Load of math help needed, come rescue me!

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.

 

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

 

OK. You've got a function [itex]x=f(y,z)=yz[/itex]. 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:

[tex]
\frac{dx}{dy}=\frac{\partial f}{\partial y}\frac{dy}{dy}+\frac{\partial f}{\partial z}\frac{dz}{dy}[/tex]

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

 

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.

 

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

 

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

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

 

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?

 

See there? You didn't say that before. The definition of a trapezoid does not require that the two nonparallel sides each make the same angle with the parallel bases.

I would let one base of the trapezoid rest on the x-axis, with its midpoint at the origin. Then I would write down coordinates for the vertices. Since the trapezoid will be symmetric about the y-axis, you'll have 3 independent variables (1 for each base and 1 for the height). The constraint that the total perimeter is 24 can be used to eliminate 1 variable, which brings you to 2.