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Differentiating with respect to time

  1. Jan 3, 2007 #1
    How do you go about differentiating rcos$ with respect to time....?

    In the book im studying from it says d$/dt d/d$ (rcos$) is the process to find the answer... but what does this mean...?
  2. jcsd
  3. Jan 3, 2007 #2


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    HAve you come across the chain rule? This states that [tex]\frac{df}{dt}=\frac{df}{dx}\frac{dx}{dt} [/tex]. Thus, applying this to the case you mention, we obtain [tex] \frac{d}{dt}(rcos\theta)=\frac{d}{d\theta}(rcos\theta)\frac{d\theta}{dt} [/tex]
  4. Jan 3, 2007 #3
    The chain rule has been applied in arriving at the result.

    [tex]\frac{df}{dt} = \frac{df}{d\theta}\frac{d\theta}{dt}[/tex], where f denotes the given function.

    Edit: Since Cristo posted while I was playing with the typeset, I'm using theta instead of the more obvious x. :biggrin: Apparently, the '$' symbol messes up Latex.
    Last edited: Jan 3, 2007
  5. Jan 3, 2007 #4


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    If [itex]\theta[/itex] is a function of time but r does not, then it is true that [itex]frac{df}{dt}= \frac{df}{d\theta}\frac{d\theta}{dt}[/itex].

    If both f depends on both r and [itex]\theta[/itex] and both r and [itex]\theta[/itex] depend upon time,
    [tex]\frac{df}{dt}= \frac{\partial f}{\partial r}\frac{dr}{dt}+ \frac{\partial f}{\partial \theta}\frac{d\theta}{dt}[/tex]
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