Leading up to the QotD
Suppose I have a function of two variables, f(x,y), and I know the following information:
    f(0,0)=A and (∂f/∂x)(0,0)=B and (∂f/∂y)(0,0)=C and
    (∂²f/∂x²)(0,0)=D and (∂²f/∂x∂y)(0,0)=E and (∂²f/∂y²)(0,0)=F
Now I will define a function g with a one-variable input by the equation g(t)=f(4t¹+5t²,6t³). 1 2 3 4 5 6 This is entirely absurd -- I just assigned the powers and the coefficients in a direct way.

Some questions and some answers

• What is g(0)?
  Since g(t)=f(4t+5t²,6t³), I know g(0)=f(4·0+5·0²,6·0³+2)=f(0,0)=A.
• What is g´(0)?
  Since g(t)=f(4t+5t²,6t³), I will use the Chain Rule and the fact that f is differentiable in two variables. That is,
      g´(t)=(∂f/∂x)(4t+5t²,6t³)(4+10t)+(∂f/∂y)(4t+5t²,6t³)(18t²).
  If we plug in t=0, then we get g´(0)=(∂f/∂x)(0,0)(4)+(∂f/∂y)(0,0)(0)=4B.
• What is g´´(0)?
  Well, this is a curious computation. I need to start with the formula for g´(t). I don't care about g´(0): that's a specific number and looking at it will not help me at all.
  Since g´(t)=(∂f/∂x)(4t+5t²,6t³)(4+10t)+(∂f/∂y)(4t+5t²,6t³)(18t²) I need to consider how to differentiate things like (∂f/∂x)(4t+5t²,6t³)(4+10t).
  I will need the Product Rule, surely (this is a product). But the first piece, (∂f/∂x)(4t+5t²,6t³), has a strange function of two variable and so I will need to use the Chain Rule for a two variable function. Let me try this piece alone:
      d/dt((∂f/∂x)(4t+5t²,6t³))=(∂²f/∂x²)(4t+5t²,6t³)(4+10t)+(∂²f/∂x∂y)(4t+5t²,6t³)(18t²)
  When t=0 this is 4D+0E, just 4D. But this is only a piece of the computation. It should be multiplied by (4+10t). And then there is the other part of the Product Rule, and then there is the other term, which similarly needs the Product Rule and the Chain Rule.

  Here is, I hope, the correct formula for g´´(t):
      ((∂²f/∂x²)(4t+5t²,6t³)(4+10t)+(∂²f/∂x∂y)(4t+5t²,6t³)(18t²))(4+10t)+
      (∂f/∂x)(4t+5t²,6t³)(0+10)+
      ((∂²f/∂x∂y)(4t+5t²,6t³)(4+10t)+(∂²f/∂y²)(4t+5t²,6t³)(18t²))(18t²)+
      (∂f/∂y)(4t+5t²,6t³)(36t)

  When t=0, many things drop out. The result is 16D+10B. I hope this is what people got in class. Oh well, maybe 5 out of 85 people completed this computation correctly.

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.

So suppose you are looking at the function Q=x³w+zy²–2z. When x=1 and w=3 and y=2 and z=–1, Q's value is 1³3+(–1)2²–2(–1) which is 2–4+2=0. Isn't that nice? Maybe, but suppose I have the task of trying to increase this function as much as possible, with the restriction being that the square root sum of the squares of the changes should be at most, say, .01 (maybe there is a restriction on various kinds of costs or material or ...). What should be done?

The gradient of this function is a 4-dimensional vector with components (x, y, z, w derivatives in order) 3x²w and 2zy and y²–2 and x³. When x=1 and w=3 and y=2 and z=–1, this is the vector ∇Q=<9,–4,2,1>: the gradient vector. A unit vector in this direction is this vector divided by sqrt(102). The direction of this vector is the direction of greatest increase. That direction is <9/sqrt(102),–4/sqrt(102),2/sqrt(102),1/sqrt(102)>. If I want something whose length is .01 which will likely increase the value of the function the most, I would bet on <.01[9/sqrt(102)],.01[–4/sqrt(102)],.01[2/sqrt(102)],.01[1/sqrt(102)]>.

This is definitely too computational. Let me try a differentiation problem which is more conceptual. It resembles some calc 1 computations, but things will get more complicated soon.

A sphere problem
The points in R³ which satisfy x²+y²+z²=R² are a sphere of radius R centered at (0,0,0). We could say that z is defined as a function of x and y by this equation. In fact, since the situation is relatively simple, we can actually solve for z in terms of x and y: z=±sqrt(R²–x²–y²). From this equation we can compute ∂z/∂x etc. But if we assume that z is defined implicitly as a function of x and y by the equation, we can compute using the Chain Rule by just ∂/∂x'ing the whole equation:
    x²+y²+z²=R² becomes
    2x+0+2z(∂z/∂x)=0 so ∂z/∂x=–2x/2z.

Some of the details of this computation need explanation. The derivative with respect to x of x² is certainly 2x. What about the next term, the 0? x and y are independent variables, and there is no way a change in x can force changes in y. So here ∂y/∂x is 0. (The notation in this subject is notoriously unhelpful -- you must keep track of the logical meaning of the symbols.) And the Chain Rule applies to z²: we are assuming that z is a differentiable function of x and y, so we apply the Chain Rule to the square of this unknown function, and the result is twice the function multiplied by the function's derivative. Last, on the right-hand side of the equation, R², in spite of its appearance, is a constant and therefore has derivative equal to 0. We take the resulting equation and "solve" for the desired ∂z/∂x.

A more complicated example
The computations can be intricate. Let me try to write something quite irritating:

Suppose z is implicitly defined as a function of x and y by the
equation z²sin(x+y²)+z³x=5e^zy. Find a formula for ∂z/∂y.

    (–z2cos(x+y2)(2y)+5ezyz)
zy= ------------------------
    (2z sin(x+y2)+3z2–5ezyy)

More abstractly ...
I could analyze a function of two variables defined by using a function of one variable (!?). O.k., here is what I mean: suppose I have a function f with one variable input and one variable output. So maybe we could write f is f(w), where w is a number. I could define F(x,y)=f(x²+y²) and then try to compute, say, both ∂F/∂x and ∂F/∂y. Since I don't know much about f, when I need to differentiate it I just will write f´ and try to go on. So here we go:
    ∂F/∂x=f´(x²+y²)2x.
    ∂F/∂y=f´(x²+y²)2y.
But then y∂F/∂x–x∂F/∂y=0 because the f´'s cancel. So if I wanted to solve the partial differential equation y∂F/∂x–x∂F/∂y=0 I have a collection of solutions:
    sin(x²+y²) and e^x2+y2 and 56(x²+y²)³³² are all solutions.
Since such equations arise often in practice, this is nice.

A more significant example
If we take a long "homogeneous" rope, and wiggle it a bit, the wiggles propagate down the rope. It turns out (neglecting units, neglecting certain other hypotheses, but keeping the central idea) that for small wiggles, if we define f(x,t) to be the height of the rope at position x and at time t, then
    ∂²f/∂x²=∂²f/∂t².
This is called the one-dimensional wave equation. I can tell you about all of the solutions. Suppose L and R are both functions of one variable. Then define f(x,t)=L(x+t)+R(x–t). I'm not going to tell you anything about the functions L and R except that they are differentiable. Then the Chain Rule again applies:
    ∂f/∂x=L´(x+t)(1)+R´(x–t)(–1) and further ∂²f/∂x²=L´´(x+t)(1²)+R´´(x–t)(1)².

No, I don't really need to "write" the 1's but I want you know that they are there: the Chain Rule was used. Now again:     ∂f/∂t=L´(x+t)(1)+R´(x–t)(–1) and further ∂2f/∂t2=L´´(x+t)(1)2+R´´(x–t)(–1)2. And clearly (no!) f(x,t) satisfies the Wave Equation. The names of the two functions, L and R, were chosen because they model the movement of two "waves", a wiggle to the Left and a wiggle to the Right. I attempted to show this in class but was frustrated because, first, the bungee cords were too darn springy -- in the wave equation's setup, there is no distortion due to internal springs, and, second, there was reflection of the waves from the ends. It is possible to analyze waves with reflections, but more complicated than the setup I described, which is for an "ideal" infinitely long string. To the right is a picture created by Maple of the function f(x,t)={1/(1+.25(x–t)2)}+{.3/(1+.4(2+x+t)2}. The picture shows this function over the interval [–12,12] on the x-axis, and t varies from –7 to 7. The picture was created using the animate command. The wave traveling to the right has profile given by R(v)=1/(1+.35v2) and the wave traveling to the left has profile given by L(v)=.3/(1+.4(2+v)2). I hope you enjoy watching them.

Why? All this is not intuitively obvious, at least to me. I bet that some engineers and physicists would assert that the equation is intuitive and clear. The equation turns out to be a very good model of physical vibrations, for at least small displacements (just like Hooke's Law does describe springs, but not if you try to stretch an ordinary rubber band 10 feet!). Here are two links to reasoning which shows how this equation follows from physical ideas:      At the University of British Columbia This uses Newton's Law and force considerations.      A Wikipedian entry This uses Newton's Law and Hooke's Law more directly. The two-dimensional wave equation is ∂2f(x,y,t)/∂x2+∂2f(x,y,t)/∂y2=∂2f(x,y,t)/∂t2. It describes vibrations in thin plates. The three-dimensional wave equation is ∂2f(x,y,z,t)/∂x2+∂2f(x,y,z,t)/∂y2+∂2f(x,y,z,t)/∂z2=∂2f(x,y,z,t)/∂t2 describes vibrations in solid objects. It is much more difficult to find simple interesting solutions for the two and three dimensional equations than for the one-dimensional equation.

Crazy computations in thermodynamics and physical chemistry
I wanted to briefly indicate some horrible computations which come up in applications. The computations are not really horrible, but the notation makes them look quite weird. First, a bit of preparation. If we have F(x,y) and both x and y are functions of one variable v (so x=x(v) and y=y(v)), then what happens if we try to differentiate F with respect to v? So basically we have, say, g(v)=F(x(v),y(v)), and I want to understand g´(v). The control from v to F's output is passed through two variables. I remember that F(x+Δx,y+Δy)≈F(x,y)+(∂F/∂x)Δx+(∂F/∂y)Δy. so that F(x+Δx,y+Δy)–F(x,y)≈(∂F/∂x)Δx+(∂F/∂y)Δy. If I now divide by Δv, we get ΔF/Δv≈(∂F/∂x)(Δx/Δv)+(∂F/∂y)(Δy/Δv). If I take limits as Δv→0, then this formula appears:
    g´(v)=(∂F/∂x)x´(v)+(∂F/∂y)y´(v).
This is another "Chain Rule" and the weird thing about it is the + which comes from the definition of differentiability in several variables. So let me show you some applications.

Implicit functions, two dimensions
I first began with a 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 a 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
Do you know 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).
By the way, in applications people frequently change ∂V/∂P to (∂V/∂P)_T to help remember that T is constant in this computation.

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. But the notation is horrible and, for me, makes things harder to state and understand.

After class a few people expreseed doubts about all this. So let me here try an explicit example. My "gas law" will just be sort of silly and not physically meaningful. Suppose P and V and T obey the following "law":     P2T+Ve3P+5T=0. So the left-hand side of that equation is G(P,V,T). Now what is ∂V/∂P (with T held constant). So ∂/∂P the equation. The result is 2PT+(∂V/∂P)e3P+5T+Ve3P+5T3=0. We solve to get ∂V/∂P=–(2PT+Ve3P+5T3)/(e3P+5T). Now I want ∂P/∂T (with V held constant). So ∂/∂T the equation. The result is 2P(∂P/∂T)T+P2+Ve3P+5T{3(∂P/∂T)+5}=0. We solve to get here ∂P/∂T=–(P2+Ve3P+5T5)/(2PT+Ve3P+5T3). Finally, I will try to compute ∂T/∂V with P held constant. We ∂/∂V the equation, and we get P2(∂T/∂V)+1e3P+5T+ Ve3P+5T5(∂T/∂V)=0. Now solving gives ∂T/∂V=–(1e3P+5T)/(P2+Ve3P+5T5). Now plug all this stuff into (∂V/∂P)T(∂P/∂T)V(∂T/∂V)P. Here we go: (–(2PT+Ve3P+5T3)/(e3P+5T) )(–(P2+Ve3P+5T5)/(2PT+Ve3P+5T3) )(–(1e3P+5T)/(P2+Ve3P+5T5) ). Check it out! Lots of things cancel, and we have (–1)3=–1. YES!!!

A large part of the exam will ask about the material we discussed in both of this week's lectures.

Exam warning!

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
I would like to compute and understand the rate of change of the temperature, T, with respect to time. It turns out that this is a significant computation. Now T is a number and t is a number and T is a function (complication: there are intermediate variables x, y, and z) of t. So the derivative of T will involve taking its value at t+Δt and looking for the linearization multiplier. That is what we declared earlier. So here we go. I will try to accompany the steps of this somewhat elaborate manipulation with explanations.

T(x(t+Δt),y(t+Δt),z(t+Δt))=
	We are "kicking" the time variable a little bit, and we would like to examine the change in the T variable.
T(x+x´(t)Δt+H.O.T.,y+y´(t)Δt+H.O.T.,z+z´(t)Δt+H.O.T.)=
	We use the fact that each of the components of the position vector are differentiable, so each function value at t+Δt can be replaced by the original value of the function, a multiplier (the derivative) which multiplies the disturbance, and higher order terms. If I were being careful, I would use different notation for each of the H.O.T.'s, but in practice people don't do that too often. You'll see why soon.
T(x,y,z)+(∂T/∂x)(x´(t)Δt+H.O.T.)+(∂T/∂y)(y´(t)Δt+H.O.T.)+(∂T/∂z)(z´(t)Δt+H.O.T.)+H.O.T.
	This is the differentiability of T. The changes in the inputs to T (there are three inputs) are passed outside. What happens? The linearization idea says that the changes are each multiplied by an appropriate partial derivative. And this isn't really exact (the numerical example showed this!) so there is also a H.O.T. from the differentiability of T. (Complicated? Sure is.)
T(x,y,z)+(∂T/∂x)(x´(t)Δt)+(∂T/∂y)(y´(t)Δt)+(∂T/∂z)(z´(t)Δt)+ (∂T/∂x)H.O.T.+ (∂T/∂y)H.O.T.+ (∂T/∂z)H.O.T.+ H.O.T.
	Now I did some multiplication and rearrangement. I pushed everything involving any of the Higher Order Terms to the end.
T(x,y,z)+(∂T/∂x)(x´(t)Δt)+(∂T/∂y)(y´(t)Δt)+(∂T/∂z)(z´(t)Δt)+H.O.T.
	Here is the important step, and it sort of asks you to think a bit about the ideas of calculus. All of the terms with H.O.T. are actually, all added together, just another big H.O.T.
T(x,y,z)+{(∂T/∂x)(x´(t))+(∂T/∂y)(y´(t))+(∂T/∂z)(z´(t))}Δt+H.O.T.
	This is the last rearrangement. Please, I hope you have the patience to see what has happened. We have the "old" unperturbed value of T(x(t),y(t),z(t)), and then we have a mess (not really, as you'll see) multiplying Δt, and then, finally, we have a whole bunch of things which logically are inside the H.O.T.

Now what? If f(v) is a function of one variable, then f(v+Δv)=f(v)+f´(v)Δv+H.O.T. identifies the derivative of f by what happens to small perturbations. The red stuff is the derivative, because it is the multiplier of the small perturbation of the input.

We carried out an elaborate analysis of how the temperature function changes. Now look at the result again:
T(x(t+Δt),y(t+Δt),z(t+Δt))=T(x,y,z)+{(∂T/∂x)(x´(t))+(∂T/∂y)(y´(t))+(∂T/∂z)(z´(t))}Δt+H.O.T.
The multiplier of Δt is dT/dt, so we see that

dT/dt=(∂T/∂x)(x´(t))+(∂T/∂y)(y´(t))+(∂T/∂z)(z´(t))

But this formula is not the point of the discussion. Look at the equation and think a bit. 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   ∂y    ∂z /    \ 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 and is sometimes also called grad T. The upside-down triangle (or upside down Δ) is sometimes called "del" or "nabla". This vector can be computed only from the nebula information.

Thinking about the temperature derivative this way "decouples" (yeah, a word that's used) the influence of the nebula from that of the spaceship.

Now I will try to make a sequence of observations which might help people understand the 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. Let me call the unit tangent vector u(t) here (in the discussion of curvature it was called T(t) but that will just be too darn confusing). Then v(t) is the same as (ds/dt)u(t) where ds/dt is the speed and u(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 the more subtle aspect is what happens as the direction changes.

Observation 3
Here I will suppose that ds/dt=1 for simplicity. Again, call the unit tangent vector, 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(θ). Since cos(θ) 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(θ) equal to +1. Therefore we need θ 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 (?)
Here is 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 baking 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. Then 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 plane 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.

Another example
Suppose I want an equation for the plane tangent to z=3x²+5xy² when x=1 and y=2. Well (curious trick!) if I consider T(x,y,z)=3x²+5xy² z then the surface z=3x²+5xy² is exactly the isothermal or level surface corresponding to T=0. Now ∇T will be perpendicular to that surface. We compute: ∇T=<6x+5y²,10xy,–1>. We know that x=1 and y=2 so that the gradient is <26,20,–1>, which is a vector perpendicular to the plane. Now we need a point on the surface. But we know z=3x²+5xy² and x=1 and y=2 so we can deduce z: z=23. A point on the plane is (1,2,23). Therefore an equation for the tangent plane is
    26(x–1)+20(y–2)+–1(z–23)=0.

Here is another way we could get the tangent plane. I could have done this last time, but the gradient is just a simpler way to get the result. 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. An equation for 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). Example Find the equation of a plane tangent to z=3x2+5xy2 when x=1 and y=2. Here z=3+5·22=23. Now for the derivatives:     zx=6x+5y2 and this is 26 when x=1 and y=2;     zy=10xy and this is 20 when x=1 and y=2; The tangent plane equation (z–z0)=∂f/∂f(x0,y0)(x–x0)+∂f/∂y(x0,y0)(y–y0) becomes (z–23)=26(x–1)+20(y–2). Below are some pictures generated by Maple. The third, showing both the surface and the tangent plane, is a bit hard to understand. The surface bends below the plane in one direction, and above in another (very "saddle-like"). The surface near (1,2,23) The plane near (1,2,23) Both near (1,2,23)

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 many implementation details to worry about, but the general strategy is valuable.

Contour map example So to the right is a made-up example of a contour map. The lines shown are at height multiplies of 100 feet. I hope that you can see (?) there is sort of one part of the map which deserves the name "peak" (in the upper left-hand corner). This map is a bit more complicated than the one I drew in class. I claim there is also a part of the map which could be named "the pits" (probably at least two of them!) and that's inside the two loops at the 300 foot height.
Hill climbing -- a way to get to the maximum Now suppose we are at the point P on the map. How should we hike in order to get uphill as fast as possible? I claim that the directions shown will do that. Notice that I am not always walking towards "the peak", but I just want to always walk perpendicular to the contour lines -- in the direction of the gradient of the height function.
The method of steepest descent -- how to get to the minimum This is a way people use to try to get to minima of functions in several variables. We start at the point Q and move in the direction opposite the gradient of the height function. In what I'm showing here, we take some sort of fixed step size (more sophisticated versions vary the step size) and after that step we keep going until we hit another contour line. Then we recompute the direction (minus the gradient) and step again. You can see, I hope, that this path is headed for one of "the pits". These methods don't always work and don't always work efficiently, but they are easy to understand, they can be done in many dimensions, and the programs run fast.
The most change I could also ask where there is the most change in height in a fixed distance. These are contour lines corresponding to equal changes in height. The heights will change most rapidly (as a function of the position on the map) when the contour lines are most closely spaced. This also corresponds, of course, to the region in the plane where ||∇ height|| is largest: the largest magnitude of the gradient vector. I think the region indicated (in magenta) is the region in this rather simple contour chart.

The QotD: tangent plane to an 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)? 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};

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 (∂/∂t and ∂²/∂x²) of F(x,y)=e^–x2/(4t)/sqrt(t). This function occurs in many applications analyzing diffusion, including heat and fluid mixing. Let's see:

• ∂F/∂t=e^–x2/(4t))(–x²(–t^–2/4)/sqrt(t)+e^–x2/(4t)(–1/2)t^–(3/2). This is e^–x2/(4t) multiplying x²/(4t^{5/2)–1/(2t3/2)}
• ∂F/∂x=e^–x2/(4t)(–2x/(4t))·(1/sqrt(t)) which is e^–x2/(4t)(–x/(2t^3/2)).
• ∂²F/∂x²=e^–x2/(4t))(–2x/(4t))(–x/(2t^3/2))+e^–x2/(4t))(–1/(2t^3/2)). This is e^–x2/(4t)) multiplying (–2x/(4t))(–x/(2t^3/2))+(–1/(2t^3/2)). The second factor is x²/(4t^5/2)–1/2t^3/2.

So F is a solution of the Heat or Diffusion Equation
    ∂F/∂t=∂²F/∂x²
This is the one-dimensional version of the equation, and describes how temperature changes over time, t, at position x in, say, a thin homogeneous rod. Similar but more complicated equations describe what's happening in a plate or a solid object, or how fluids diffuse (heat and fluid diffusion have very similar mathematical descriptions). There are more than three and a half million Google links for the phrase "heat equation".

To the right is an animation of the temperature for time ranging from t=.2 to t=2. Near .1, a bunch (??) of heat is injected into the bar at x=0 (strike a match, etc.). I started the picture a bit later than 0, because the match is infinitely hot (?) at t=0. The amazing thing is that this function does describe or model a real situation nicely. The heat diffuses quite rapidly. Please note that I have not mentioned units either here or in the lecture. This was intentional, although the situation does get distorted a bit. The units etc. can get quite complicated.

A betting game (?)
Consider this situation: suppose f(x,y)=(sin(y⁴)x–7)³. Then I flip a (fair) coin. If it lands "heads", I ∂/∂x this function. If the coin shows "tails", I ∂/∂y the function. What's going to happen? I asked students to speculate about this. Almost everything that students said was correct. I sometimes tried to distract people from the real question by making interesting and true assertions. For example, ∂¹⁰⁰f/∂y¹⁰⁰ yields a mess with 342 terms. And 200 y derivatives gives an algebraic mess with 680 terms. These computations were not done by hand, but with Maple. The expressions begin to swell (get larger and larger). I asked students if they thought that this sort of growth would be likely under the conditions of the experiment. Some students kept remarking about ∂/∂x and I kept "distracting" with facts about ∂/∂y. Here is an important and relevant result for this game.

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 (you sort of take two double differences and analyze them, much as the differences in rows and columns of a spreadsheet). 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. The partial derivatives of this function are 0 at (0,0), so one might expect and hope that the linear perturbation near (0,0) would be 0 *because both Constant1 and Constant2 are 0. But, in fact, if we increase x and y both by the same about (so change x from 0 to 0+tiny and y from 0 to 0+tiny, then the resulting value of f is exactly tiny, and this is first order and is not 0. This function is not differentiable at (0,0) even though both partial derivatives (tangent lines to the curves in the slices!) exist there: an irritating example. There is no local linearity at the origin -- no matter how much you magnify it, the "surface" looks more and more like a sort of cone, not like a piece of a plane.

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+Constant₂h+Error, where the Error term→0 faster than |h|+|k| (so, faster than first order -- it is H.O.T.).

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)≈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" (up to 10 decimal places!) of f(2,3) is 4.147314409.
As I declared in class, in 1910 this would be nearly amazing! Accuracy to almost 4 decimal places. But, yuh'see, we have programs to compute this. The idea of linear approximation is what will be important.

QotD
If f(x,y)=x^y, find ∂f/∂x and ∂f/∂y.

The first one is easy: ∂f/∂x=yx^y–1 (this is just x to a "constant" power after all). For the first, the best way is to rewrite the function using exp and ln: x^y=(e^ln(x))^y=e^(y ln(x)) and then ∂/∂y of e^(y ln(x)) is e^(y ln(x))ln(x). This is a fine answer, or it can be rewritten x^yln(x).

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 in this course will be all inputs for which the function makes sense. The usual restrictions that will concern us are:

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

The range will be the collection of all possible outputs. You may remember from calculus that while determining precise domains is often possible but tedious, precise descriptions of ranges can be quite difficult (this can involve exact determinations of max and min values).
Here are some examples chosen to illustrate the likely possible restrictions. I certainly did not have time to discuss all of these in class.

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 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.

Here is one of the variations which can be produced with the plot3d command. It has the three-dimensional (the two-dimensional picture of the three-dimensional graph!) plot with the contour lines shwon at the correct places on the graph. There are all sorts of views which can be obtained using the options of plot and plot3d . Sometimes these variants can be useful. When this picture is rotated so we are looking down from the positive z-axis, a view similar to the contour lines shown previously is obtained.

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? If x=3, then f(3,y)=9–y2. In the plane perpendicular to the x-axis given by the equation 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 may be 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!
This saddle point is stranger than you might think. 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!), then we actually get a value on the surface z=x2–y2 for every value of t. 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.

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>0,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.

Continuity Almost all of the functions we will consider in this course will be continuous (in fact, much nicer than just continuous -- they will be smooth in a way that I'll explain later. But I should at least remark on what continuity means. The definition should not be too surprising. A function f(x,y) is continuous at a point (a,b) in its domain if lim(x,y)→(a,b)f(x,y) exists and is equal to f(a,b). Of course this doesn't help too much if we don't know what lim(x,y)→(a,b)f(x,y) means. So I will discuss this very briefly. Limits in one dimension In one variable, limits are relatively simple. To define limx→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 This function has only two values, 0 and 1. Certainly if you "walk" towards 0 on a straight line approach in the second, third, and fourth quadrants, the function values are all 0 and therefore the limit is 0. What's not so obvious perhaps is the behavior of the function on straight line approaches in the first quadrant. Look at y=x. This line intersections y=x2 only at x=0 and x=1. So if we "walk" towards the origin on this line from some large x>0 considering the values of the function f(x,y), the function will be 0 at every point except x=1 where it will be 1. Certainly the limit as x→0 will exist, and it will be 0. In fact, the limit exists on every straight line approach to (0,0), and the value of the limit is 0. But the real, two-variable limit should not exist, because the values of f(x,y) do not get close to 0 as (x,y)→(0,0). If you want more precise definitions of limits, here they are. 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 δ when 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 ε and 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.

QotD I introduced ∂ by just using it. This was an interesting pedagogical exercise.
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 compute ∂F/∂a, ∂F/∂b, and ∂F/∂c if F(a,b,c)=a²e^(7b–5c3).
So ∂F/∂a=2ae^(7b–5c3), ∂F/∂b=a²e^(7b–5c3)7, and ∂F/∂c=a²e^(7b–5c3)(–15c²).

It turns out that ∂FUNCTION/∂VARIABLE, when FUNCTION depends on many variables, means that all variables except the variable specified are thought of as constants, and then the derivative of the function with respect to the remaining variable is computed with the help of the usual algorithms.

Let's define the unit tangent vector, T, to be a unit vector in the direction of r´(t), the velocity vector. 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, the speed, is the length of r´(t). Now differentiate the formula for r´(t) using one of the product rules we had stated earlier.

We get r´´(t)=(d²s/dt²)T+(ds/dt)dT/dt. Let me show you one of the most interesting math/physical consequences of vector calculus. Since T is always a unit vector, its length is 1 so T·T=1 (length squared!). Differentiate this equation and use the product rule for dot products. The result is dT/dt·T+T·dT/dt=0. But scalar product is commutative, so the two terms are the same, and we can divide by 2 and get T·dT/dt=0. That means dT/dt is perpendicular to T.

But T is cos(θ)i+sin(θ)j, and dT/dt=(–sin(θ)(dθ/dt)i+(cos(θ)(dθ/dt)j= (dθ/dt)[(–sin(θ)i+(cos(θ)j]. Now dθ/dt=(dθ/ds)(ds/dt). But we called dθ/ds curvature, κ, last time: the rate of change of the angle with respect to arc length or with respect to speed reset to be uniformly 1. So dT/dt=κ(ds/dt)[–sin(θ)i+(cos(θ)j]. The term [–sin(θ)i+(cos(θ)j] is a unit vector perpendicular to T and is usually called the unit normal vector and written N. The equation r´´(t)=(d²s/dt²)T+(ds/dt)dT/dt turns into
    r´´(t)= (d²s/dt²)T + κ(ds/dt)²N.

We have decomposed acceleration into the normal and tangential directions. This can be significant physically and can help to understand physical and geometric situations. Here are two "extreme" cases. Realize, as you consider them, that since F=ma and m, mass, is a positive scalar, force is zero exactly when acceleration is zero and also that force and acceleration have the same direction since multiplication by a positive scalar doesn't change direction.

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? For a straight line κ=0 always, 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. There is never any transverse force.

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? The normal component of the acceleration (a scalar multiple of the force acting on the object) is κ(ds/dt)². 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. There is always a normal force.

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.

The geometry of space curves, curves in R³, 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), computer graphics: amazingly useful formulas and ideas!

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. Again, 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 something used repeatedly in this course:

Any non-zero(!) vector is the product of its magnitude multiplied by a unit vector in its direction.

Please remember the basic information about a right circular helix. Click on the link and reread what we did if you need to.

Here r(t)=a cos(t)i+a sin(t)j+btk, the position vector. The velocity vector is r´(t)=–a sin(t)i+a cos(t)j+bk. The length of this velocity vector is sqrt([–a sin(t)]²+[a cos(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 just did, if we differentiate this we should get a vector perpendicular to T(t). Here we go: 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, for curvature, we need 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 direct computation from the definition is too 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.
  When a gets large, we have a¹ on top and essentially a² 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 the plots package 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 help you understand what curvature represents. 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´(t)xr´´(t)||/(||r´(t)||)³.
Even though this formula looks weird, it is much better to use than trying to work through 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, after this {ludicrous|wonderful} computation, I do admit that I get almost nothing out of it. Here is this ridiculous formula, and what does it tell me? 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 it does have some interesting information about physical quantities as we saw in two dimensions for the straight line and the circle.

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: 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 right-handed triple T, N, B is called the Frenet frame. Describing how this "frame" changes in time tells a lot about the curve. The curvature says how much the airplane is being pulled from a straight line. It measures how much the T is changing in the N direction. The green 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 airplane is twisted out of the two-dimensional plane determined by T and N. The rate of change of B is called the torsion, and 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. Let me do things with a bit more detail.

I apologize to any pilots, because what's above is not exactly correct. Actually, they would be thinking, "That's not correct at all". I am certainly simplifying. Reality is more complicated.
Also, right now I'm losing the advantage I got on the first day compared to the standard syllabus because I'm using an extra lecture for this material. Oh well.

How does B(t) change? Since B(t)·B(t)=1, differentiation results in 2B´(t)·B(t)=0, so B´(t) is orthogonal to B(t). But differentiation of B(t)=T(t)xN(T) results in B´(t)=T´(t)xN(t)+T(t)xN´(t). Since T´(t) is parallel to N(t) (because of our definition of N(t)!), the first product is 0 (another property of cross-product!) so that B´(t) is a cross-product of T(t) with something. Therefore B´(t) is also perpendicular to T(t). So B´(t) is perpendicular to both T(t) and B(t), and since only one direction is left, B´(t) must be a scalar multiple of N(t). The final important definition here for space curves is: dB/ds is a product of a scalar and N(t). The scalar is –τ. That is supposed to be the Greek letter tau, and the minus sign is put there so that examples (the most important is coming up!) will work out better. This quantity is called torsion, and is a measure of "twisting", how much a curve twists out of a plane (the particular plane is the plane determined by T and N).

If a space curve does lie in a plane, and if everything is nice and continuous, then B will always point in one direction (there are only two choices for B, "up" and "down" relative to the plane, and by continuity only one will be used) so that the torsion is 0 since B doesn't change. The converse implication (not verified here!) is also true: if torsion is always 0, then the curve must lie in a plane!

Let me apply this to the right circular helix with a direct computation. I emphasize that this is by far the simplest example, and almost all curvature and torsion computations I've done in the last decade or so have used computer algebra systems.

So for our helix with r we found     T(t)=1/sqrt(a2+b2))(–a sin(t)i+a cos(t)j+bk);  N(t)=–cos(t)i–sin(t)j;  ds/dt=sqrt(a2+b2). So B(t)=T(t)×N(t). This is the scalar 1/sqrt(a2+b2)) multiplying the cross product of –a sin(t)i+a cos(t)j+bk and –cos(t)i–sin(t)j: | i j k | det |–a sin(t) a cos(t) b | = b sin(t)i –(–b)(–cos(t))j+a(sin2+cos2)k | –cos(t) –sin(t) 0 | Yes, there really are 3 minus signs and a trig identity. So B(t) is the vector (1/sqrt(a2+b2))(b sin(t)i–b cos(t)j+ak). A few questions: Is B(t) a unit vector? Its length is (1/sqrt(a2+b2))sqrt(b2sin(t)2+b2cos(t)2+a2) which is (1/sqrt(a2+b2))sqrt(a2+b2)=1. Is B(t)⊥T(t)? The dot product of (1/sqrt(a2+b2))(b sin(t)i–b cos(t)j+ak) and 1/sqrt(a2+b2))(–a sin(t)i+a cos(t)j+bk) is ab–ab=0. Is B(t)⊥N(t)? The dot product of (1/sqrt(a2+b2))(b sin(t)i–b cos(t)j+ak) and N(t)=–cos(t)i–sin(t)j. is (–b)–(–b)=0. Now for the torsion we need to find dB/ds. I compute dB/dt and get (1/sqrt(a2+b2))(b cos(t)i+b sin(t)j). To get dB/ds we multiply by 1/(ds/dt) which is 1/sqrt(a2+b2). The result is (1/(a2+b2))(b cos(t)i+b sin(t)j). This is supposed to be –τN(t) where N(t)=–cos(t)i–sin(t)j. Now notice that the darn minus signs cancel (that's one of the reasons for the minus in the definition of τ) and we see clearly that for a right circular helix, τ=b/(a2+b2). If b=0 this is a circle in the plane, and the torsion is 0. Intuition about torsion is hard to get. Curvature tells how T changes, and torsion tells how B changes. What about N? In fact, if we look at the expression N=BxT and differentiate, using the product rules again, no new quantities are needed (this is done in detail below). The result is called the Frenet-Serret equations (also called the Darboux equations in mechanics): dT/ds = 0 + κN + 0 dN/ds =–κT + 0 + τB dB/ds = 0 –τN + 0 These are three vector differential equations, so they actually represent nine (9) scalar differential equations and there is considerable significance in the way these equations look. If you want to move robot arms or analyze molecular backbones, then ... you may need to deal with these equations. The first and third equations are used to define κ and τ. Their information content is that the rate of change of both T and B is in the direction of N which is maybe a bit surprising. The middle equation is the one that, to me, is really startling: the rate of change of N can be predicted from the quantities already defined -- nothing new needs to be considered. Getting the middle equation Since N=BxT, we can differentiate with the appropriate product rule, using the other two equations. If we d/ds, then dN/ds=dB/dsxT+BxdT/ds=–τNxT+BxκN. Remember that T,N,B is a right-handed coordinate system. Therefore TxN=B so that NxT=–B. And NxB=T so that BxN=–T. Therefore dN/ds=–τNxT+BxκN=τB–τT. This is the middle equation in the collection above.

Stay calm!
Consideration of torsion is not officially part of the course, and neither are the Frenet-Serret (Darboux) equations. And also no binormals. So stop worrying. There is more about all this in the textbook if you are interested. And also in books about motion, molecules, materials, ...

QotD
Here is a formula for torsion from Wikipedia. I think it is correct. The first version of the formula that I wrote (with the exponent=2 in the bottom!) was correct, darn it! I suggested 3 which was incorrect and I am very sorry. I will "correct" the answers taking this into accout. The management, looking sorry, is shown to the right.

If r´, r´´, and r´´´ are the first three derivatives of the position vector (velocity, acceleration, and what is sometimes called jerk), then

    (r´×r´´)·r´´´   
τ= ---------------
     ||r´×r´´||2

Answer r´×r´´ turns out to be <1,–2,–4> and then τ=–5/21.

Here we suppose that v₁(t)=<x₁(t),y₁(t),z₁(t)> and v₂(t)=<x₂(t),y₂(t),z₂(t)> are both differentiable (this is logically the same as asking that all 6 of the scalar functions x₁(t), ... ,z₂(t) be differentiable.)

• Differentiation and vector addition
  If w(t)=v₁(t)+v₂(t), then w(t) is a differentiable vector-valued function, and w´(t)=v₁´(t)+v₂´(t).
• Differentiation and dot product
  If f(t)=v₁(t)·v₂(t), then f(t) is a differentiable scalar-valued function, and f´(t)=v₁´(t)·v₂(t)+v₁(t)·v₂´(t).
• Differentiation and cross product
  If w(t)=v₁(t)×v₂(t), then w(t) is a differentiable vector-valued function, and w´(t)=v₁´(t)×v₂(t)+v₁(t)×v₂´(t).
• Differentiation and scalar multiplication
  Suppose c(t) is a scalar-valued differentiable function. Then w(t)=c(t)v₁(t) is a vector-valued differentiable function, and w´(t)=c´(t)v₁(t)+c(t)v₁´(t).

Use the cross-product formula with some care since cross-product is not commutative. The order in the terms matters.

What will we do?
Today I want to use calculus as a way of investigating the geometric properties of curves. This is a bit difficult because calculus applies to the parameterizations of the curves, and since there can be many sometimes wildly different parameterizations of the same curve, we'll need to be clever. This material was mostly developed between 1750 and 1850. It has been found useful recently in such subjects as molecular biology (describing how long molecules might twist) and computer graphics (what curves should shadow and light make?).

Arc length
If speed is the magnitude of the velocity vector, then since distance=rate·time, then, 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. What I'll show you is an analysis that was apparently first done by Euler in about 1750. First we'll consider 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 (dFROG/ds)(ds/dt)=dFROG/dt if FROG is any quantity of interest, such as θ. So to get dFROG/ds from dFROG/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.

The right circular helix
This will be our 3-dimensional "test case". We'll consider a right circular helix.

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.

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.

Vector functions of a real number
Now the course moves on: we will deal with functions whose domain is all or part of R¹ and whose range is R³. These are vector-valued functions of a scalar variable. The motivation is really the motion of points in space, and the analysis of the resulting paths (curves). There will be interactions between motion and geometry.

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? 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 ideal three-dimensional situation. What can we see? 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? 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). Is this somehow "proof" that this was (iii)? 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! This non-smooth behavior is officially called a cusp.
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
People who want 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). To the right is shown a possible picture of the situation. The the secant vector, r(t+h)–r(t) is in magenta. It is close to the curve, with its head and tail at two nearby points on the curve. 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! This direction is a vector, drawn in red. It is called the velocity vector or the tangent vector.

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.

So I had considered parameterizations of various portions of a straight line earlier in this lecture. Let's look at the speed of the parameterizations. <t,t,t> has derivative <1,1,1> with speed therefore sqrt(3): a constant. So this is a uniform speed parameterization. <t3,t3,t3> has derivative <3t2,3t2,3t2> with speed therefore sqrt(3)3t2. For t large negative or positive, the speed is large. It decreases near 0 and the speed is 0 at 0. This is what we observed when considering the graph. Also the direction is always (except t=0!) <positive,positive,positive,>. Now consider <t2,t2,t2>. The velocity vector is <2t,2t,2t>. For t<0 it points towards the origin, and for t>0 it points away from the origin, as we saw earlier. And the speed is sqrt(3)2|t| (remember that the square root of t2 is |t|, not t!). So this is large when |t| is large. And finally <sin(t),sin(t),sin(t)> has speed sqrt(3)|cos(t)|, which varies periodically also as predicted.

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 2Π, the curve passes around one loop. That means the change in t is Π, so the change in z is 8Π.

Let's find the parametric equations of a line tangent to this helix when t=Π/2. The line must pass through r(2)=<3cos(2[pi/2]),3sin(2[Π/2]),8[Π/2]>=<–3,0,4Π>. A vector in the tangent direction can be gotten from the velocity vector. So: r´(t)=<–6sin(2t),6cos(2t),8> which, when t=Π/2, gives <0,–6,8>. Therefore the parametric equations for the line are:
x=–3+0t
y=0–6t
z=4Π+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. I will try to remember to write them on the board before the start of the next class.

QotD
Suppose r(t)=<5t,3e^(t2),2+cos(t)>.

1. Find r´(t)
   Answer r'(t)=<5,3e^(t2)(2t),–sin(t)>.
2. Find r(0) and r´(0).
   Answer r(0)=<5·0,3e⁽⁰²⁾,2+cos(0)>=<0,3,3> and r´(0)=<5,3e⁽⁰²⁾(2·0),–sin(0)>=<5,0,0>.
3. Write parametric equations for the line tangent to this curve when t=0.
   Answer x=0+5t
   y=3+0t
   z=3+0t.

I wrote a short list of properties of dot product and cross product on the board. I wanted a summary for use during the lecture. For us, the dot product is useful for quickly and efficiently checking when two vectors are orthogonal. For most of the course, the most important application of the cross product is producing a vector perpendicular to a pair of vectors. We will use both of these later in this lecture and repeatedly in the course. The magnitude of the cross product will be used in more advanced applications.

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 15
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

   | i  j k |
det|–8 –1 4 |=–i–(–8–36)j–(–9)k=–i+44j+9k 
   | 9  0 1 |

 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. 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). So 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. Here are some quantities which are vectors and some which 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 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. 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.

Start reading chapter 12, please. Do the problems.

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Maintained by greenfie@math.rutgers.edu and last modified 1/19/2010.