)x¹(+Stuff₂)x²+(Stuff₃)x³
where each of the "Stuff" terms is some function only involving y. An x derivative, ∂/∂x, lowers the degree in x. And four x derivatives will leave us with 0. If you toss a coin a large number of times, it is overwhelmingly likely that there will be at least 4 heads, and therefore, in the differentiation choices, at least 4 x derivatives. So since we can reorder these mixed partials in any way we want, we could put those four derivatives first. And the result will be 0. So, almost surely, if we toss a coin many times, and follow the directed sequence of derivatives, the result will be 0.

What does differentiable mean in 1 variable?
What does f´(x)=Q mean? The definition we all tried to memorize (for a while, anyway) went something like this:

limw→0(f(x+w)-f(x))/w=Q

This definition is difficult to compute with because it has division and subtraction. People frequently "unroll" it to the following:

f(x+w)=f(x)+Qw+Error

Here f(x) is the old, unperturbed output or response of f to the input of x. We perturb (kick?) the function box with a small w. The response of f to x+w can be decomposed (if f is differentiable!) into f(x), the old response, a linear or first-order disturbance, Qw, and "Error". The Error term is very complicated. Of course it will depend on f and x and w (and maybe the phase of the moon). But what is most important about the error term computationally is that it approaches 0 faster than first order. Frequently in applications the Error term is thought of or labeled, H.O.T. for "higher order terms". A function is differentiable in one variable exactly when its response to a small kick can be described as above. This corresponds geometrically to a well-known phenomenon that can be demonstrated nicely using graphing calculators. If you take a point on the graph of a differentiable function and zoom in repeatedly on the graph (centered at the point) within usually a few "zooms" the graph begins to look like a straight line (this is certainly not true of f(x)=|x| at x=0!). Therefore the graph of y=f(x) is (approximately) locally linear exactly when the function is differentiable. It is the property of being approximately locally linear which turns out to be important in higher dimensions. The derivative is the multiplier effect of a small change in the input.

A really silly (?) two-variable example
I gave a very artificial example to show that things can be quite strange in more than 1 variable. I defined a function, f(x,y) to be 2x if y=0, 3y if x=0, and 0 otherwise. To the right is an effort to show the graph, z=f(x,y), of this strange function. The green line is supposed to be tilted above the x-axis (the tilt or slope is 2), the red line is tilted 3 over the y-axis, and the remainder of the graph is flat at z=0, the xy-plane.

Certainly if you slice the graph by y=0 (this is called a vertical trace in the textbook) then you get z=2x, and the partial derivative of f with respect to x exists and is 2. That is, ∂f/∂x(0,0)=2. Similarly, if you slice the graph by x=0 then you get z=3y, and the partial derivative of f with respect to y exists and is 3. That is, ∂f/∂y(0,0)=3.

This is a seriously silly function, but let me analyze it from the point of view of differentiability at the point (0,0). Suppose that h is a small perturbation in the first coordinate and k is a small perturbation in the second coordinate. I would like to understand f(0+h,0+y) in terms of the unperturbed value of f (that is, f(0,0)=0) and also the first-order part (this should be ∂f/∂x(0,0)h=2h and also ∂f/∂y(0,0)k=3k). So I ideally would like
     f(0+h,0+y)=f(0,0)+2h+3k+H.O.T.
But look: what if both h and k are not both 0? Then we are on the grey part of the graph, and the equation above becomes 0=0+2h+3k+H.O.T. whenever h and k are not both 0. We also want H.O.T. to be truly "higher order" so it might have stuff like h^3/2 or k² or even a combined higher order term like hk. None of these will cancel out the 2h+3k. This extremely silly (but rather simple) function does not have the approximation property which we would like in differentiable functions. And it is this approximation property which is needed in many applications -- this is not "theory".

Repeating, looking at slices is not good enough to consider limits and continuity. Slices, even collections of slices in two perpendicular directions, just do not contain enough information about the function. In the case of f(x,y) and its partial derivatives, the key idea, both abstractly and computationally, turns out to be the two dimensional analog of (approximate) local linearity. If we "kick" the input to f in both x and y, we need to understand how the function "responds". The nicest response, similar to one dimension, would be the unperturbed response, f(x,y), then something proportional to h plus something proportional to k, and, finally, a higher-order error term.

By the way, if you don't like piecewise defined functions, you could consider the function defined by this formula: f(x,y)=xy/sqrt(x²+y²). It has properties which are similar to the piecewise function, but it is a bit harder to analyze. You could look at it using Maple, and then you might get some insight.

Differentiable in two variables
In mathematics, when something is wrong, one way to help is by making a definition. The functions we want to consider are called differentiable and have exactly the property that they can be approximated nicely.
f(x,y) is differentiable at (x,y) if there are numbers Constant₁ and Constant₂ so that for h and k small, f(x+h,y+k)=f(x,y)+Constant₁h+Constant₂+Error, where the Error term→0 faster than |h|+|k| (so, faster than first order).

Important results
Before hysteria strikes, here are two results which are verified in the text. They are not difficult to check (again, the essential key is the 1-variable Mean Value Theorem), but we just don't have time in class.

Theorem If f(x,y) is differentiable, then the partial derivatives of f(x,y) exist, and Constant₁=∂f/∂x(x,y) and Constant₂=∂f/∂y(x,y).

Theorem If ∂f/∂x and ∂f/∂y are both continuous then f(x,y) is differentiable (in the approximation sense defined above).

Please realize that essentially all functions we will meet in Math 251 will satisfy the hypotheses of the preceding theorem, so these functions will be differentiable and will have suitable approximation properties. In fact, here is a sort of easy example, although there is some computation involved.

Linear approximation: an example
Here we looked at something like f(x,y)=sqrt(x⁴-y²+2xy-3). Notice that f(2,3)=sqrt(2⁴-3²+2·2·3-3)= sqrt(16-9+12-3)=4. This is an example in a calculus class, and it was chosen so that f(2,3) was nice.

Then ∂f/∂x=(1/2)sqrt(x⁴-y²+2xy-2)^-1(4x³+2y) and ∂f/∂y=(1/2)sqrt(x⁴-y²+2xy-2)^-1(-2y+2x). We can evaluate these derivatives at (2,3):
∂f/∂x(2,3)=(1/2)(1/4)(4·2³+2·3)=(38)/8 and ∂f/∂y(2,3)=(1/2)(1/4)(-2·3+2·2)=-2/8.

If we want a linear approximation to f(2.03,2.98), then we may use the following formula:
f(2.03,2.98) is approximately f(2,3)+∂f/∂x(2,3)(.03)+∂f/∂y(2,3)(-.02).
Here the change in x from 2 to 2.03 means that h is .03 and the change in y from 3 to 2.98 means that k is -.02. The linearized approximation gives us 4+(38/8)(.03)+(-2/8)(-.02) which is 4.1475. The "true value" (well, up to 10 decimal places) of f(2,3) is 4.147314409.

Tangent planes
We can get a bit more out of the slicing picture. The vector i+∂f/∂xk is tangent to the curve in R³ gotten by fixing y on the surface z=f(x,y), and the vector j+∂f/∂xk is tangent to the curve in R³ gotten by fixing x on the surface z=f(x,y). If the surface is nice and smooth (that is, if the function f(x,y) is differentiable) people agree that the two vectors determine a plane which is tangent to y=f(x,y). To write the equation of a plane, we need a point and a normal vector.

Suppose we're at the point (x₀,y₀,f(x₀,y₀)). The normal vector will be perpendicular to both i+∂f/∂x(x₀,y₀)k and j+∂f/∂x(x₀,y₀)k. So we need to compute the cross product: [i+∂f/∂xk]x[j+∂f/∂xk]. So:

   ( i  j     k )
det( 1  0  ∂f/∂x)=-[∂f/∂x]i -[∂f/∂y]j +k=a normal vector
   ( 0  1  ∂f/∂y)

Then if you remember the data needed to write a plane (a point and a normal vector) we can write the equation of a tangent plane.

If f(x,y) is differentiable, then the equation of a plane tangent to z=f(x,y) when x=x₀ and y=y₀ is (z-z₀)=∂f/∂f(x₀,y₀)(x-x₀)+∂f/∂y(x₀,y₀)(y-y₀).

I did begin the nebula discussion in this class, but I decided to put it all in "one place" so what I did on this day is glued together with the stuff for Wednesday's class (October 1).

Two half planes
I defined a piecewise function to exercise "our" intuition.

        ( y  if x>0
f(x,y)= (
        (2x if x<=0

The behavior on the two sides of the yz-plane (where x=0) is different. More globally, the "rear" halfplane (where x<0) is a half of a plane whose equation is z=y. This equation gives a plane tilted at 45^o to the y-axis. The front half, where x>0, was a plane which was tilted up as y increased. The graph that is shown is Maple's version. There are some real problems with how the program shows this graph. The jumps, which we investigated carefully, are connected because what Maple does is just connect dots. So it connects the jumps even when these are not part of the graph of the function (in 1 variable graphs, this "connecting" idea can be turned off, but I don't know how to turn it off in several variables). There are no vertical line segments in this graph! I strongly recommend that you try to graph this function yourself, and rotate the Maple plot systematically to understand what's going on.

How to define this function
Use the following, please:
     >f:=(x,y)->piecewise(x>y,y,2*x);
This command tells Maple that if x>0, then the value of f(x,y) is y. If the statement is false, the value is 2x. Then I loaded the package plots and just used the commands plot3d and contourplot.

The contour "lines"
The contour lines are also not sketched too well. Most particularly, the contour "line" f(x,y)=0 is very peculiar. It actually consists of the y-axis together with the positive x-axis. Maple doesn't want to draw anything like that, so it actually omits a line segment in this T-shaped contour line. I tried various options with contourplot but I could not get the T contour (C=0) drawn correctly. The other contours are again for integer level sets. The level sets for C>0 are horizontal half lines in the first quadrant. The level sets fo C<0 have two pieces. One part is a horizontal half line in the fourth quadrant, and one part is a whole vertical line in the left halfplane. This may be hard to visualize. I urged people to try to educate their intuition. The left lines are closer together than the horizontal halflines. This is all very very peculiar, but the worst is to come.

The suicidal bug
Now comes some of the harder stuff. I asked people to imagine that some bugs were "walking" on the graph of z=f(x,y). The green bug, whose path is shown to the right, strolls along in a path which is roughly circular around the origin. This bug runs into trouble at any point on the positive y-axis, where there's a drop. It also has problems along the negative y-axis, where again there is a big difference in heights. This is a very small bug. I have tried to indicate this by a sort of light reddish color surrounding these half-lines. The blue bug walks from the right halfplane to the left halfplane. It is careful to cross only at the origin. The blue bug is totally safe, and never comes across any severe height differences. So I would like to discuss (and name [define], since it is a math course!) the differences the bugs encounter more precisely.

Limits in one dimension
In one variable, limits are relatively simple. To define lim_x→af(x) we look at how x gets close to a from both sides. There are some standard pictures and standard examples of bad situations. Below are a few, to remind you.

Bad limiting behavior in dimension 1
A jump (y=x+7 for x<3, and 2x otherwise.)	Many wiggles (y=sin(1/x) for x positive, y=0 otherwise.)

Several variables
In several variables limiting behavior can be quite complex, much more than with one variable. I tried to give a few examples.

Many straight line limits exist
I asked students to consider the function
f(x,y)=xy/(x²+y²)
This is an algebraic formula which behaves is a strange fashion for (x,y) near (0,0). We could try some values, but we can also take advantage of the appearance of x²+y². Almost always that's a signal to at least attempt to understand things in polar coordinates -- that is, to take advantage of circular symmetry.

Since x=r cos(θ) and y=r sin(θ), we know that x²+y²=r² and xy=r²cos(θ)sin(θ). Therefore
f(x,y)=xy/(x²+y²)=cos(θ)sin(θ)
The value of f(x,y) only depends on the angular part of the polar coordinate representation of (x,y) and not at all on the radial component. The graph is made up of a bunch of half lines all parallel to the xy-plane, radiating out from the z-axis. These halflines, since cos(θ)sin(θ)=(1/2)sin(2θ), all have height between -1/2 and +1/2.
A Maple graph of the surface over the first quadrant (x>0 and y>0) is shown to the right. I also attempted, with the help of a stalwart student accomplice, to "draw" the surface kinetically. The student "volunteer" held one end of a bungee cord under some tension (both in the student and the cord!) while the calculus instructor held the other end and walked around the student. The calculus instructor raised and lowered the cord twice and the student was asked to keep the end of the cord at the same level as the instructor's end. Therefore along every angle a limit existed, but as the angle changed, the limits changed. There were infinitely many different limits possible along straight line approaches to (0,0).

Always 0 on a straight line approach
The final example of misbehavior was the following function:

        ( 1  if y=x2 and x>0
f(x,y)= (
        ( 0 otherwise

QotD I introduced ∂ by just using it.
Example 1 If F(a,b,c)=a²b-3bc³ then ∂F/∂a=2ab and ∂F/∂b=a²-3c³ and ∂F/∂c=-9bc².

Example 2 If F(a,b,c)=a³sin(7b-5c²) then ∂F/∂a=3a²sin(7b-5c²) and ∂F/∂b=a³cos(7b-5c²) and ∂F/∂c=a³cos(7b-5c²)(-10c).

Limits, 1 dimension Here we've got a function of one variable, and we want to define and understand limx→af(x)=L. The actual definition, frequently stated but rarely stressed in calc 1 classes, is the following: (and, yes, the Greek letters ε and δ are almost always used) Given any ε>0, there is some δ>0 so that if 0<|x-a|<δ, then |f(x)-L|<ε. One way to possibly understand this is uses the model of a "function box" as I did in class: a box labeled "f" which has input and output. In this model, the ε is an output tolerance. We'd like our outputs to be within ε of the ideal output (for this problem) L. Then the limit definition states that there is some input tolerance, δ, which when applied to stuff going into the machine (only allowing inputs within δ of the initial input) then the output tolerance will be satisfied. The definition itself may be difficult to understand for several reasons. First, it is a complicated logical statement, Second, it provides no structure for computing or even estimating δwhen actually given an ε. To me, this is a bit distressing. But some understanding of the input/output model and its approximation properties is fine right now. Limits, 2 dimensions We looked at several examples which were not continuous and did not have limits at (0,0). Let me show you the actual mathematical definition of lim<x,y>→<a,b>f(x,y)=L. It is very analogous to the 1 dimensional definition quoted above: Given any ε>0, there is some δ>0 so that if 0<|<x,y>→<a,b>|<δ, then |f(x,y)-L|<ε. Again the εand δare output and input tolerances, respectively. The interesting feature to me is |<x,y>-<a,b>|. This means the distance from <x,y> to the point <a,b>. This is distance in any direction, along any path. The examples we saw last time only considered approaches to <a,b> along straight line segments. This turns out not to be enough. You've got to allow any paths, and, in fact, allow consideration of all points close to <a,b> (a sort of blob completely surrounding <a,b>). I think this makes limits much more "strict" in several dimensions. Derivative, 1 dimension What does f´(x)=Q mean? The definition we all tried to memorize (for a while, anyway) went something like this: limw→0(f(x+w)-f(x))/w=Q This definition is difficult to compute with because it has division and subtraction. People frequently "unroll" it to the following: f(x+w)=f(x)+Qw+Error Here f(x) is the old, unperturbed output or response of f to the input of x. We perturb (kick?) the function box with a small w. The response of f to x+w can be decomposed (if f is differentiable!) into f(x), the old response, a linear or first-order disturbance, Qw, and "Error". The Error term is very complicated. Of course it will depend on f and x and w (and maybe the phase of the moon). But what is most important about the error term computationally is that it approaches 0 faster than first order. Frequently in applications the Error term is thought of or labeled, H.O.T. for "higher order terms". A function is differentiable in one variable exactly when its response to a small kick can be described as above. This corresponds geometrically to a well-known phenomenon that can be demonstrated nicely using graphing calculators. If you take a point on the graph of a differentiable function and zoom in repeatedly on the graph (centered at the point) within usually a few "zooms" the graph begins to look like a straight line (this is certainly not true of f(x)=|x| at x=0!). Therefore the graph of y=f(x) is (approximately) locally linear exactly when the function is differentiable. It is the property of being approximately locally linear which turns out to be important in higher dimensions.

Slicing and partial derivatives
Now we can define partial derivatives. Please realize that everything we are doing can be done in any number of variables (want a picture of 703 dimensions?) but I'll stick with 2 dimensions here because I can draw pictures and I like pictures. So look at a graph of z=f(x,y). We can slice this in various ways. For example, we could slice this by a plane perpendicular to the y axis with y fixed. This will give sort of an z-x curve. We could "lift" that curve up and just consider it as a function of one variable, x, and then look at the derivative. That's ∂f/∂x. This notation, with a sort of curly d, is not terrific, and it can be (as we will see) quite confusing. But it is what almost everyone uses. Similarly, we could slice by a plane perpendicular to the x axis with x fixed and consider the derivative of the resulting curve or function. That will be ∂f/∂y. Here are the formal definitions if you would like them:

limh→0(f(x+h,y)-f(x,y))/h=∂f/∂x limk→0(f(x,y+k)-f(x,y))/k=∂f/∂y

I'll use h for little changes in the first variable and k for little changes in the second variable.

I did some silly examples and then actually computed some partial derivatives of 1/sqrt(2πt)e^-x2/t. This function occurs in many applications analyzing diffusion, including heat and fluid mixing.

This is lecture #7, and I've "covered" up to section 14.3, so I am already (!) a section behind the syllabus.

Here r(t)=a*cos(t)i+a*cos(t)j+btk, the position vector. The velocity vector is r´(t)=-a*sin(t)i+a*sin(t)j+bk. The length of this velocity vector is sqrt([-a*sin(t)]²+[a*sin(t)]²+b²). This simplifies because we know that sin²+cos²=1. There are very few other curvature computations which are so simple.

So ds/dt=(sqrt(a²+b²)) (the length of the velocity vector, which is the speed) and we get the unit tangent vector by dividing the components of r´(t) by ds/dt. SO T(t)=(1/sqrt(a²+b²))(-a sin(t)i+a cos(t)j +bk). According to what we did last time, if we differentiate this we should get a vector perpendicular to T(t). So d/dt(T(t))= =(1/sqrt(a²+b²))(-a cos(t)i-a sin(t)j+0k). If you take the dot product of this with T(t) you will get 0 (the strange-looking signs make that true). But this is d/dt(T(t)) and I want for curvature d/ds(T(t)). The Chain Rule suggests that I divide d/dt(T(t)) by ds/dt to get d/ds(T(t)). If I do this, the result is (1/(a²+b²)(-a cos(t)i-a sin(t)j+0k). Wow. We are not done yet. This should be κN: that is, it should be the curvature, a positive number, multiplying a unit vector, and this is the unit normal vector. If you stare at what we have you should eventually see the following:

dT/ds=(1/(a²+b²)(-a cos(t)i-a sin(t)j+0k)=[a/(a²+b²)](-cos(t)i-sin(t)j)
Here what is in blue is a scalar, multiplying what is in green which is a unit vector. Therefore what is in blue is κ, the curvature, and what is in green is the unit normal vector. By the way, this is why everyone (and me too, darn it, me too!) uses formulas such as those in the book to compute these things, because going directly from the definition is too darn complicated.

If κ=[a/(a²+b²)] for the helix, how does this match up with our forecasts? Here is the scorecard.

• If a→∞, then κ→0⁺. The curve gets flatter.
  Well, when a gets large, basically we have a¹ on top and essentially a³ on the bottom. The result certainly does →0.
• If b→∞, then κ→0⁺. The slinky (?) gets stretched out more.
  Here if a is fixed, and b gets really large, again the result ;→0 because b only appears on the bottom.
• If a=0, then κ=0. The curve is a straight line.
  True -- set a=0 in the formula.
• If b=0, then κ=1/a. The curve is a circle.
  Again, if b=0 in the formula, then the result is a/a² and this is 1/a.

Here are some pictures of various helices produced by Maple (the plural of "helix" is "helices"). The pictures below were produced using the command spacecurve([a*cos(t),a*sin(t),b*t],t=0..6*Pi, axes=normal,color=black,thickness=2,scaling=constrained,numpoints=180); The procedure spacecurve is loaded as part of plots using the command with(plots);. I used the option scaling=constrained in order to "force" Maple to display the three curves with similar spacing on the axes. Otherwise the x and y variables would be much altered in each image. I hope that these pictures give some idea of what the curvature and torsion represent. Some helices: x=a cos(t) & y=a sin(t) & z=bt a=100 & b=5 κ=.01 a=10 & b=10 κ=.05 a=5 & b=100 κ=.0005

The curvature of the twisted cubic
O.k., I tried to compute the curvature of r(t)=ti+t²j+t³k. I used a formula from section 13.4:
     κ=||r´xr´´(t)||/(||r´(t)||)³.
Even though this formula looks weird, it is much better to use than trying the definitions (I have tried using the definitions with this example, and the computations are terrible).

So r´(t)=1i+2tj+3t²k and r´´(t)=0i+2j+6tk. Now for the cross product computation:

   | i  j  k | 
det| 1 2t 3t2| = det|2t 3t2| i– det|1 3t2| j + det|1 2t|k
   | 0  2 6t |      | 2 6t |       |0 6t |        |0  2|

Therefore (as they write in textbooks) the curvature of the twisted cubic is
     sqrt(36t⁴+36t²+4)
     ---------------------
      (1+3t²+9t⁴)^3/2

Now, having done this ludicrous computation, I will admit that I get almost nothing out of it. Here is this ridiculous formula, and what does it tell me? Well, maybe a little bit. If t is very large positive or very large negative, it seems to say that the curvature gets small (look at the "net" power of t on top, maybe a t², and on the bottom "net" a sort of t⁶). I guess this means that the curve gets flatter as |t|→∞. Maybe this is interesting. (It sort of resembles y=x² that way.)

How to think of all this?
What would I like "you" (especially the engineer and physics "yous") to take away? Another formula from the textbook (section 13.5) resembles exactly what we already saw in two dimensions:
     r´´(t)= (d²s/dt²)T + κ(ds/dt)²N
This is a decomposition of the acceleration vector into tangential and normal components, and actually has some interesting information about physical quantities. Remember that F=ma, and mass is a scalar, so that the Force an object "feels" is exactly a scalar multiple of its acceleration. So let me look at two special cases.

Motion in a straight line
Imagine a ball bearing (?) moving in a straight tube. I claim that the tube never "feels" any force from the walls of the tube. Why? Well, since for a straight line κ=0, the coefficient of the normal component of the acceleration, κ(ds/dt)², is always 0, for any motion, and therefore, since force is a scalar multiple of acceleration, the force also must have zero normal component. So the ball bearing can be pushed along or back in the tube, but it is never pushed against the wall if you insist that the ball bearing move in a straight path.

Motion in a circular arc
Now a more complicated thought experiment. Imagine the ball bearing constrained to move in a piece of a circular tube. All we know is that the object is moving, and its motion is part of a circular arc. I claim that then/ no matter what, the ball bearing is "feeling" a transversal (normal) force from the "walls" of the tube. Why? Well, again the normal component of the acceleration (a scalar multiple of the force acting on the object) is κ(ds/dt)². Well, κ is not 0 since it is 1/(radius of the circle). And, since the object is moving, we also know that ds/dt is not 0. Therefore the product isn't 0, and the normal component of the acceleration isn't 0.

This is, of course, related to the wonderful primitive experiment of quickly spinning a bucket of water on a rope -- and the water is "pushed" into the bucket by the "centrifugal force". Hey, that force is a scalar multiple of the normal component of the acceleration of the bucket's motion. It really works.

Just a little bit more ...
I hate to leave this subject since there is much more to be told, and everything turns out to be amazingly useful in a wide variety of applications. I hate to leave the subject, but the course is very dense. So: just a little bit more. Think of an airplane flying along a curve in the sky (the dashed red line). Then the unit tangent and normal vectors are as shown. There's another vector, B, called the binormal vector, which is TxN. Since T and N both have length 1 and they are perpendicular, then B also has length 1 (sine of a right angle is 1). And B is perpendicular to both T and N. The path is determined by two functions. One of these functions is curvature which says how much the plane is being pulled from a straight line. It measures how much the T is changing in the N direction. Thegreen surface on the tail or the rudder of the plane is the major object determining κ. But the motion is three-dimensional. The way the binormal, B, changes, says how much the plane is twisted out of the two dimensional plane determined by T and B. The rate of change of B is called the torsion, and is usually denoted by τ, the Greek letter tau. It is determined mostly by the angles that the blue control surfaces have with the wings. When a plane takes off or is landing, these surfaces have relatively high angle to the wings.
This is all a big simplification, but it is mostly correct.
There is more about all this in the textbook if you are interested.

We move on to one of the major topics of the course. The word "several" is almost technical in mathematics, and means "more than 1". We will start with an almost ludicrously simple function.

x²+y²
Here f(x,y)=x²+y². This is a function defined by a formula (essentially all of the functions we'll consider in this course will be defined by formulas). The notation means that the input to the function is an ordered pair of numbers, (x,y), and the output is one number. Here the output for the ordered pair (-2,3) is 13.

Formalities: domain and range
The domain will be the collection (the "set") of all possible inputs. Just as in calc 1, if the function is defined by a formula, then the domain will be all inputs for which the function makes sense. The usual restrictions that will concern us are:

• Don't divide by 0.
• Square roots of non-negative numbers only (same for other even roots).
• Logarithms only for positive numbers.

f(x,y)=x2+y2
Domain I think all pairs (x,y) of real numbers, all of R2.	Range Since squares are non-negative, certainly the values of this function are non-negative. And f(0,0)=0, and f(sqrt(A),0)=A for A positive. I am just verifying precisely that the range is all non-negative real numbers.
f(x,y)=1/(y-x2)
Domain So this example is chosen to illustrate the restriction about not dividing by 0. The domain is all pairs (x,y) of real numbers for which y is not equal to x2. Geometrically, this means all points of R2 which are not on the parabola y=x2.	Range Well, 0 isn't in the range (it isn't the reciprocal of any number). But everything else is: check this by just looking at what happens to (0,A), which gives 1/A for all non-zero A's.
f(x,y)=sqrt(y-x2)
Domain So this example is chosen to illustrate the restriction about square roots. The parabola y=x2 divides R2 into two pieces. One piece contains, say, the point (3,4) ("below" the parabola). This point has y-x2=4-32=-5<0, so (3,4) is not in the domain of this function. The domain is the "other" piece of R2 and also those points which are on the curve y=x2.	Range The range is all non-negative numbers. Again, to check this you could look at what happens to (0,A) for A>=0.
f(x,y)=ln(y-x2)
Domain I still must "throw out" the part of R2 which is below the parabola. But here inputs to ln must be positive, so the domain does not include the curve y=x2. The domain is all of the points in R2 which are above the parabola.	Range The range is the range of ln, which is all real numbers.

Kinds of graphs

Let me return to the simplest of the functions I just considered: f(x,y)=x2+y2. There are various graphs which are commonly used. Maybe the simplest is to consider the points (x,y,z) in R3 which satisfy the equation z=x2+y2: this is usually called the graph of the function. A Maple representation of this graph is shown to the right, and the procedure which produced it is plot3d, part of the plots package. This is rather a simple function, and I hope you can see the shape of this surface. It is a cup, axially symmetric around the z-axis. It is called a paraboloid. I looked at this graph and studied curves for fixed values of x and y (called traces in the textbook. These are just (?) parabolas opening up. Piecing them together to get this parabolic cup is not totally obvious.

Another kind of plot, or, anyway, some geometric clue to the nature of the function, can be gotten by looking the contours of f(x,y). There are topographic maps (say, used by hikers) which give a two-dimensional representation of the information in the surface picture above. Pick a constant, C, and look at the (implicitly defined) "curve" f(x,y)=C. I put quotes around the word "curve" because maybe it doesn't have to be a neat nice curve. (An example was discussed in class, and is below.) To the right is a collection of contours for f(x,y)=x2+y2. These contours correspond to the positive integers 1, 2, 3, 4, 5, and 6. Please notice how these contours, which are at evenly spaced "heights", get closer together as the three-dimensional graph gets steeper. Of course, if the contours are not labeled with the values of the constants, I can't tell if the function is increasing or decreasing! This picture was made with contourplot, another part of the Maple package, plots.

The origin in this paraboloid graph is a local and absolute minimum. I briefly sketched a graph of f(x,y)=-x²-y², which is an upside-down version of what was just drawn. So it has a local and absolute minimum. That's not too new. But let's consider something weird and wonderful.

A saddle Consider the function f(x,y)=x2–y2 (I'm trying to exaggerate the minus sign typographically in this since that's the most interesting part). The traces are again "just" parabolas. But here, when, say, y=2 (a plane transversal or perpendicular to the y-axis), we get f(x,2)=x2-4. So the curve z=x2-4 is a parabola opening up. As we change the y's in this sort of slice, the bottom of the parabola moves down for big |y|. What about the other traces, with x=a constant? Well, if x=3, then f(3,y)=9-y2. In the plane perpendicular to the x-axis given by the equaltion x=3, the curve is z=9-y2. a parabola opening down. And by thinking we can see the top of these parabolas moves up when |x| is large. It is difficult for a novice to see how to put these curves together. To the right is a Maple graph of this function for -3≤x≤3 and -3≤y≤3. When you look at this on the computer (please try!) you can rotate it and magnify it, and things might become more clear.
The contour curves of this function are shown to the right. Consider x2-y2=3, for example. One point on this curve is x=2 and y=1, so look at the curve which goes through the point (2,1). This curve is a hyperbola. The hyperbolas (each has two pieces) corresponding to positive values of the function open left and right. On the other hand, there is a family of hyperbolas opening up and down. These correspond to negative values of the function. And there's a special number. If x2-y2=0, then, since x2-y2=(x+y)(x-y), the points on this "curve" correspond to the two straight lines x=y and x=-y. (The strange little box near (0,0) in the contour plot is because Maple thinks that very small +/- numbers which it samples should also be 0. I am sorry about that. It may not be obvious looking at the curvy surface above that there are two straight lines on the surface. But there indeed are.
The point at the origin has a new kind of behavior, not found in one variable calculus. In certain directions it is a maximum. In other directions it is a minimum. This sort of point will be called a saddle point (not too strange a name) or a minimax. You could imagine that there will be a whole range of behaviors in three hundred variables!
Actually there are a heck of a lot of straight lines on this surface, many more than you might think. For example, if x=7t+4 and y=-7t+5 and z=112t-16 (these are parametric equations for some sort of straight line!) is actually on the saddle z=x2-y2. This isn't obvious. I could insert the formulas are check algebraically that (7t+4)2-(-7t+5)2 is the same as 112t-16. Or I could just show you a picture, to the right. After loading plots I typed >A:=plot3d(x^2-y^2,x=-10..20, y=-10..20,axes=normal,color=green); >B:=spacecurve(<7*t+3,-7*t+5,112*t-16>, t=-2..2,color=red,thickness=3); >display3d({A,B}); and I got a version of the picture displayed to the right. On the computer, you can rotate it and play with it, and sort of convince yourself that the straight line actually is on the graph. Sort of amazing. Don't believe me! You try it.

Arc length
If speed is the magnitude of the velocity vector, then since distance=rate·time, well, with variable speed, we need to chop up the time interval and compute and add up pieces of distance. This is the basic idea behind the definite integral, so it makes sense to compute the distance along a curve from time t₁ to time t₂ by this: ∫_t1^t₂||v(t)|| dt.
O.k., this is good (all math is good!) but maybe some examples ... will show how silly it is.

Textbook example
Here is an example taken from a textbook. Suppose r(t)=<12t,8t^3/2,3t²>. What is the length of the curve from t=0 to t=1?
This is a typical artificial problem in a textbook. We know that r´(t)=<12,8(3/2)t^1/2,6t> so that the speed is ||r´(t)||=sqrt{12²+12²t+36t²}. Now we "remember" that distance=rate·time, and that the magnitude of the velocity vector is ds/dt, the speed. So the total distance traveled is ∫₀¹sqrt{144+144t+36t²} dt. Now look at the coincidences. We can pull out the 36 from under the square root, and the result is 6∫₀¹sqrt{4+4t+t²} dt, and, what a coincidence, 4+4t+t² is (2+t)² and the integral becomes 6∫₀¹(2+t) dt=6(2t+(1/2)t²]₀¹ and this is 6[2+(1/2)].

More realistically ...
Almost every combination of two or three functions, even rather "simple" functions that you try to use as a position vector, will yield something that can't be antidifferentiated in terms of familiar functions. For example, look at the following mess:

> a:=t->t^3;
                                            3
                                 a := t -> t

> b:=t->cos(t);
                               b := t -> cos(t)

> c:=t->exp(t);
                               c := t -> exp(t)

> int(sqrt(a(t)^2+b(t)^2+c(t)^2),t=0..1);
bytes used=4000036, alloc=3407248, time=0.14
bytes used=8007244, alloc=5241920, time=0.30
                         1
                        /
                       |     6         2         2 1/2
                       |   (t  + cos(t)  + exp(t) )    dt
                       |
                      /
                        0

> evalf(%);
                                  1.971185471

Curvature is a number which measures how a curve bends
I tried to analyze the idea of how a curve bends. Curvature will be a measure of this bending. I began an analysis that was apparently first done by Euler in about 1750. I'll study plane (2-dimensional) curves. A curve is a parametric curve, where a point's position at "time" t is given by a pair of functions, (x(t),y(t)). Equivalently, we study a position vector, r(t)=x(t)i+y(t)j. Here x(t) and y(t) will be functions which I will feel free to differentiate as much as I want.

There are some special test cases which I will want to keep in mind.

#1: a straight line
A straight line does NOT bend, so it should have curvature 0.

#2: a circle
A circle should have constant curvature, since each little piece of a circle of radius R>0 is congruent to each other little piece, and, in fact, the curvature should get large when R gets small (R is positive), and should get small when R gets large (and looks more like a line locally).

#3: "the" parabola
Even y=x² might be a good test to keep in mind, since there the curvature should be an even (symmetric with respect to the y-axis) function of x, and should be bell-shaped, with max at 0 and limits 0 as x goes to +/-∞. The curve actually bends most near the origin, and far away, when x is large positive or large negative, even though the curve is very steep, it gets really flat.

The problem of understanding or defining curvature is to somehow extract the geometric information from the parameterized curve. That is, if a particle moves faster, say, along a curve, it might seem like the same curve bends more. So what can we do? Somehow the geometry of the curve should be separated from the dynamics (kinetics?) of motion along the curve.

We looked at θ, the angle that the velocity vector r´(t) makes with respect to the x-axis. How does θ change? After some discussion it was suggested that we look at the rate of change with respect to arclength along the curve: that is the same as asking for the rate of change with respect to travel along the curve at unit speed, and therefore somehow the kinetic information will not intrude on the geometry. This is the same as asking a bug to travel along the curve at unit speed, and to describe the rate of change of θ (the turning of r´(t), the velocity vector) as the bug strolls along the curve.

The quantity s and the quantity t
Arc length, s, on a curve is computable with a definite integral: sqrt(x´²+y´²) integrated from t₀ to t with dt is the arc length. As shown before, this is rarely exactly computable with antidifferentiation using the usual family of functions. But ds/dt is just sqrt(x´²+y´²) by the Fundamental Theorem of Calculus. And the Chain Rule suggests that d*/ds(ds/dt)=d*/dt if * is any quantity of interest, such as θ. So to get d*/ds from d*/dt, multiply the latter by 1/(ds/dt) which is 1/sqrt(x´²+y´²).

By drawing a triangle we see that θ is arctan of y´/x´: θ=arctan(y´/x´). We can find dθ/dt: it is (y´´x´-x´´y´)/(x´²+y²)². This uses the formula for the derivative of arctan, the Chain Rule, and the quotient rule. Then the previous results say that

Back to the test cases

The straight line We compute for some constants A and B and C and D: x(t)=At+B x´(t)=A x´´(t)=0 y(t)=Ct+D y´(t)=C y´´(t)=0 so that the top of the curvature formula is 0. κ=0 for any straight line.
A circle Suppose we have a circle of radius R centered at the origin. Then parametric equations are not too difficult to get: x(t)=Rcos(t) x´(t)=-Rsin(t) x´´(t)=-Rcos(t) y(t)=Rsin(t) y´(t)=Rcos(t) y´´(t)=-Rsin(t) Let me look first at the bottom of the curvature formula. This is (x´2+y´2)3/2. Notice that x´2+y´2=R2([-sin(t)]2+[cos(t)]2)=R2, so the 3/2´s power is R3. The top is y´´x´-x´´y´ and this is -Rcos(t)(-Rsin(t))--Rcos(t)Rcos(t)=R2·1. The whole curvature formula is then R2/R3 which is 1/R. When R is large, this is near 0. When R is close to 0 and positive, this is large positive. And the curvature is constant which it should be since the circle has the same local geometry at every point.
A parabola Here I want to look at y=x2. A simple parameterization is good. So let's try: x(t)=t x´(t)=1 x´´(t)=0 y(t)=t2 y´(t)=2t y´´(t)=2 Now the bottom of κ, (x´2+y´2)3/2, becomes (1+4t2)3/2, and the top, y´´x´-x´´y´, is just 2. So κ=2/(1+4t2)3/2. Here the "local geometry" definitely changes from point-to-point. The most curvy (?) part of the graph is at the origin. As x→+/-∞, although the graph gets steeper and steeper, the curve locally actually gets more and more flat: the curvature→0.	The smaller figure is supposed to be the curvature of the parabla.

I defined the unit tangent vector, T, to be a unit vector in the direction of r´(t). If θ is the angle that r´(t) makes with the positive x-axis, then T must be cos(θ)i+sin(θ)j. Also r´(t)=(ds/dt)T, where ds/dt is the length of r´(t), the speed. Now differentiate the formula for r´(t) using one of the product rules we had stated earlier. The result is r´´(t)=(d2s/dt2)T+ (ds/dt)d/dt(T). But T is cos(θ)i+sin(θ)j, and differentiation with respect to t is the same as differentiation with respect to s multiplied by ds/dt. But differentiation with respect to s gives (-sin(θ)i+cos(θ)j) multiplied by the derivative of θ with respect to s, and this is κ. All this put together is: r´´(t)= (d2s/dt2)T + κ(ds/dt)2N where N is (-sin(θ)i+cos(θ)j), a unit vector normal to T (check this by dot product!), where N is called the unit normal. We have decomposed acceleration into the normal and tangential directions. This can be significant physically and can help to understand physical situations. Here are two "extreme" cases. Straight line motion: no transverse forces I used this to show that motion in a straight line (where κ= 0) had no normal component, and therefore a particle moving in a straight line had no force needed transverse to its motion (remember force and acceleration are vectors related by the scalar m, according to Newton's second law). Moving in a circle: always normal force On the other hand, in our circular situation, the curvature was a positive number, and as long as the particle was moving (ds/dt not equal to 0) a force was needed to keep it on the circle. This is because the curvature κ was non-zero, and so were the other terms. This is not at all "intuitively clear" to me. Silly example A racetrack: this racetrack consists, as nearly as possible, of two straight line segments, each about 3 miles long, connected with a semicircular track. Since the racetracks were specified to be 2 miles apart, we "deduce" that the radius of the semicircle is 1 mile. The racetracks certainly have curvature=0, and the semicircle, which is Pi units long, has curvature 1. What would a graph of the curvature of this racetrack look like? An answer I attempted to be "correct" both qualitatively and quantitatively in this graph. The blue curves are my attempts to draw smooth transitions between the curvatures of the circles.

Now let's look at space curves. The geometry of these curves, as seen from the point of view of calculus (called "differential geometry of space curves") is a subject which originated in the 1800's. The material presented here was stated in about 1850-1870. When I learned about this material in college, it mostly seemed rather abstract and useless -- stuffy formulas no intelligent person would care about. Like a number of other judgments about merit that I've made in life, this is completely wrong. The geometry of curves has within the last few decades become very useful in a number of applications: robotics, material science (structure of fibers), and biochemistry (the geometry involving the structure of big molecules such as DNA): amazingly useful formulas and ideas!

The right circular helix
I'll carry along is a right circular helix as a basic example.

x(t)=a cos(t)
y(t)=a sin(t)
z(t)=b t

How should curvature behave for the helix? We discussed this and decided κ should have these properties:

• If a→∞, then κ→0⁺. The curve gets flatter.
• If b→∞, then κ→0⁺. The slinky (?) gets stretched out more.
• If a=0, then κ=0. The curve is a straight line.
• If b=0, then κ=1/a. The curve is a circle.

Now let's analyze space curves
If r(t)=x(t)i+y(t)j+z(t)k (the position vector), then r´(t)=x´(t)i+y´(t)j+z´(t)k=(ds/dt)T(t) is called the velocity vector. Here T(t) is called the unit tangent vector and is a unit vector in the direction of r´(t). ds/dt is the speed, and is sqrt(x´(t)²+y´(t)²+z´(t)²), the length of r´(t). We use ds/dt also to convert derivatives with respect to t to derivatives with respect to s, as in two dimensions using the Chain Rule.

Since T(t)·T(t)=1 differentiation gives T´(t)·T(t)+T(t)·T´(t)=0. But the dot product is commutative, so this is 2T´(t)·T(t)=0 or just T´(t)·T(t)=0. This means that T´(t) and T(t) are always perpendicular. In fact, we are interested in dT/ds, which is the same as (1/(ds/dt))T´(t) (it is usually easier to compute T´(t) directly, however, and "compensate" by multiplying by the factor 1/(ds/dt)).
Think about this sentence, which states a fact I'll use again and again in this course:
     Any non-zero(!) vector is equal to the product of its magnitude times a unit vector in its direction.
For dT/ds, the magnitude is defined to be the curvature, κ, and the unit vector is defined to be the unit normal N(t). This essentially coincides with what's done for plane curves, when curvature was defined to be d(θ)/ds.

I will continue with this material next time. I confess that I am currently about 1/2 lecture behind the syllabus, but the formulas and examples and concepts involving curvature are important and complicated. Please read the textbook carefully and try the homework problems in sections 13.3, 13.4, and 13.5. This is all work, but please do this! I will compute the curvature for the helix next time, and discuss some of the consequences of all of this stuff before continuing with the syllabus.,

Parametric curves: a nuisance?
So I began with some irritating examples.

1. The curve r(t)=<t,t,t>. This curve is a straight line. It goes through the origin, (0,0,0), and the path described is all of the line. The motion described by this vector function (a triple of standard real-valued functions) is uniform rectilinear (straight-line) motion.
2. The curve r(t)=<t³,t³,t³> is all of the straight line, but the kinetic aspect is very different. Motion is not uniform, and gets faster as |t| gets larger both positive and negative.
3. Now the curve r(t)=<t²,t²,t²> is not too surprising. It is a half line or ray. Dynamically this represents the motion of a point starting way out in the first octant where all of x and y and z are positive, then coming in (slower and slower) to (0,0,0). Then the point turns around (?) and starts out into the octant again, retracing its path and going faster and faster.
4. The curve r(t)=<sin(t),sin(t),sin(t)> is quite nasty psychologically if discussed immediately after the line. All of the coordinates are equal so every point on this curve is on the straight line. Some thought is needed to force yourself to agree that the "curve" is only a line segment from (-1,-1,-1) to (1,1,1) because sine's range is [-1,1]. The motion described by this vector function oscillates endlessly between the two points named.

Distinguishing between geometric aspects of the path of a particle and the dynamic aspects of the particle will take up most of our next class. This won't be obvious.

Two problems in the textbook
I tried to discuss two problems from the textbook. These pictures are copied from page 743 of the text.

Problem 5 of section 13.1
Match the space curves in Figure 8 with their projections onto the xy-plane in Figure 9.
Answer
How can we "solve" this problem? Well, I actually think most people can solve the problem, but I think describing the process is somewhat difficult. It is difficult because, for example, the pictures in Figure 8 are two-dimensional. And they are images of some idea three-dimensional situation. What can we see? Well, in Figure 8 (A) I see a straight line. I believe that the straight line will not be made "curvy" if we project into the xy-plane. Therefore the only candidate which matches is the non-curvy (!) picture (ii) in Figure 9. What else? Well, maybe the picture in (B) seems to be helical (more precise information about a helix later). That's sort of a motion around a cylinder whose axis of symmetry is, in this case, the x-axis. If we push this down and neglect the z coordinate, I think what we get is picture (i). This leaves (C) matching up with (iii). If you think about a particle moving around the path indicated in (C), maybe you see the xy-"squashing" of it moving back and forth along the parabolic path in (iii). I was asked if there was somehow "proof" that this was (iii), and, well, we're told that the triples of pictures match up, and (C) and (iii) are what's left. It is certainly true that something like (C) could have a more complicated squashing than what is in (iii).

Problem 6 of section 13.1
Match the space curves in Figure 8 with the following vector-valued functions:
(a) r₁=<cos(2t),cos(t),sin(t)>
(b) r₂=<t,cos(2t),sin(2t)>
(c) r₃=<1,t,t>
Answer
Well, the linear formula is (c), and if we just consider the first two coordinates, the 1 and the t, that would seem to describe (ii), so I think that (c) is the algebraic version of (A). As for (b), the first two coordinates are t and cos(2t) which seem to describe (i). And if you consider the second and third coordinates of (b), they are cos(2t) and sin(2t). The sum of the squares of these is 1 (cos²+sin²=1) and the first coordinate in (b) just moves the point along, so we get the helix in (B). That leaves r₁ in (a) to describe (C) and (iii). Let's consider the first two coordinates of (a), which are cos(2t) and cos(t). A trig identity you may/may not remember is cos(2t)=2(cos(t))²-1. If we take x as cos(2t) and y as cos(t), this becomes the equation x=2y²-1 which surely looks like (iii). And the third coordinate, sin(t), of (a) makes the image go up and down, up and down. So (a) is consistent with both (C) and (iii). I think this third curve and formula are a bit interesting and strange.

Pictures of a point in motion
Twisted cubic
The twisted cubic is the curve r(t)=<t,t²,t³>. It is one of the beginning curves shown to students in this subject, with the warning that here is the pictures can be deceptive. A Maple command created a 3-dimensional viewing box of the curve. I rotated this box in various ways and exported the images shown below.
spacecurve(<t,t^2,t^3>,t=-5..5,color=black,axes=normal,thickness=2);

The view from the z-axis Here we look down on the curve from high up on the z-axis. This has the effect of suppressing (?) or deleting the z-coordinate from the triple, and we just see the geometry of the first two coordinates: (t,t2). Of course, this is the parabola y=x2.
The view from the y-axis Now delete the central variable. In the xz-plane, the curve is the collection of points (t,t3) and this is z=x3, which is, I hope, a fairly familiar cubic. And, indeed, if you orient the twisted cubic properly, then Maple shows you the displayed picture.
The view from the x-axis This is picture which makes life difficult. Projected only the yz-plane, the collection of points (t2,t3) is the same as those points satisfying y3=z2. Now since z2 must be non-negative, any y's on this curve had better be in the right half of the yz-plane. And for each y>0, there are two z's (the positive and negative square roots). But the worst part is the behavior at the origin. z=+/-y3/2 has a horizontal tangent at (0,0): the curve has what's known as a cusp at the origin, a type of corner. It looks sharp, not smooth!
An oblique view This "curve" actually does not have a corner in three-dimensions. The pictures already shown do not make this easy to see. To the right is sort of an oblique view of the curve as drawn by Maple. Although this picture seems to show a loop, the curve does not actually have any loops. The loopy look is a result of the angle I picked. The difficulties I discussed here are one reason the instructors would like you to be familiar with Maple -- use of it will help your intuition. A perspective on real life Frequently, similar difficulties have occurred in dissections of bodies, when slides of different cells are prepared. Depending on the angles of the slices, very different pictures are seen.

Robots and molecules
I also remarked that people who wanted to understand the conformation (geometry) of biologically interesting macromolecules will need to learn how to look at curves in R³. Also, people who want to describe motion of robot arms necessarily need the ideas we will describe.

The derivative
If r(t)=<a(t),b(t),c(t)> is a vector function of t, we can consider (1/h){r(t+h)-r(t)}. Inside the {} is a difference of vectors, and that's therefore a vector. Then multiplying by 1/h: that's a scalar multiple. So the result is a vector. It may happen that the limit as h-->0 of this quotient may or may not exist. If it does, we'll say that r(t) is differentiable and that the limit, which is labeled r´(t), will be called the derivative. It is also called the velocity vector.

It doesn't take many examples to convince yourself that the (vector) derivative will exist exactly when all of the scalar functions a(t) and b(t) and c(t) are differentiable, and that the components of the resulting vector derivative will be the derivatives of these functions. That is, if a(t) and b(t) and c(t) are differentiable, then the vector function will also be differentiable, and r´(t)=<a´(t),b´(t),c´(t)>. Computing the derivative of such a vector function when the component functions are defined by familiar formulas involves nothing essentially new. I did a few examples.

Meaning of the derivative
The derivative balances out a large scalar, 1/h (when h is small), with a very small secant vector, r(t+h)-r(t). That the limit exists is rather neat fact, that somehow the shrinking of the vector stabilizes with the increasing of the scalar amount, and that the direction tends to a fixed direction -- this is not obvious, and one should really not expect it!

The magnitude: speed
The magnitude of the velocity vector, ||r´(t)||, is called the speed.

Velocity vector, tangent vector
I tried to argue, looking at a local picture of the path of a particle, that the direction of the velocity vector is tangent to the path of the particle. So the velocity vector is a vector tangent to the path.

Tangent line to a helix
Look at the curve r(t)=<3cos(2t),3sin(2t),8t>. What is this?
The first two variables describe uniform circular motion. The radius of the circle is 3, and that's the distance from the central axis of this curve, which turns out to be a helix. The 2 changes the angular velocity of the curve, and doubles it. The 8 affects the "pitch", the angle of the helix, and also the distance between "loops". When 2t changes by 2Pi, the curve passes around one loop. That means the change in t is Pi, so the change in z is 8Pi.

Let's find the parametric equations of a line tangent to this helix when t=Pi/2. The line must pass through r(2)=<3cos(2[pi/2]),3sin(2[Pi/2]),8[Pi/2]>=<-3,0,4Pi>. A vector in the tangent direction can be gotten from the velocity vector. So: r´(t)=<-6sin(2t),6cos(2t),8> which, when t=Pi/2, gives <0,-6,8>. Therefore the parametric equations for the line are:
x=-3+0t
y=0-6t
z=4Pi+8t
To the right is a picture of both the helix and the tangent line just specified.

Various formulas, especially product formulas
The derivative of u(t)·v(t) is u´(t)·v(t)+u(t)·v´(t). The derivative of u(t)xv(t) is u´(t)xv(t)+u(t)xv´(t). And there's even another product rule (multiplying a vector function by a scalar function and then differentiating). Since the cross product is not commutative, the order is important. We will use these formulas next time. They are very important in describing the dynamics of motion of a point in space.

This course is like ...
I get confused about the meaning of metaphor and simile. One reference I found declares

Metaphor is often confused with simile, the difference being that the metaphor draws a parallel between concepts, while the simile points to poetic similarities.

I did problems 15 and 28 in section 12.4 since I wanted to show some direct examples of cross product computations.

Problem 13
What is the cross product of <(1/3),1,(1/3)> and <-1,-1,2>? We must compute:

   |  i   j   k  |
det|(1/3) 1 (1/3)|=det| 1 (1/3)|i-det|(1/3) (1/3)|j+det|(1/3) 1 |i=
   | -1  -1   2  |    |-1   2  |     |  -1    2  |     |  -1 -1 |   
= [(1·2-(1/3)·(-1)]i-[((1/3)·2-(1/3)·(-1)]i+[((1/3)·(-1)-1·(-1)]i=(7/3)i-j+(2/3)k

 x |  i  |  j  |  k
--------------------
 i |  0  |  k  | -j  
--------------------
 j | -k  |  0  |  i
--------------------
 k |  j  | -i  |  0

       | i j k | 
vxw=det| a b c |
       | d e f |

I remark that the distance defined is called the Euclidean distance because there are other ways of defining "good" distances, and some of them are useful in other contexts. For example, students who study digital signal processing will find out that the Euclidean distance is perhaps not always the most useful distance when comparing signals. In this course, the Euclidean distance will be the only one used, so I will almost always omit the adjective "Euclidean". What is nice about the Euclidean distance, though, is that it is very computationally neat, and leads to the method of least squares, which is a very commonly used method of fitting "curves" (function descriptions) to data points.

A sphere of radius 3
What algebraic condition on a point (x,y,z) is equivalent to the geometric statement that the point lies on a sphere of radius 3 centered at (0,0,0)? This means that the distance from (x,y,z) to (0,0,0) should be 3, or sqrt((x-0)²+(y-0)²+(z-0)²)=3 which of course is the same as x²+y²+z²=9.

Completing the square to "recognize" a sphere
We could of course "reverse" the process if we're given an equation (or at least an equation of the correct form). For example, the equation
x²-3x+y²+4y+z²-7=0
represents a sphere. What is its center and its radius? The key algebraic maneuver here is completing the square, and everything works very much as in two dimensions.
x²-3x --> x²+2(-3/2)x --> x²+2(-3/2)x+(-3/2)²-(-3/2)² --> (x-3/2)²-(-3/2)²
y²+4y --> y²+2(2y) --> y²+2(2y)+2²-2² --> (y+2)²-2²
We don't have any term with z to the first power. So the equation becomes:
(x-3/2)²-(-3/2)²+(y+2)²-2²+z²-7=0
which is the same as
(x-3/2)²+(y+2)²+(z-0)²=(-3/2)²+2²+7
and we can "read off" that the center is (3/2,-2,0) and the radius is sqrt((-3/2)²+2²+7). It is easy to make mistakes with minus signs when identifying the center. It is also easy to forget to take the square root when identifying the radius. Oh well.
If we had different (but positive) numbers in front of the squares (x² and y² and z²) then we'd get an egg-shaped object, an ellipsoid. We'll see more about this later.

Vectors A vector is a directed line segment. It is used as a mathematical model for any quantity which has both magnitude and direction. For example, some of the following quantities are vectors and some are not:

Some vector quantities Acceleration, velocity, displacement For velocity, for example, the length of the vector corresponds to speed: "I am traveling at 25 miles per hour" (in metric, I think that is 78 hectares per wombat). Of course the direction of the vector corresponds to the direction of motion: "I am traveling at 25 miles per hour North by Northwest" (in the metric system, this direction is called "Colder"). Force Another basic example of a vector. The magnitude is the amount of oomph in the force (I am giving up on units) while the direction is the direction the force is pushing in. Moment, torque, etc. You'll meet lots of vector quantities.

Things which aren't vectors Mass (=energy, according to our good friend Albert). Temperature Quantities which are not vectors have the strange adjective, scalars. So both Mass and Temperature are scalar quantities, and are measured by numbers (both positive and negative) and have no intrinsic direction.

Equal vectors
A vector can also be specified by its head and tail. Two vectors will be called equal (maybe the official word is equivalent) if when the tail of one of them is moved to the tail of the other, then their heads are in the same place.

Vectors are used to do algebra in more than one dimension. This is because algebra has really been successful, and the interaction between algebra and geometry has paid off well in both directions. Therefore we'll need to add and multiply vectors. Efforts to multiply run into serious difficulties, as we will see next time. Addition is neat and "everything" works well. As I mentioned in class, the generally accepted definition for vector addition models some situations which can be experienced and measured in "real life" with forces, velocities, etc. The word resultant is sometimes used in connection with this definition.

Definition of vector addition

First suppose there are two vectors, v and w, which are roaming, free and happy in R3.	We take the vector w and drag it so that the tail of w is at the same point as the head of v.	The vector v+w is now defined using the geometric display we just arranged. v+w the vector whose tail is at the tail of v and whose head is at the head of w.

Properties of vector addition

0. Reality Addition (superposition, resultant) of vector quantities in physical "reality" is modeled by vector addition. As I mentioned in class, there probably are many ways we could define vector addition. This way is used because it accurately models certain physical situations. Please keep this in mind!
1. Commutativity v+w=w+v for all pairs of vectors v and w.
   This can be proved from the geometric definition by drawing a parallelogram which has one pair of sides translates of v and the other pair of sides translates of w. Anyway, the order in which you add vectors doesn't matter.
2. Associativity (v+w)+u=v+(w+u) for all triples of vectors v, w, and u.
   I think this can be proved by drawing some parallelopided (three-dimensional analogue of a parallelogram) with sides gotten form v and w and u: but the diagram would almost be more annoying than just thinking about it. The important consequence is that you can group adding vectors in any way you want, and the result will be the same.

Zero and minus
The zero vector has its head equal to its tail. If v is a vector, then -v is the vector whose length is v's but whose direction is the reverse: so the head of -v is the tail of v and the tail of -v is the head of v. Huh. Say that fast.
I am especially proud of the image of the zero vector in the picture to the right. I wanted to show the special beauty of the zero vector. I tried several different angles. (This is a joke!)

Then v+(-v)=0, and v+0=0 etc. for all vectors.

Scalar multiplication
In this course the scalars will be real numbers. Scalars are things that multiply vectors. Many of the students in the course will see later situations where the scalars are complex numbers. In the case of the vectors arising in computer science and electrical engineering, collections of scalars occur which are much less familiar.

If c is a scalar and v is a vector, then the vector cv is defined by the following:
If c>0 the direction of the vector cv is the same as the direction of v, and the length of cv is the length of v multiplied by c.
If c=0 then cv is the zero vector.
If c<0 the direction of the vector cv is the same as the direction of -v and the length of cv is the length of v multiplied by |c|. (The absolute value makes sure that lengths stay positive.)
Here are pictures of v and -v and 3v and -2v.

Important vectors Some vectors are more important than others, especially if you have a coordinate system. The vectors i and j and k are vectors of length 1 (such vectors are called unit vectors) which are parallel to the coordinate axes x and y and z, respectively.
Now take any vector v and move it so that the tail of v is at the origin, (0,0,0). Then the head of v will be at some point, (a,b,c). If you think about the geometry I hope you can convince yourself that v=ai+bj+ck. We have written v as a sum of its components. The textbook also uses the notation v=<a,b,c> for this.

Vector addition in terms of components
Since vector addition is commutative and associative, and certainly (c₁+c₂)(any vector)=(c₁)(that vector)+(c₂)(that vector), we see clearly that if v=ai+bj+ck and w=di+ej+fk, then v+w=(a+d)i+(b+e)j+(c+f)k.
The components of the sum are the sum of the respective components. If you know the components, it is easy to compute vector sums and scalar multiples of vectors.

The head and the tail produce the vector (algebraically)
Suppose the point P has coordinates (x₁,y₁,z₁) and the point Q has coordinates (x₂,y₂,z₂). Then the vector from P to Q (so P is the tail and Q is the head) is just (x₂-x₁)i+(y₂-y₁)j+(z₂-z₁)k. I hope that you can draw a picture convincing yourself of that, or you can look in the text.
Example The vector from P=(1,2,-3) to Q=(-5,6,2) is (-5-1)i+(6-2)j+(2-[-3])k=-6i+4j+5k. It is easy to make mistakes with signs in this computation.

How long is a vector?
I will use length and norm and magnitude to mean the same thing about a vector: its length. If v=ai+bj+ck, then we can think of the tail of v as sitting at (0,0,0) and the head of v, at (a,b,c). The length formula we got then says that the length must be sqrt(a²+b²+c²). So if you know the components, the magnitude can be computed. In the textbook we are using the magnitude is written ||v||. Please note that in some books and some situations, this quantity is just written |v|.

NO! NO, NO, NO!!!
Suppose that v=<1,2,3> and w=<-2,3,2>. Then v+w=<-1,1,5>. And ||v||=sqrt(14) (approximately 3.74), ||w||=sqrt(17) (approximately 4.12), and ||v+w||=sqrt(27) (approximately 5.19). In general, these numbers need not be related in any simple fashion, aside from the Triangle Inequality (see the textbook, please).

Next time I will explore how these lengths can be explained, as we discuss multiplication of vectors.

After the class ...
I think the classroom is badly suiting to teaching math. Most of the blackboards are barely visible to most of the chairs. Next time I will try to teach at the narrow end of the room. I did investigate if any other class rooms are available. Exactly one other room on College Avenue is vacant at each class time, and both of those two rooms are the same geometry as the one we have. They are both classrooms in a river dorm! The physical plant at Rutgers is overused.

Start reading chapter 12, please. Do the problems.

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Maintained by greenfie@math.rutgers.edu and last modified 9/5/2008.