Implicit differentiation wrt time

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SUMMARY

The discussion focuses on implicit differentiation with respect to time, specifically analyzing the horizontal distance water travels from a lawn sprinkler described by the equation x = v²sin(2θ)/32, where θ varies between 45 and 135 degrees. The user initially misapplied the chain rule in their differentiation attempt, leading to confusion about the relationship between dx/dt and the derivative of sine and cosine functions. The correct expression for dx/dt was clarified as dx/dt = [2v*sin(2θ) + 2cos(2θ)(dθ/dt)v²]/32², emphasizing the importance of including dθ/dt when differentiating trigonometric functions.

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Homework Statement



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

x=v^2sin(2\theta)/32

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?

Homework Equations



The Attempt at a Solution



I took the derivative of the function getting,

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

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

Incidentally, isn't v=\frac{dx}{dt}
 
Oh! right, v=dx/dt . So now I have \frac{dx}{dt}=[2v*sin2\theta + 2cos2\theta \frac{d\theta}{dt}v^2]/32^2<br />

Is that right? Because we are taking the derivative of cos wrt theta we must multiply by \frac{d\theta}{dt}.
 
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