1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differentiating with multiple variables

  1. Mar 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Find dw/dt. Check result by substitution and differentiation

    w = (x^2 + y^2)^1/2, x = e^2t , y = e^-2t

    2. Relevant equations

    3. The attempt at a solution
    dx/dw = x/(x^2 + y^2)^1/2 dy/dw = y/(x^2 + y^2)^1/2

    Dont really know where to go with it
    Last edited: Mar 8, 2008
  2. jcsd
  3. Mar 8, 2008 #2
    One way would be to substitute for x and y, which would result in a function of t alone.
  4. Mar 8, 2008 #3


    User Avatar
    Science Advisor
    Homework Helper

    That's the way they suggest to check the answer. They actually do want you to use partial derivatives, mattb8818. Use the chain rule.
  5. Mar 9, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    Hi matt! :smile:

    First, it's not dx/dw and dy/dw, it's t'other way up!:

    dw/dx = x/(x^2 + y^2)^1/2 dw/dy = y/(x^2 + y^2)^1/2

    Does that help? :smile:
  6. Mar 9, 2008 #5
    since w is a function of x and y
    and x and y are functions of t
    the chain rule is

    dw/dt = dw/dx dx/dt + dw/dy dy/dt

    compute everything above and you're set
    an easy way to remember the chain rule is to draw a tree diagram

    x y
    t t
  7. Mar 9, 2008 #6
    Thanks for the replies

    I did it with the chain rule and got

    (2xe^2t)/(x^2 + y^2)^1/2 + -(2e^-2t)/(x^2 + y^2)^1/2

    The answer is 2(s)^1/2(sinh4t)/(cosh4t)^(1/2)

    I dont really understand this solution
  8. Mar 9, 2008 #7


    User Avatar
    Science Advisor
    Homework Helper

    Hi matt!

    I don't know what you did to get 2(s)^1/2(sinh4t)/(cosh4t)^(1/2). :confused:
    Start from that line (which is correct, except you missed out a "y");
    then re-write it as:
    [(2xe^2t) - (2ye^-2t)]/√(x^2 + y^2)
    (this is both to simplify it, and to lessen the risk of making a mistake)
    and just substitute for x and y …

    so what is the next line? :smile:
    [size=-2](btw, if you type alt-v, it prints √ )[/size]​
  9. Mar 9, 2008 #8
    Ok thanks all

    I am glad I atleast did it right. I looked it up and e^x-e^-x/2 = sinh or something like that, but I guess that was my only mistake so Im happy.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?