# Implicit differentiation wrt time

1. Oct 28, 2007

### rook_b

1. The problem statement, all variables and given/known data

"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?

2. Relevant equations

3. 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.

Last edited by a moderator: Oct 28, 2007
2. Oct 28, 2007

### cristo

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

Incidentally, isn't $$v=\frac{dx}{dt}$$

3. Oct 28, 2007

### rook_b

Oh! right, v=dx/dt . So now I have $\frac{dx}{dt}=[2v*sin2\theta + 2cos2\theta \frac{d\theta}{dt}v^2]/32^2$

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

Last edited by a moderator: Oct 28, 2007