# Load of math help needed, come rescue me!

First one up is a chain rule problem
x=y*z and y=2sin(y+z), find dx/dy. I've never seen a chain rule problem set up like that before and it's driving me insane. I've done all kinds of solving for various variables and subbing, and I can't get the right damned answer(which I do know from the book, at least)

Then there's a differential problem
the thin lens formula is 1/i +1/o =1/f

if i=15 when o=10, use differentials to find i when o=10.1

Again, it's different from all the others ones and it bugs me. do I go ahead and solve for f and am I assume f is the same or something for the +.1 case?

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yup, it's dx/dy

Tom Mattson
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schattenjaeger said:
First one up is a chain rule problem
x=y*z and y=2sin(y+z), find dx/dy.
Are those two equations supposed to be a system? If so then I would eliminate z in the second one and subsitute in into the first.

Eliminating z is what I tried first, except the answer in the book is z-y+tan(y+z)

Tom Mattson
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OK, one other question: Are you familiar with partial derivatives?

Yah

that's what this chapter is, in fact

Tom Mattson
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OK. You've got a function $x=f(y,z)=yz$. What makes this wierd is that in the chain rule you usually find dx/dt, where t is not the same as either of the two variables y or z. Only this time, it is.

So the Chain Rule for your problem would be written as:

$$\frac{dx}{dy}=\frac{\partial f}{\partial y}\frac{dy}{dy}+\frac{\partial f}{\partial z}\frac{dz}{dy}$$

Note that all I've done is put "y" where "t" is would normally be. Give that a shot.

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Tom Mattson
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schattenjaeger said:
Again, it's different from all the others ones and it bugs me. do I go ahead and solve for f and am I assume f is the same or something for the +.1 case?
f is the focal length, and it is a constant for a given lens. It is i and o that vary. So yes, find out what the value of f is and use it for the next part of the problem. You have a function of a single variable, and you can use Calculus I methods on it.

Ok, thanks! That's what I figured for the second one but wasn't sure

back to the first one, so you end up with

z+y(1/2sqrt(1-y^2/4))? Does that simplify to the given answer?

Tom Mattson
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I did indeed get the answer that you posted, but I didn't get any square roots along the way.

HallsofIvy
Homework Helper
I don't even see HOW you could get a square root out of that!

x=y*z so
$$\frac{dx}{dy}= z+ y(\frac{dz}{dy})$$ and you can get $$\frac{dz}{dy}$$ by using "implicit differentiation" on y=2sin(y+z).

oh, haha, I solved for z and then took the derivative of some arcsin

Ok, yet annooother question!

I'll spare you the nitty details of the word problem, but I've got a piece of material 24 cm long, and if you bend the ends up it makes a trapezoid, and I need to maximize the area of that trapezoid

so I guess I need to express the area in terms of two variables, then take the partial derivatives in terms of each, and set the two equations =0 and solve for the variables?

Tom Mattson
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You might have to lay the details on us. If all you have is a trapezoid, then there can be up to 5 independent variables by my count. Is anything else said about the trapezoid?

Well you bend the sides of the line/string/whatever up by an angle theta on each side, so I was able to express the area in terms of w(the length of the side that got bent up)and and sins and cosines of theta

Tom Mattson
Staff Emeritus