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Homework Help: Load of math help needed, come rescue me!

  1. Sep 20, 2005 #1
    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?
     
  2. jcsd
  3. Sep 20, 2005 #2
    yup, it's dx/dy
     
  4. Sep 20, 2005 #3

    Tom Mattson

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    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.
     
  5. Sep 20, 2005 #4
    Eliminating z is what I tried first, except the answer in the book is z-y+tan(y+z)
     
  6. Sep 20, 2005 #5

    Tom Mattson

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    OK, one other question: Are you familiar with partial derivatives?
     
  7. Sep 20, 2005 #6
    Yah

    that's what this chapter is, in fact
     
  8. Sep 20, 2005 #7

    Tom Mattson

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    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.
     
    Last edited: Sep 20, 2005
  9. Sep 20, 2005 #8

    Tom Mattson

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    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.
     
  10. Sep 20, 2005 #9
    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?
     
  11. Sep 20, 2005 #10

    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.
     
  12. Sep 20, 2005 #11

    HallsofIvy

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    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).
     
  13. Sep 20, 2005 #12
    oh, haha, I solved for z and then took the derivative of some arcsin
     
  14. Sep 20, 2005 #13
    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?
     
  15. Sep 20, 2005 #14

    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?
     
  16. Sep 20, 2005 #15
    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
     
  17. Sep 20, 2005 #16

    Tom Mattson

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    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.
     
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