# Math Help - Application! Sprinklers and water speed.

1. ## Application! Sprinklers and water speed.

Here is the exact problem:

Lawn Sprinkler: A lawn sprinkler is constructed in such a way that $\frac{dy}{dt}$ is constant, where y ranges between 45(degrees) and 135(degrees). The distance the water travels horizontally is:

x = (V^2 sin(2y))/32

where v is the speed of the water. Find $\frac {dx}{dt}$ and explain why the water does not water evenly. What part of the lawn receives the most water?

Here is the image.

If anyone could walk me through this, that would be awesome!

2. ## Re: Application! Sprinklers and water speed.

Originally Posted by TylaYuhh
Here is the exact problem:

Lawn Sprinkler: A lawn sprinkler is constructed in such a way that $\frac{dy}{dt}$ is constant, where y ranges between 45(degrees) and 135(degrees). The distance the water travels horizontally is:

x = (V^2 sin(2y))/32

where v is the speed of the water. Find $\frac {dx}{dt}$ and explain why the water does not water evenly. What part of the lawn receives the most water?

Here is the image.

If anyone could walk me through this, that would be awesome!
start by taking this derivative w/r to time ...

$\frac{v^2}{32} \cdot \frac{d}{dt}[\sin(2y)]$

3. ## Re: Application! Sprinklers and water speed.

Thanks I am going to give that a shot. I'll let you know my results!

4. ## Re: Application! Sprinklers and water speed.

Originally Posted by skeeter
start by taking this derivative w/r to time ...

$\frac{v^2}{32} \cdot \frac{d}{dt}[\sin(2y)]$
w/r = water/rate?

5. ## Re: Application! Sprinklers and water speed.

Originally Posted by TylaYuhh
w/r = water/rate?
with respect to

6. ## Re: Application! Sprinklers and water speed.

Originally Posted by skeeter
with respect to
Oh okay.
I think I did this right:

x' = (32(2v sin(2y) + 2v^2 cos(2y))) / 32^2

7. ## Re: Application! Sprinklers and water speed.

Originally Posted by TylaYuhh
Oh okay.
I think I did this right:

x' = (32(2v sin(2y) + 2v^2 cos(2y))) / 32^2
v is a constant

8. ## Re: Application! Sprinklers and water speed.

Originally Posted by skeeter
v is a constant

Oops! I hope I got it right this time.

x' = 32(2v^2 cos(2y)) / 32^2

I think I messed up here too though. If it is dx/dy, that means I've got to use implicit differentiation on the sin(2y)?

9. ## Re: Application! Sprinklers and water speed.

$x = \frac{v^2}{32} \cdot \sin(2y)$

$\frac{dx}{dt} = \frac{v^2}{16} \cdot \cos(2y) \cdot \frac{dy}{dt}$

$\frac{v^2}{16}$ and $\frac{dy}{dt}$ are both constant ... what happens to $\frac{dx}{dt}$ as the angle $y$ of the sprinkler varies?

10. ## Re: Application! Sprinklers and water speed.

Originally Posted by skeeter
$x = \frac{v^2}{32} \cdot \sin(2y)$

$\frac{dx}{dt} = \frac{v^2}{16} \cdot \cos(2y) \cdot \frac{dy}{dt}$

$\frac{v^2}{16}$ and $\frac{dy}{dt}$ are both constant ... what happens to $\frac{dx}{dt}$ as the angle $y$ of the sprinkler varies?
The water is spread out further?

Sorry, I'm just having trouble with this one D: ...Thanks for helping me on this.

11. ## Re: Application! Sprinklers and water speed.

The fact that, as skeeter says, $\frac{v^2}{16}$ and $\frac{dy}{dt}$ are both constant, tells you that the entire $C= \frac{v^2}{16}\frac{dy}{dt}$ is a constant and so $\frac{dx}{dt}= C cos(2y)$. That is not a constant because y is not a constant and so cos(2y) is not a constant.

12. ## Re: Application! Sprinklers and water speed.

Originally Posted by HallsofIvy
The fact that, as skeeter says, $\frac{v^2}{16}$ and $\frac{dy}{dt}$ are both constant, tells you that the entire $C= \frac{v^2}{16}\frac{dy}{dt}$ is a constant and so $\frac{dx}{dt}= C cos(2y)$. That is not a constant because y is not a constant and so cos(2y) is not a constant.

Thank you for the input. It helped =D