Calculus problems

This is not homework. These questions are from a sample final (first year calc), and are fairly straightforward, but I’m having some difficulties.

Here are two questions:

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Here’s what I did with the first problem:

http://iupload.net/042003/question1.jpg

What I’m not sure from the first question is how to “expand” the operator on the derivatives. My answer differs from the solution set on the last term expansion.

Also, I know the definition of a Jacobian, but what exactly is it? Why do we use it?

The Jacobian is, in a certain sense, the derivative of a function from R[sup]m[/sup] to R[sup]n[/sup]. It’s good for changes of variables in multiple integrals, among other things.

Yeah, that’s the definitation they gave us. And I know we use it for transformations for change of basis and to see how the size of a function in one space compares to its size in a different space.

What I don’t really understand is the math behind it, like why do we choose this number? Why do we take the determinant of it? Why do we take the absolute value of it for change of variable? Why does the transpose of the Jacobian matrix not affect the Jacobian determinant (I know why in the linear algebra sense, but not why in the calculus sense since I don’t know why we take the determinant)? I’m sure there are more questions if I can think of them.

One thing Jacobian matrices can be used for is in solving systems of simultaneous nonlinear equations.

In Newton’s method, you use the iteration:

x[sub]i+1[/sub] = x[sub]i[/sub] - f(x[sub]i[/sub]) / f’(x[sub]i[/sub])

When there’s a system of equations, you use:

x[sub]i+1[/sub] = x[sub]i[/sub] - f(x[sub]i[/sub]) J[sup]-1/sup

Wow, I didn’t know that. Our cruel teacher had us solve nonlinear equations by other methods :(. But anyways… Why can you use that equation?? :smiley:

Why can you use that equation? Because the Jacobian matrix is analogous to the first derivative.

Multiplying by J[sup]-1[/sup] is the same as dividing by f’.

Ok… now how about the other questions? :slight_smile:

Sigh… why don’t you tell us what the solution set says?

I just got the answers from a classmate, presumably from the prof. No worked solutions.

Okay, so you have it thirdhand. Then it’s all the more important that you tell us what it is. :smiley: The thing is, your answer looks right to me. So I can’t tell you why you didn’t get the answer you wanted, if you don’t say what that answer is.

http://iupload.net/042003/problem1-.jpg

The answer is on the top with my work on the bottom, outlining which derivative he used the derivative operator on. What I don’t get is why he chose it to do it that way… would the answer really be different if he chose to differentiate (partial y - partial v) wrt z instead of (partial x - partial y) wrt z?

And what about the second question? The answer is 2/5.

Okay, I think I see what the deal is. Yes, that would be different; you can’t do it that way. It’s not a matter of choosing one or the other. The thing is, x is a function of z. y isn’t a function of z. Let’s look at that term, as an example:

d/dz(dx/dy dy/dv)

Now, y does not depend on z, and so dy/dv does not depend on z. So you can pull it out. It’s a constant. So you get this:

d/dz(dx/dy) × dy/dv

Does that make it any clearer? I’ll work on the second question while you consider that.

That’s just the type of explanation I was looking for on the first question. Thanks.

There is probably more than one good way to do the second problem. My way of choice is implicit differentiation, but if you’ve studied another way, you should probably use that. Anyway, here’s how I did it. Do implicit differentiation on the three equations:

dx = vCos[w]du + uCos[W]dv - uvSin[w]dw
dy = vSin[w]du + uSin[w]dv + uvCos[w]dw
dz = udu - vdv

Since you’re holding x and z constant, you can set dx = 0 and dz = 0. This leaves you with three equations and four unknown differentials. Eliminate dv and dw. (Hint: multiply the first equation by Cos[w] and the second one by Sin[w], and add them together.) Anyway, here’s my result:

vSin[w]dy = (v[sup]2[/sup] + u[sup]2[/sup])du
du/dy = vSin[w] / (v[sup]2[/sup] + u[sup]2[/sup])

Then just substitute the values, and you should get the correct answer.

Ok I gotcha. I wouldn’t really have thought to take the total differential of each variable {x,y,z}. I think we did it this way in class, but I also think that there’s a formula using Jacobians. Do you know of such a formula?

Yes. Take your Jacobian matrix, and invert it. The result will be a matrix that has the following values:

[ du/dx du/dy du/dz ]
[ dv/dx dv/dy dv/dz ]
[ dw/dx dw/dy dw/dz ]

The answer will simply be the 1,2 element.

Oh, one more thing, in case you don’t have a symbolic algebra tool, and you want to avoid lots of work. For this problem, since you just want the numerical value, it’s okay to substitute (u, v, w) = (1, 2, pi/2) into the Jacobian before you invert it. So you’ll get the Jacobian equal to this:

[ 0  0 -2 ]
[ 2  1  0 ]
[ 1 -2  0 ]

This should be much easier to invert, and will give you the same final answer.

Ok thanks. It looks like I have some hope for my final exam :).