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Basic implicit differentiation question

  1. Feb 9, 2016 #1
    So it has been quite a few years since I learnt about implicit differentiation so the content is a bit rusty in my mind.

    x=rcos(θ)

    How do you find dx/dt?

    I know the answer but I am trying to figure out why. I mean dx/dt can be written as (dx/dθ)*(dθ/dt) so why is the answer not just (-rsinθ)*(dθ/dt)?
     
  2. jcsd
  3. Feb 9, 2016 #2

    blue_leaf77

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    You have to know which variables in the RHS which depends on ##t##. By the way, you don't need implicit differentiation in this case since what you want to find is ##dx/dt## and ##x## in your equation has been expressed explicitly in terms of the other variables.
    So you know the right answer? It will be helpful to post it as well.
     
  4. Feb 10, 2016 #3

    Svein

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    Since you have not specified otherwise, I assume that you really have [itex]x(t)=r(t)\cdot \cos(\theta (t)) [/itex]. Then [itex]\frac{dx(t)}{dt}=\frac{dr(t)}{dt}\cdot \cos(\theta(t))+r(t)\cdot\frac{dcos(\theta(t))}{dt}=\frac{dr(t)}{dt}\cdot \cos(\theta(t))+r(t)\cdot(-\sin(\theta(t)))\frac{d\theta(t)}{dt} [/itex]
     
  5. Feb 10, 2016 #4

    HallsofIvy

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    For x a function of two variables, r and [itex]\theta[/itex], where r and [itex]\theta[/itex] are functions of t, the "chain rule" is
    [tex]\frac{dx}{dt}= \frac{\partial x}{\partial r}\frac{dr}{dt}+ \frac{\partial x}{\partial \theta}\frac{d\theta}{dt}[/tex]
    That gives Svein's answer.
     
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