Application! Sprinklers and water speed.

**TylaYuhh** · September 25th 2011, 12:25 PM

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.

Application! Sprinklers and water speed.-sprinkelrs.jpg

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

**skeeter** · September 25th 2011, 02:51 PM

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.

Click image for larger version.

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

start by taking this derivative w/r to time ...

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

**TylaYuhh** · September 25th 2011, 03:31 PM

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

**TylaYuhh** · September 25th 2011, 03:48 PM

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?

**skeeter** · September 25th 2011, 03:55 PM

Originally Posted by TylaYuhh

w/r = water/rate?

with respect to

**TylaYuhh** · September 25th 2011, 06:08 PM

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

**skeeter** · September 25th 2011, 06:30 PM

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

**TylaYuhh** · September 25th 2011, 06:44 PM

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)?

**skeeter** · September 25th 2011, 06:47 PM

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

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

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

**TylaYuhh** · September 25th 2011, 07:02 PM

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.

**HallsofIvy** · September 26th 2011, 09:01 AM

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.

**TylaYuhh** · September 26th 2011, 02:20 PM

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

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