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Implicit differentiation wrt time

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data

    "A lawn sprinkler is constructed in such a way that d[itex]\theta[/itex]/dt is constant, where [itex]theta[/itex] ranges between 45 degrees and 135 degrees. The distance the water travels horizontally is


    where v is the speed of the water. Find dx/dt and explain why this lawn sprinkler does not water evenly. What part of the lawn receives the most water?

    2. Relevant equations

    3. The attempt at a solution

    I took the derivative of the function getting,

    [tex]dx/dt=(2v*sin2\theta(dx/dt) +2v^2cos2\theta(dx/dt))32[/tex]

    I made a mistake because I can factor out dx/dt, then dividing both sides by dx/dt sets the equation to 1 but I don't want that. Where am I making my mistake? It seems to me that my differentiation was correct but I must have made a mistake. If you want I'll walk through it. This answer isn't the same as the one in the back of the book either, I avoid looking at the back usually and forget the answers are even there.
    Last edited by a moderator: Oct 28, 2007
  2. jcsd
  3. Oct 28, 2007 #2


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    Staff Emeritus
    Science Advisor

    Yes, you did make a mistake. This may help; [tex]\frac{d}{dt}(\sin\theta)=\cos\theta\cdot\frac{d\theta}{dt}[/tex]

    Incidentally, isn't [tex]v=\frac{dx}{dt}[/tex]
  4. Oct 28, 2007 #3
    Oh! right, v=dx/dt . So now I have [itex]\frac{dx}{dt}=[2v*sin2\theta + 2cos2\theta \frac{d\theta}{dt}v^2]/32^2

    Is that right? Because we are taking the derivative of cos wrt theta we must multiply by [itex]\frac{d\theta}{dt}[/itex].
    Last edited by a moderator: Oct 28, 2007
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