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Related rate problem

  1. Nov 2, 2012 #1
    1. The problem statement, all variables and given/known data
    For a related rate problem I am only given the two rates say dx/dt=1 and dy/dt=2 and a the function which depends on both x and y say f(x,y)=4xy^2 find how it is changing at T=3


    2. Relevant equations
    Chain rule


    3. The attempt at a solution

    So I can take both partial derivatives with respect to the chain rule and sub in 1 and 2 for the dx/dt and dy/dt respectively I just have no idea how T comes into play and what to sub in for x and y. My thoughts are that possibly since dx/dt=1 and dy/dt=2 then y=2x and I can isolate for the one variable but then I still dont know what to do form there.
     
  2. jcsd
  3. Nov 2, 2012 #2
    Those two tell you quite a bit about x and y as functions of time :)
     
  4. Nov 2, 2012 #3
    A bit more of a pointer would be much appreciated.
     
  5. Nov 2, 2012 #4

    HallsofIvy

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    Staff Emeritus
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    The chain rule says that
    [tex]\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}[/tex]

    If [itex]f(x,y)= 4xy^2[/tex], what are [itex]\partial f/\partial x[/itex] and [itex]\partial f/\partial y[/tex]?

    Now, if you are only told dx/dt= 1 and dy/dt= 2, then you do NOT have enough information to determine what x and y are at t= 3 and so could NOT evaluate the partial derivatives at t= 3.

    If you have a particular problem in mind then please tell us the exact statement of the problem.
     
  6. Nov 2, 2012 #5

    It states a cone grows in height starting from zero height dy/dt = 1 radius dx/dt = 2 how fast is the volume growing at t=3
     
  7. Nov 2, 2012 #6
    You need to find df/dt at t=3, so the first step is to perform the differentiation, as stated above. You'll find the result has x and y in it and to find a value for df/dt at a specific time, you will need to know x and y as functions of time.
     
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