Example #1
Here the constraint is x²+xy+y²=1, and the function to be maximized, the objective function, is x²+y². The picture corresponding to this situation is shown to the right. The bigger circles correspond to larger values of the objective function.
Suppose that T(x,y)=x²+y² were the temperature in a thin metal plate with shape the interior of x²+xy+y²=1, where will the plate be hottest or coldest? I remind you that in this "heat" language the level curves or contour lines are called isothermals.
Well, local extrema only occur at critical points, and only (0,0) is a c.p. That, easily, is the coldest point in the plate. But where is the hottest point? It must be on the edge, and it will NOT be a local extremum, but only an extremum for a constrained maximization. We seek therefore the extrema on the boundary using Lagrange multipliers.
Compute the gradients, etc. Then the multiplier equations and the constraint equation are:

2x+y=()(2x)
2y+x=()(2y)
x2+xy+y2=1

Fan mail for the Lagrange multiplier method
I think it is wonderful that a relatively small amount of algebraic effort can produce such a lovely geometric result (the specific circles centered at (0,0) which are also tangent to the ellipse). This reassures me that things algebraic and geometric both reflect the same reality.

Example #2 Find the maximum and minimum values of 3x-4y+5z on the unit sphere x2+y2+z2=1. Here is perhaps a more complicated picture, with the constraint (the unit sphere) and five planes representing where f(x,y,z)=3x-4y+5z=-8 and -3 and 1 and 5 and 9. The picture is supposed to help you understand that max/min occur where the planes will be tangent to the sphere. The system of Lagrange multiplier equations (three of them here, since we are in R3) together with the constraint follows. 2x=3() 2y=-4() 2z=5() x2+y2+z2=1 The left-hand sides are the components of (x2+y2+z2) and the right-hand sides are multiplying the components of (3x-4y+5z). You can solve for x and y and z in terms of , and substitute these values in the constraint equation, getting =+/-(2/sqrt(50)). Then 3x-4y+5z turns out to be (for the two choices of , generating two candidates for where extreme values take place) sqrt(50) and -sqrt(50). Here a final picture of the constraint and the two planes given by 3x-4y+5z=+/-sqrt(50).

Proofs, etc.: the dual (?) nature of math
I learned and "liked" Lagrange multipliers in a several variable calculus course, just as I hope you are. The justification for the method was more or less what I have shown you. So I knew it was "true". But I never saw a "proof" of the Lagrange multiplier method until my second year of grad school. Sigh. It really isn't that difficult to prove. Maybe I didn't (even as an apprentice professional mathematician!) feel the need to prove such a lovely idea.

• The function x³+y³-3xy
  This function has a saddle at (0,0) and a local min at (1,1.) This is not obvious and needs to be confirmed by computation.
  Ms. Orani and Mr. Nelan and Ms. Lai and Mr. Bonilla contributed to the solution.
• The function (x/y)+(8/x)-y
  This function has a local max at (-4,2). This is not obvious and needs to be confirmed by computation.
  Mr. Boxer and Mr. Ericson and Mr. Faiwiszewski and Mr. Yen contributed to the solution.
• The function xe^-x3+y3
  This function has a critical point at (3^-1/3,0) and the second derivative test gives no information. However, if you think of x as fixed, then we have (Constant)e^y3 and this is an increasing function of y with no minima or maxima. Therefore the critical point has to be like a saddle, with points close by having both bigger and smaller values of the function.
  Mr. Comito and Ms. Curcio and Mr. Baig and Mr. Shtabnoy contributed to the solution.

We studied the following problem from 1 variable calculus:
Consider the ellipse x²+5y²=1. Find the rectangle of largest area inscribed in this ellipse with sides parallel to the coordinate axes. Of course this turns into: maximize 4xy (the objective function) subject to x²+5y²=1 (the constraint). Consideration of the geometry (varying rectangles) suggests that there is indeed a "biggest" rectangle, somewhere.
How the "heck" does a calc 1 student solve this problem since the function to be maximized, 4xy, has two variables. I suggested the following methods of solution:

1. Reduction of dimension, simple version
   Since x²+5y²=1, we know that y=sqrt((1-x²)/5). Then the area is F(x)=4x·sqrt((1-x²)/5). The domain for this function is 0=<x=<1. General theory from one variable calculus states that max/min are obtained at end points or critical points. But F(0)=F(1)=1. So the max is gotten where F'(x)=0. We computed this. Of course, in a random situation, it may be very difficult to solve (effectively!) for one of the variables in terms of the other.
2. Reduction of dimension by parameterization
   We could make an inspired "guess": try x=cos(theta) and y=sin(theta)/sqrt(5). Then the pair (x,y) is on the ellipse, and since the max is obtained somewhere in the first quadrant, we are left with maximizing 4xy=(4/sqrt(5))cos(theta)sin(theta)=(2/sqrt(5))sin(2theta) for theta between 0 and Pi/2. This can be solved almost "by inspection": just take theta to be Pi/4. The max value is then 4/sqrt(5). Of course, in a "random" situation it may be very difficult to get nice parameterizations.
   Now we move from the prosaic
      1. like prose, lacking poetic beauty.
      2. unromantic; dull; commonplace ("took a prosaic view of life").
   to the more recondite
      1. (of a subject or knowledge) abstruse; out of the way; little known.
      2. (of an author or style) dealing in abstruse knowledge or allusions; obscure.
   
3. A weird way to do the problem
   

   I had Maple sketch some level curves of 4xy, the objective function, and compare them with the constraint curve x2+5y2=1. Here is the result of these Maple commands. A:=contourplot(x*y,x=-1.1..1.1,y=-1.1..1.1,color=red, thickness=2,scaling=constrained,grid=[50,50], contours=[.02,.05,.08,.2,.3,.5,-.02,-.05,-.08,-.2,-.3,-.5]): B:=implicitplot(x^2+5*y^2=1, x=-4..4,y=-4..4,color=blue, thickness=2, scaling=constrained, grid=[80,80]): display({A,B}); The picture is shown to the right.
   A close-up view Suppose you consider a level curve of the objective function that crosses the constraint curve, as shown. One math word which applies to this situation is that the two curves are transversal. So we have 4xy=C crossing x2+5y2=1. What happens if we "wiggle" C a little bit, so we consider 4xy=C+epsilon and 4xy=C-epsilon. Now it seems reasonable (4xy is certainly continuous, so its values don't hop around or break or anything) that these level curves are close to 4xy=C. These level curves must also cross the constraint curve. That means the function 4xy has values C+epsilon and C-epsilon on the constraint curve. (The level curves are exactly where that function takes on its values!) Since there are both larger and smaller values of 4xy on the constraint curve, C can't be an extreme value (either max or min) for 4xy on x2+5y2=1.	Local picture near a level curve corresponding to a non-extreme value
   Another close-up view This seems to imply, if you examine the picture closely, that the largest (and the smallest) values of 4xy will be at points on the ellipse where the ellipse will be tangent to level curves of the constraint, x2+5y2=1. If the level curves of the objective function are not tangent, then we will be able to vary the values of the constant generating that contour and get bigger and smaller values of the objective function on the constraint curve. If the level curves are tangent then the normal vectors of the constraint curve ( f at that point) and the objective function ( g) at that point will both be perpendicular to the same line (in three dimensions it would be a tangent plane). These gradient vectors may not be exactly the same vector, but one of them must be a scalar multiple of the other.	Local picture near a level curve corresponding to an extreme value

   Now the algebraic side
   If f(x,y)=4xy and g(x,y)=x²+5y²=1, then at such points (extreme values of the objective function on the constraint curve), there is some real number so that g= f (everyone uses this Greek letter) because the tangent lines are the same, and therefore the normal vectors must be parallel: one must be a scalar multiple of the other. This one vector equation in R² gives two scalar equations, one for each component of the vectors:

    2x=()4y
   10y=()4x

   This, together with the constraint equation g(x,y)=x²+5y²=1 gives a system of 3 equations in 3 unknowns. We can solve this by, for example, solving for in each of the first two equations and setting them equal. We need to watch out for spurious solutions or evasions of solutions. These may occur when we divide by certain variables. This gave us another way to solve the maximization problem, a method which is more in the spirit of several variable calculus. It turns out that this strange idea is actually quite useful in "real world" problems. The method is called Lagrange multipliers What kind of pliers? and is discussed in section 14.8 of the text. The method is used extensively in economics and in many areas of engineering.

34 of the 140 students who took the exam did not come to the next class to pick up their graded exams. I take the course seriously. Therefore I flew across the continent Saturday night to read your papers and return exams as soon as possible. I tried to grade your work carefully. Please: you should take your education seriously also. Unsurprisingly, the average grade of the exams which were not picked up was substantially lower than the overall course average.

HOMEWORK
Please hand in 14.7: 6, 12 tomorrow in recitation.
Be prepared to discuss problems in 14.7 and to go on in section 14.8.
You will have a quiz.

Max/min in several variables
I remarked that that finding maximums and minimums (usually called together, "extreme values") can be both theoretically and computationally very difficult. In 251, we "scratch the surface". Some straightforward definitions are given, and a few techniques are explored. Mostly we will look at functions of two variables, sometimes in three, and rarely in more variables. Even in two variables, as I hope to show you, things can be very different from one variable and can be computationally very complicated.

Review of 1 variable
I try not to work hard, so I thought maybe a quick review of extreme value material from 1 variable calculus would be useful. The names of ideas to recall include these:
maximum, minimum, absolute maximum, absolute minimum, local maximum, local minimum.

Fermat's fact
What I called "Fermat's fact" was the following wonderful observation in one-variable calculus:
If f is differentiable at x₀ and if f´(x₀) is not 0, then f does not have an extreme value at x₀.

The picture shows a "proof" (well, I hope fairly convincing to a picture person). If there is a tilt in the tangent line, then there are both higher and lower values near x₀. If x₀ is either kind of extreme value (max/min), then we see that f´(x₀) cannot be 0.

Critical number
Therefore the following definition is written.
x₀ is a critical number of the function f if either f is not differentiable at x₀ or f´(x₀)=0.
For simplicity in this discussion I'll assume that f is defined in some interval that has x₀ inside it (in the interior).

Consequence
Here is the result which has made calculus famous and earned big bucks:
If f has an extreme value at x₀ then x₀ is a critical number of f.
So you can search for extreme values by looking first for critical numbers.

Conversely
I asked if the converse of the preceding statement was true. Maybe you should have some acquaintance with logic words. Here I'm asking whether x₀ being a critical number means that the function f must have an extreme value at x₀. The answer is certainly "No." I hope you find the pictures will be persuasive and/or help your memory.

By the way, the type of functions illustrated in the first and third pictures are called piecewise linear. In lots of variables, finding extreme values of these functions is rather difficult. The problems are wildly varied, and can deal with topics like scheduling and allocation. You could begin with an undergrad course to study the math of these problems (Linear Optimization) and then look into various aspects of Operations Research or Industrial Engineering.

Identifying ("classifying") the type of critical point
Well, suppose you "have" a critical number. How can we discover if the function has extreme (max or min) behavior there? I'll make a very restrictive assumption, which is that the functions we'll consider will always be differentiable, so corners and jumps and other very poor local behavior won't occur. Then there are various observations which can be made, and some will guarantee a (local) max or a (local) min.

The 68^th derivative test
I stated the following result and remarked that almost surely this was in everyone's calc 1 course:
Theorem Suppose that f´(x₀)=0 and f´´(x₀)=0 and f´´´(x₀)=0 and f⁽⁴⁾(x₀)=0 and ... and f⁽⁶⁷⁾(x₀)=0. (The first 67 derivatives of f at x₀ are all 0).
Then
If f⁽⁶⁸⁾(x₀)>0, f must have a local max at x₀.
If f⁽⁶⁸⁾(x₀)<0, f must have a local min at x₀.
If f⁽⁶⁸⁾(x₀)=0, the 68^th derivative text supplies no information.

Huh?
Well, I don't know many calc 1 courses which do teach the "68^th derivative test". Maybe this is because the "test" is silly, or maybe because the best way to understand why it is true is to use Taylor's Theorem. I had mentioned way near the beginning of the course that Taylor's Theorem is a calc 2 result which turns out to be very useful. Here if we use Taylor's Theorem centered at x₀ up to degree 67, we would see something like:
f(x)=f(x₀)+lots of 0's+[f⁽⁶⁸⁾(near x₀)/68!](x-x₀)⁽⁶⁸⁾.
Since 68 is even, if the sign of the 68^th derivative is positive, f will have the predicted local min (the graph will look like a really narrow parabola locally!). And a negative sign gives the other implication.

I mentioned this absurd "test" so that when you see what happens in, say, 2 variables, you won't be as horrified or startled as you might be. The whole process is to decide what algebraic conditions on derivatives at a critical point can guarantee extreme behavior. The second derivative test in 1 variable is just the tip of the iceberg (the snout of the shark?).

Now several variables
Today's word: morass. A neat word, which I didn't make up but which I always laughed at early in life. It means:

1. an entanglement; a disordered situation, esp. one impeding progress.
2. [literary] a bog or marsh.

The simple pictures with simple formulas
In class I presented the pictures initially and then the formulas. I will do both together here.

Discussion and formulas	The pictures
Min A function defined on all of R2 with a local (and absolute) minimum is f(x,y)=x2+y2. The graph of this function is a surface called a paraboloid. It is a nice, smooth "cup" opening up. Vertical slices through (0,0) are all parabolas opening up and the contour lines are circles. The red dot is the critical point and the brown plane is the tangent plane at that point (the xy-plane).
Min The simplest local and absolute strict maximum is, of course, just the reflection of the previous example, done with minus signs algebraically. So here f(x,y)=-x2-y2, and (0,0) provides a strict maximum. The graph is a paraboloid whose axis of symmetry is again the z-axis. This graph opens "down".
A saddle The function f(x,y)=-x2+y2 gives a nice example of a saddle point. The xz-slice (where y=0) shows the curve z=-x2 and the yz-slice (where x=0) shows z=y2. Each has a (strict) extreme point at 0. One is a max and one is a min. Such behavior is called a saddle point. Perhaps the behavior most similar in one variable calculus would be that of the function x3 (an inflection point). But in 2 and more variables the local situation can be much more complicated. Here the surface is more complicated, and my picture is certainly not so good. But the tangent plane and critical point are the same. The tangent plane cuts through the surface (similar to the way a tangent line at an inflection point in 1 variable calculus cuts through the graph of a curve).

Ridiculous interesting (?) fact
The surface z=-x²+y² is what's called a ruled surface. The word 'rule" comes from an older version of English, and here means "straight line". Every point on the surface is on a straight line which sits entirely in the surface. I didn't believe this when I was told it for the first time, and maybe you don't either. For example, the point (1,2,3) is on this surface (since 3=-1²+2²) and the line whose parametric equation is x=1+t, y=2-t, and z=3-6t goes through that point and sits entirely on the surface. Such surfaces have applications in computer graphics.
You can verify algebraically that the line whose parametric equations gare given above is on the surface, or you can look at the picture to the right, which shows the saddle together with the line: nearly silly.

Definition of critical point
Suppose f is a function of n variables. Then f has a critical point at p in Rⁿ if either f doesn't exist at p (so at least one of the first partial derivatives fails to exist) or f(p)=0 (the zero vector, remember!).
In this course, almost all the functions we'll consider will be differentiable. This doesn't mean that non-differentiable functions )functions with jumps or corners) are not important or interesting in mathematics and its applications (again: linear optimization, shock waves in physical phenomena). Just learning to use the tools for higher dimensional analysis of differentiable functions is a big enough task.
Suppose z=f(x,y), and f is differentiable. What is the geometric meaning of "(x₀,y₀) is a critical point of f"? Since f(x₀,y₀)=0, both of the first partial derivatives are 0. Therefore z=f(x₀,y₀) (that is, z=a constant) is the tangent plane to z=f(x,y) at the point (x₀,y₀,f(x₀,y₀)). The "flat" plane through the point, parallel to the xy-coordinate plane, is tangent to the surface. This can be difficult to "see" in a graph, though.

Using Fermat's fact here
If a point is a local extreme point of some function f in several variables, and if that function is differentiable at that point, then all of the first partial derivatives of the function must be 0 at that point. If that's not true, just "slice" the function at that point in the direction of the derivative which is not 0. The one variable Fermat fact implies that the function does not have an extreme value (max or min) at the point in one variable, and therefore the function in several variables has both higher and lower values near the point. Therefore (whew!):
An extreme point must be a critical point.
Our functions will almost always be differentiable, so our functions will have their extreme values where f=0.

Monkey saddle
The examples already shown are the standard critical points for functions of two variables. But there are many, many other kinds of critical points. The graph z=x³-3xy² shows one of them. Again the origin, (0,0), is the only critical point, and the xy-plane is the tangent plane at the origin. This critical point's local behavior is up/down repeated three times (at equally spaced 120^o angular intervals) if you walk around the surface in a small circle centered at the origin. The critical point is called a monkey saddle because, presumably, a monkey could sit on it with spaces for two legs and a tail to hang down.
Critical points of more than one variable can have many, many different local pictures, and there has been a great deal of effort expended trying to understand them.

Two book problems
There are two amazing and disconcerting problems in section 14.7. At least, to me these problems are both amazing ("surprise greatly; overwhelm with wonder" -- well, at least the first) and disconcerting ("disturb the composure of; agitate; fluster" -- certainly they show me I don't understand too well what can happen in "space").

Problem #35, section 14.7
The function f(x,y)=-(x²-1)²-(x²y-x-1)² is given. This is not the world's most horrible function. It is "only" a polynomial of degree 6. First, the text asserts that this function has two critical points. We can check that easily:
Hey! I was going to do this "by hand" when I realized that I could have a friend (?) do it. So here:

> f:=-(x^2-1)^2-(x^2*y-x-1)^2;
                               2     2     2           2
                       f := -(x  - 1)  - (x  y - x - 1)
> solve({diff(f,x),diff(f,y)});
                        {x = 1, y = 2}, {x = -1, y = 0}

Problem #36, section 14.7
Here f(x,y)=3xe^y-x³-e^3y. I think even I showed that this has one critical point. My friend replies:

> f:=3*x*exp(y)-x^3-exp(3*y);
                                           3
                        f := 3 x exp(y) - x  - exp(3 y)

> solve({diff(f,x),diff(f,y)});
                               2                                 2
 {x = 1, y = 0}, {x = RootOf(_Z  + _Z + 1), y = ln(-1 - RootOf(_Z  + _Z + 1))}

The left graph below is a local picture of the critical point. This seems to convincingly support the textbook's assertion that (1,0) is a local strict maximum of the function. (We can verify this assertion with the second derivative test to be stated later.) In the graph on the right, x goes from -5 to 5 and y varies just between -.05 and .05: therefore y is just about 0, and 3xe^y-x³-e^3y is just about 3x-x³-1. Certainly this shows that the function has no absolute max or min.
The situation in 1 variable calculus is considerably different.
Suppose we have a function defined on all of the real numbers (o.k., a differentiable which has exactly one critical point and that critical point is a local maximum. Then that local maximum is a global, absolute maximum for the function. What the heck is happening in several variables? I just don't understand, really understand.

Now a second derivative test in two variables
There's one second derivative test which is usually "given" to students in a third semester calculus course. It is a bit complicated. The test essentially results from computing the second directional derivative at the critical point and seeing how to ensure that this result is always positive (or always negative or ...). That together with results from one variable calculus (on concavity) will insure some kinds of local behavior near the critical point.

and the chain rule more

Now the transition to something

The statement of the second derivative test

Hessian: http://en.wikipedia.org/wiki/Hessian_matrix QotD

(Also an animal, about which is written: "... Sloths move
only when necessary and then very slowly: they have about
half as much muscle tissue as other animals of similar weight.")

Today's lecture discusses the most important single application of the chain rule, and maybe the most important single idea in vector differential calculus. I'll introduce it with a silly but still almost reasonable "story".

The spaceship in a nebula
My online dictionary states that a nebula is "a cloud of gas and dust, sometimes glowing and sometimes appearing as a dark silhouette against other glowing matter." So we could pilot a spaceship through a nebula. We might be concerned about the physical effects of the nebula, for example, the temperature. I'll assume that the spaceship measures temperature at the tip of its front. A point in the nebula will be located with rectangular coordinates, (x,y,z). The temperature at that point will be T(x,y,z). The rocket will fly a path so that at time t its location will be <x(t),y(t),z(t)>.
From this we can see that the temperature measured at the rocket at time t is T(t)=T(x(t),y(t),z(t)), and this is a composition. First we find out where the spaceship is at time t, and then we compute the temperature at that point.

Computing dT/dt
Well, the chain rule applies, so

dT   T  dx    T  dy   T  dz
-- = --- -- +  --- -- + --- -- 
dt   x  dt    y  dt   z  dt

Recognition
Now the luck and glory is recognizing that the mess on the righthand side is a dot product. In fact, look:

dT    / T   T   T \    / dx   dy   dz \
-- =   --- , --- , ---  ·   -- , -- , --
dt    \ x   x   x /    \ dt   dt   dt /  

Right
The vector on the right-hand side is one we've looked at when discussing curves. It is the derivative of r(t), the position vector, so it is v(t), the velocity vector. This vector deals with the spaceship and its motion.

Left
This vector seems to be "new": it is the vector of all the first partial derivatives of T in order. This is called the gradient of T and is frequently written T. The upside-down triangle (or upside down ) is sometimes called "del". This vector can be computed only from the nebula information.

So we have separated ("decoupled", where one definition of "decouple" is "disconnect or separate") the nebula and the spaceship. The nebula information is T and the spaceship information is v(t).

Now I tried to make a sequence of observations which might help people understand the actual excitement I feel thinking about gradient.

Observation 1
Let's imagine two spaceship trips through the nebula. Now these trips (voyages?) may be completely different except that at the time the two spaceships pass through the point (x,y,z), the spaceships have the same velocity vectors: that is, the spaceships are heading in the same direction and at the same speed. Their v(t)'s are the same. Then the rate of change of the temperature, dT/dt, that the two spaceships measure is exactly the same.

I asked students if they could deduce this from the physical and geometric aspects of the "scenario". I don't think I can. As a math fact goes, this is nearly obvious: since the v(t)'s are the same, the right-hand side doesn't change, and the nebula's temperature function is the same, so the left-hand vector (T) doesn't change. Therefore the dot product, which computes dT/dt, is the same. But ... but ... what the heck ... can you "see" this physically? This is not the temperature at the point, but the rate of change of the temperature: the rate of change is the same if the velocity vectors are the same.

Observation 2
Now r´(t)=v(t), the velocity vector. It is the same as (ds/dt)T(t) where ds/dt is the speed and T(t) is the unit tangent vector. In the formula T·r´(t) the ds/dt effect just "filters out" of the dot product. If you travel twice as fast on the same path, then the rate of change of the temperature with respect to time is just doubled. So this is easy to understand. But, as several students in both lectures observed, the more subtle aspect is what happens as the direction changes.

Observation 3
Here I will suppose that ds/dt=1 for simplicity. Also in order to keep my notation sane (how many uses of T can I have in the same problem?) I will replace the unit tangent vector by u, for unit vector. Then what can we say about T·r´(t)? It is (ds/dt)T·u, or just (since I'm assuming unit speed) T·u. But, hey, the dot product is also |T| |u|cos(theta). Since cos(theta) is between -1 and +1, I now know that dT/dt is between -|T| and +|T|.

How could we choose u so that dT/dt is largest? We need to make cos(theta) equal to +1. Therefore we need theta to be 0, and u should be a unit vector in the direction of T. That is, choose u to be T/|T|. To make the rate of change as much negative as possible, choose u to be -T/|T|, and then dT/dt will be -|T|.

An example (?)
Students' patience for abstraction began to wear out. So I computed an example. If T(x,y,z)=x²e^yz-5z3 then since T=<T/x,T/y,T/z>, we compute:
T=<2xe^yz-5z3,x²e^yz-5z3(z),x²e^yz-5z3(y-15z²)>
As far as I know this function and this computation has no great or special "meaning".

A better example (!)
Im my kitchen I have just finished backing my famous chocolate brownie pie and I left the oven door slightly open. Also I managed to forget to close the refrigerator. As a result, the contour lines of temperature could like what is shown to the right. In what direction should I go (I am the little green man in the picture!) to most rapidly increase the temperature? In the direction of the gradient, which will point towards the oven. I will most rapidly decrease the temperature by traveling in the opposite direction, towards the source of the cold.

Observation 4
I could imagine that spaceship travels through the nebula on an isothermal surface. An isothermal is a collection of points where the temperature is all the same. We have seen this already: T(x,y,z)=C is a level surface (dimension 3) or level curve (dimension 2) or a contour {surface|curve}. But if the spaceship travels on such a surface, then the rate of change of the temperature must be 0. But then T·v=0. This means that the velocity vector is perpendicular to the gradient. But then in turn this means that the gradient vector is perpendicular to the level surface, and it is perpendicular to the tangent plane of the level surface. In the kitchen, I would walk perpendicular to the contour lines to increase or decrease temperature most rapidly. I would walk along the contour lines if I wanted no rate of change of temperature.

Back to the example
Let me look more closely at the example with T(x,y,z)=x²e^yz-5z3 when x=3 and y=2 and z=1. Well, T(3,2,1)=9e^-3. And T=<2xe^yz-5z3,x²e^yz-5z3(z),x²e^yz-5z3(y-15z²)> becomes T(3,2,1)=<6e^-3,9e^-3(1),9e^-3(-13)>=<6e^-3,9e^-3(z),-117e^-3>

Now forget all that, and solve the following geometric problem:
What is the equation of a line tangent to the surface x²e^yz-5z3=9e^-3 at the point (3,2,1)?
This could be, indeed, I claim, this is a hard problem. But if we now disobey my urging ("forget all that") I can tell you that T(3,2,1) is perpendicular to the surface and to its tangent plane at (3,2,1). So I can write the answer, since I know a point and a normal vector to the plane requested:
6e^-3(x-3)+9e^-3(y-2)+-117e^-3(z-1)=0.
I think that solving such a problem so efficiently is really remarkable.

Topographic maps
A topographic map shows contour lines. Frequently while hiking people mind want to find the most direct route to the "top" (a mountain peak) or to the "bottom" (a creek?). They know by experience that the most direct route, only looking at the map, that is, only the geometry of the situation, would be to walk as nearly as possibly perpendicular to the contour lines.
This can be adapted into computational strategies for finding maxes and mins. If you can readily compute your function's gradient, then find maximums by going in the direction of the gradient. This is hill climbing. Find minimums by going opposite the direction of the gradient. This is the method of steepest descent. Of course these computational ideas don't always work, and there are a great deal of implementational matters to worry about, but the general strategy is valuable.

Ellipsoid
Here's a neater example. Consider the ellipsoid (egg) x²+2y²+3z²=9. The point (2,1,1) is on this ellipsoid. What is the equation of a plane tangent to the ellipsoid at (2,1,1)? Well, the gradient of the function x²+2y²+3z² is <2x,4y,6z> and at (2,1,1) this is <4,4,6>. The equation of the tangent plane is 4(x-2)+4(y-1)+6(z-1)=0.

To the right is a Maple picture made by the commands which follow. I hope that the picture helps to convince you that the plane is the tangent plane. A:=implicitplot3d(x^2+2*y^2+3*z^2=9,x=-5..5,y=-5..5,z=-5..5,grid=[20,20,20], axes=normal,labels=[x,y,z],color=green,style=hidden); B:=implicitplot3d(4*(x-2)_4*(y-1)+6*(z-1)=0,x=-5..5,y=-5..5,z=-5..5,axes=normal, labels=[x,y,z],color=green,style=hidden); display({A,B};

Directional derivative
If u is a unit vector, then the directional derivative of T at (x,y,z) in the direction u is the rate of change of T at unit speed in the direction u (at the point). The textbook's notation for this is D_uT(x,y,z) and the preceding discussion should convince you that the directional derivative's value is T(x,y,z)·u.
more notation, more words ... this is so terrific!!! (so academic)

QotD
I asked people to look at a hyperboloid, a surface given by the equation x²+y²-z²=12. The point (3,2,1) is on this surface. Since x²+y²=12+z² I know that slices with z fixed are circles. The slice on the xy,plane is a circle with center the origin and radius sqrt(12). As z increases, the slice is still a circle with center (0,0) (on the z-axis) but the radius is increasing and is sqrt(12+z²). The picture to the right is supposed to show some representative circles and "suggest" part of the surface to you.
The Maple command
implicitplot3d(x^2+y^2-z^2=12,x=-5..5,y=-5..5,z=-8..8,grid=[20,20,20], axes=normal,labels=[x,y,z],color=green,style=hidden);
will get you a nice picture which you can rotate and examine.

I asked for the equation of a plane tangent to this hyperboloid at (3,2,1) and also for the three parametric equations of a normal line to the hyperboloid at (3,2,1). I tried to emphasize that almost no computation would be needed. Since the gradient of x^2+y^2-z^2 is <2x,2y,-2z> we can just evaluate this at (3,2,1) and get <6,4,-2> and this vector is normal to the hyperboloid at (3,2,1). Here are the answers:
The tangent plane: 6(x-3)+4(y-2)-2(z-1)=0
The normal line:
  x=3+6t
  y=2+4t
  z=1-2t

HOMEWORK

1. Do the workshop problem.
2. Look at the formula sheet and the review problems. Please send in solutions! Sincere statement: I would very much like students to do well on the exam, and for that to occur, students should practice. Most of the exam will cover material which has been emphasized in lectures and in homework.
3. Do problems in 14.5 and 14.6. The content is quite important.
4. There are no textbook problems due in recitation Wednesday (only the workshop problem!).
5. You may try the spade suit problem from the handout. The answer is available here.

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 the 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.

But what happens?
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
Stuff₀x⁰+Stuff₁x¹+Stuff₂x²+Stuff₃x³
where each of the "Stuff" terms is some function of y alone. 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.

The handout
Then with the help of a student volunteer (frowning at a student creates a volunteer!), each student received a copy of data about some functions.

Do we understand the handout?
I remarked that I used unusual variable names and some new notation. The first table on the handout. f and g are declared to be differentiable functions of two variables.

  M     N  f(M,N) D1f(M,N) D2f(M,N)  g(M,N) D1g(M,N) D2g(M,N)  
 -1    -2     6      4        0         3      8       1
 -1     2     2     -2        1        -5      7       6  
  1    -1    -2     -5        4        -2      9       4 
  1     2     5     -7        6        -1     -2       7
  2     1     0     -1       -2        -3      7       4

The second table on the handout referred to values to two differentiable functions of one variable.

  V  h(V)  h´(v)  k(V)  k´(v)
 -2   5     2      3     5
  0   0     2     -2     7
  1   1     3      2    -1
  2  -1     4      4    -2

Doing the problems
A sequence of four problems were given. Let's try them.

Club suit
If S(t)=h(k(t)), compute S(1) and S´(1).
This is one variable calculus. But let me try to think about it a bit first:
Thinking about it
I can think about the function S as a sort of box, which takes inputs and processes them in some fashion, and produces outputs. The S box also has some internal structure. The input first is sent to box representing the k function, and then the output from that (sub?)box is sent to the h function. If we follow through (using values from the second table) we can see that 1 "changes" to 2 and then to -1.

What about the derivative? The derivative is a multiplier of a tiny change in the input. It signals the first order change in the output compared to the input. In the case of S, if we "kick" the input by c (think of c as a small number) then 1+c is fed into the k box. The output will be approximately (neglecting H.O.T., higher order terms) 2 (the old output) +k´(1)c, which is 2+(-1)c. Now feed in 2+(-1)c. If c is small, (-1)c will be small. The output from the h box will be -1 (the old output, what h "does" to 2, plus a change. The first order part of the change will be a proportionality constant, h´(2), multiplying the kick that is passed to the h box. The kick passed to the h box is (-1)c, so the compounded effect is that h's new output (approximately, first order) is -1 (the old value of h's output) plus h´(2)(-1)c=4(-1)c.

Now go "up" a logical level. The input, 1, to S was kicked to 1+c. The S output, to first order, is -1 (the old output) plus 4(-1)c. Therefore the derivative of the S box at 1 is 4(-1), since the derivative is the multiplier of the kick.

The diagram below is supposed to be visual "support" of the preceding discussion.

A formula
If S(t)=h(k(t)), the one variable chain rule states that S´(t)=h´(k(t))k´(t), so S´(1)=h´(k(1))k´(1)=4(-1). Formulas are good!

Diamond suit
If W(t)=f(h(t),k(t)), compute W(1) and W´(1).
Thinking about it
Now the W box has a different structure. The input is split (bifurcated -- what's the point of being in an academic environment if a silly, uncommon word isn't used in place of one that would be understood!) into two, and each is fed separately into h and k boxes. The outputs, now in order, are carefully put into f. The output from the f box is then pushed outside as the value of W.
To compute W(1), we find h(1)=1 and k(1)=2, and then compute f(1,2)=5. Easy (?)..

What about the derivative? Suppose we kick 1 to 1+c. The response of the one variable boxes, h and k, should not be difficult to understand: the outputs, linearized, are 2+k´(1)c=2+(-1)c and 1+h´(1)c=-1+3c, respectively. It is important to remember which output is which! Now feed this into f. The multiplier for perturbations in the first variable is D₁f(2,1), so the effect of the change in the first variable adds on D₁f(2,1)(-1)c to the output. The second variable contributes in proportion to its perturbation, with the constant of proportionality being D₂f(1,2), so D₂f(1,2)(3)c gets added on. If we look up the numbers and do arithmetic, we can see that the total (linearized) effect (neglecting higher order errors!) is 5 (the old output) plus 25c. Therefore the output of the W box seems to indicate that the derivative is 25.

A formula
O.k., if W(t)=f(h(t),k(t)), we will label the variables in f: the first variable is x and the second variable is y. Then we follow through the changes and use the chain rule:
W´(t)=(f/x)h´(t)+(f/y)k´(t).
This is a fine result, but if we need to evaluate it, we'd better remember that
W´(t)=(f/x)(h(t),k(t))h´(t)+(f/y)(h(t),k(t))k´(t).
and now you should see the numbers that appeared above.

Heart suit
If Q(x,y)=f(h(x),g(x,y)), compute Q(1,2) and D/x(1,2) and D/y(1,2).
Thinking about it
Certainly Q(1,2)=f(h(1),g(1,2))=f(1,-1)=-2: easy enough. Now to get the derivative with respect to "x", the first variable, let's perturb or kick 1 to 1+c. The effect filters through h as (linearized!) 1+h´(1)c which is 1+3c. If we kick 1 in g but hold the second variable constant at 2, then the output, to first order, is -1 (the old output) plus D₁g(1,2)c. This is -1+(-1)c.

Now the input to f is, in order, 1+3c and -1+(-1)c. There are changes to both variables. So we need to use a linear approximation in both variables. The output from f (which is what is reported as the output from Q) will be -2+D₁f(1,-1)(3)c+Df(1,-1)(-1)c=-2+(-5)(3)c+4(-1)c=-2+(-19)c. Therefore the proportionality factor is -19, and this is the requested Q/x(1,2).

A formula
So if Q(x,y)=f(h(x),g(x,y)), then I think that Q/x=(f/x)h´(x)+(f/y)(g/x). There's still the question about how to get values, and in fact, in more detail, this chain rule reads:
Q/x(x,y)=(f/x)(h(x),g(x,y))h´(x)+(f/y)(h(x),g(x,y))(g/x)(x,y).

If we want the y derivative, then we could compute this: Q/y(x,y)=0h´(x)+(f/y)(h(x),g(x,y))(g/y)(x,y). The 0 is there because there is no y involvement in the first variable of Q. Now insert x=1 and y=2, and read off from the information given that the value is 4·7=28.

This is all horrible. I will admit to you that I usually try to use formulas and only rarely (like most other human beings) try thinking. But sometimes thinking is needed. For example, those who like formulas might contemplate this task:
What is the partial derivative with respect to x of f(h(x),g(y,x))? Notice that I "swapped" the variables in g. I think the partial derivative with respect to x will be 0h´(x)+(f/y)(h(x),g(y,x))(g/x)(y,x). Some people might find this notation very objectionable: look, two y derivatives multiplied are one x derivative!

ODE's
I solved an ordinary differential equation:
Find all solutions of d²y/dx²=x⁵
This is certainly rather simple, and just two antidifferentiations gets the following answer: y=(1/42)x⁷+Cx+D where C and D are any real constants. So the "family" or collection of solutions of this ODE is a two-(real)parameter family of functions.

A simple PDE
Here I sort of went backwards (in fact, the going forwards is much more difficult!). I started with a differentiable function of one variable, which I called f. Then I looked at this function of two variables, F(x,y)=f(x²+y³) (slightly different from what I did in class -- I think this is better because there isn't symmetry in x and y which maybe is distracting). Then
F/x=f´(x²+y³)2x and F/y=f´(x²+y³)3y²
from chain rule arguments similar to what we just did. But then:
(3y²F/x=(3y²)f´(x²+y³)2x and (2x)F/y=(2x)f´(x²+y³)3y²
so "clearly" F(x,y)=f(x²+y³) is a solution of the partial differential equation
(3y²F/x-(2x)F/y=0.
So we now have a family of solutions of this PDE. The solutions include sin(x²+y³) and e^x2+y3 and tanh(x²+y³) (well, no one asked, but that's hyperbolic tangent which is interesting in some fluid flow problems). PDE's may have lots and lots of solutions. As I mentioned above, what I've just done is a classroom example. So if you came up to me with a PDE modeling some "real" phenomenon, I may not be able to go backwards too easily and get a family of solutions described so simply.

Implicit functions, two dimensions
Here's another application of the several variable chain rule. Again, return to a 1 variable calculus situation:
Suppose F(x,y) is a differentiable function of 2 variables, and the equation F(x,y)=0 defines y implicitly as a function of x. What is dy/dx in terms of F and "things" related to F?
So take the equation F(x,y)=0 and d/dx this equation. The right-hand side is 0, and the left gives you:
F/x(dx/dx)+F/y(dy/dx) by the chain rule.
Certainly dx/dx is 1, and dy/dx is what we want, so we can "solve" for it in the equation F/x+F/y(dy/dx)=0. This means:
A formula!

dy     F/x
-- = - ------
dx     F/y

Example
I think an example is needed here before we go on. Let's look at:
Calc 1 problem: find dy/dx if y³-7xy²+4x⁵-6=0.

Calc 1 solution to Calc 1 problem We d/dx everything, being careful to remember that y=y(x) mysteriously. Then:
3y²y´(x)-7y²-(7x)2yy´(x)+20x⁴=0, and now we solve for y´(x). We get:
y´(x)(3y²-(7x)2)-7y²+20x⁴=0 so that y´(x)=-(-7y²+20x⁴)/(3y²-(7x)2).

New technology (?) solution to Calc 1 problem We will use the formula above. Here F(x,y)=y³-7xy²+4x⁵-6 so that
F/x=-7y²+20x⁴ and F/y=3y²-(7x)2y+0 and the formula gives
dy/dx=-(F/x)/(F/y)=-(-7y²+20x⁴)/(3y²-(7x)2y+0) which is of course the same answer! And you can look at see the same pieces occurring, so the world is not so crazy.

The darn formula, though, is a bit mysterious. If you try to understand the form (?) of the formula, the x and y might seem in the wrong place and there might be an extra minus sign ... and ... and ... the notation is terrible!

P and V and T
I asked if people knew about gas laws. For a gas, there are the quantities P (pressure) and V (volume) and T (temperature). A gas law might be a function of three variables which relates these quantities:
G(V,P,T)=0.
If we assume that the function is differentiable and that each one of the quantities is implicitly defined as a function of the other two by the function, something funny happens. Let me show you.

Suppose that G(V,P,T)=0 implicitly defines V as a function of P and T. Let's compute V/P. Here T is constant, and sometimes in thermodynamics the quantity is called (V/P)_T just to remind people that T is constant. We will /V the equation G(V,P,T)=0.
I use the chain rule, and the result is:
(G/V)(V/P)+(G/P)(P/P)+(G/T)(T/P)=0.
But P/P must be 1 (the derivative of something with respect to itself) and T/P must be 0 (because T is constant!). Therefore we can solve for P/V just as we got dy/dx before and get:
V/P=-(G/P)/(G/V).

So far so good. But in fact we can find other partials in a similar way:
P/T=-(G/T)/(G/P)
T/V=-(G/V)/(G/V). Now clearly (NOT AT ALL CLEARLY!):
(V/P)_T(P/T)_V(T/V)_P=-1
because when we multiply all these expressions together the fractions all cancel and we are left with -1. Why is this true physically and what does it mean? Take physical chemistry, take thermo, etc., and find out (and maybe report back some time to me, please).

QotD (should have been!)
I wanted to ask people to compute the Spade suit problem. If you would like to try it on your own, here is the answer.

Formulas, reviews, etc. for the first exam
There's an exam a week from Friday. Please look at the formula sheet and let me know about errors. Please look at the review problems, and contribute generously to the welfare of Math 251 by sending me a solution to the problem whose initial is your last name's initial, if no solution is yet shown.

HOMEWORK
Sane human beings would do the assigned workshop problem, work on textbook problems, look at the review material and formula sheet, and maybe even glance at the next section (the last section to be tested on the exam).

Nuclear magnetic resonance (NMR)
NMR is a way of investigating the structure of molecules, relying on the spin of protons and how this spin changes in high magnetic and electric fields. Several "antiques" (objects 10 years old!) were exhibited in class. These were contour plots, outputs of NMR experiments. The lecturer believes students should be aware that real data can be much more complicated than the examples likely to be seen in this course!

Limits, 1 dimension
Here we've got a function of one variable, and we want to define and understand lim_x-->af(x)=L. The actual definition, frequently stated but rarely stressed in calc 1 classes, is the following: (and, yes, the Greek letters epsilon and are almost always used)

Given any epsilon>0, there is some >0 so that if 0<|x-a|<, then |f(x)-L|<epsilon.

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 epsilon is an output tolerance. We'd like our outputs to be within epsilon of the ideal output (for this problem) L. Then the limit definition states that there is some input tolerance, , which when applied to stuff going into the machine (only allowing inputs within ) 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 when actually given an epsilon. 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 last time 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 epsilon>0, there is some >0 so that if 0<|<x,y>--><a,b>|<, then |f(x,y)-L|<epsilon.

Again the epsilon and are 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. Similarly, we could slider 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.

Aiming for first-order understanding
We've already seen that 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 y, and, finally, a higher-order error term. Let's see another example.

An example
Here is an example related to one I gave last time. Let's look at f(x,y)=xy/sqrt(x²+y²). Again, things are "better" (at least for me!) in polar coordinates. Then the bottom becomes just r (much better!) and the top is r cos(theta) r sin(theta). The function is f(x,y)=[r cos(theta) r sin(theta)]/r=r cos(theta) sin(theta). Certainly on the x and y axes f(x,y) must be 0, because there theta is 0 or Pi (and the sine factor is 0) or Pi/2 or 3Pi/2 (and the cosine factor is 0). The linearization formula at (0,0) is
f(0+h,0+k)=f(0,0)+f/x(0,0)h+f/y(0,0)+higher order error.
Surely r cos(theta) sin(theta) is 0 when r=0, so f(0,0)=0. And since f is 0 on all of the x and y axes, f/x(0,0)=0 (the slice is a horizontal line) and f/y(0,0)=0 (this slice is also a horizontal line). So all of the linearization (the constant term and the first order stuff) is 0.

But ...
Look at f(0+w,0+w). In the original formula, this is (0+w)(0+w)/sqrt((0+w)²+(0+w)²) and the result, after some algebra, is (1/sqrt(2))w. This is first order. But the linearization is 0. So something is seriously wrong with this function. To the right is a picture of the graph of this function (at least the part which is over the first quadrant). Again, I used a stalwart (stalwart means "1. strongly built, sturdy. 2. courageous, resolute, determined") student volunteer to help. I used the bungee cords again (is their cost tax-deductible). The process was similar but not identical to the demonstration last time. The calculus instructor raised and lowered the cord twice as he walked around the student. The student this time was asked to keep the end of the cord always at the same level, not changing at all. This represented height 0.

Differentiability in 2 dimensions
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, 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).

The function f(x,y)=xy/sqrt(x²+y²) has very lousy (sorry: discontinuous) partial derivatives at (0,0). It is tedious to check this by hand, but if you plug everything into Maple then look at graphs ... well, to the right is a graph of f/x in the first quadrant. I hope you can see that the limit as (x,y)-->(0,0) along the x-axis and along the y-axis seem very different. So the hypotheses of the previous theorem do not apply to this function.

Linear approximation: a numerical 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)

QotD
I think it was something like this: write the equation of a plane tangent to z=x²y when x=2 and y=3.

HOMEWORK
Almost surely there will be an exam two weeks from today's lecture, in class, on Friday, February 24. More definite information will be available on Tuesday. Now you should work on your Maple assignments and other homework.
Answers to these problems should be handed in at the recitation on Wednesday, February 15. Please try to read and learn these sections of chapter 14 by then: up to 14.5, doing the appropriate problems.
14.3: 21, 48; 14.4: 6, 17; 14.5: 35, 46;

Tuesday from 1:40 to 3:00
Thursday from 2:00 to 4:00

Maple Data for another assignment has been sent to students. Any student who did not get data should send me e-mail, please!

Functions of "several" variables
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.

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.

Two half planes
I then turned our attention to a function which initially looks simpler, but whose graph is perhaps more bizarre.

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

I also tried to sketch the contour lines. What's shown to the right is a picture of the contour lines of this f (that is, f(x,y)=C in R²) for C integer valued ranging from -4 to 4. The contour lines are lines perpendicular to the x-axis (indicating the lack of dependence on y in the function). The contour lines are spaced twice as far apart in the right halfplane as in the left halfplane, indicating that the function f(x,y) in the left halfplane (where x<0) increases faster than in the right. Indeed, that's a reflection of the 2 in the definition.

Incidentally, Maple allows such functions to be defined with the piecewise command. As we'll see, the graphing routines don't always display the graphs correctly, because there are some assumptions of continuity.

How the bug increases
I remarked that if a bug were placed in R² and wanted to move to increase its value of f(x,y) as much as possible, it would move perpendicularly to the contour lines.

Another two half planes
I defined another piecewise function. It seems to be only slightly different from the previous one, but the graph is, to me, much "worse".

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

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.

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(theta) and y=r sin(theta), we know that x²+y²=r² and xy=r²cos(theta)sin(theta). Therefore
f(x,y)=xy/(x²+y²)=cos(theta)sin(theta)
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(theta)sin(theta)=(1/2)sin(2theta), 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 this lecture 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).

The QotD was to find F/a and F/b and F/d is F(a,b,c) was something like (a³-5b²)/(bc³-4a).

HOMEWORK
Answers to these problems should be handed in at the recitation on Wednesday, February 15. Please try to read and learn these sections of chapter 14 by then: up to 14.5, doing the appropriate problems.
14.3: 21, 48; 14.4: 6, 17; 14.5: 35, 46;

Maintained by greenfie@math.rutgers.edu and last modified 2/13/2006.