# Thread: 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

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

Every solution I see says the most water will fall where $\frac{dx}{dt}= C cos(2y) = 0$, which means $y = 45^o or 135^o$. Since $x$ is the horizontal distance the water travels, this is the angle at which the horizontal distance the water travels is maximized/minimized. But shouldn't most water fall where the velocity is minimized? Shouldn't we be looking at where $\frac{dx}{dt}$ is minimized? In other words, shouldn't we be looking at where $\frac{d^2x}{dy^2} = 0$?

If that is the case, then I get $\frac{d^2x}{dy^2} = -2C sin(2y) \frac{dy}{dt}$ and $y = 90^o$. This seems correct because in my experience these types of sprinklers water most at the sprinkler and least out farthest from the sprinkler.

But then every other person's solution that I have seen is wrong. I'm not trusting that! Am I missing something?

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

x represents the position, or point where the water falls on the ground relative to the sprinkler, not a distance

dx/dt is the rate of change of that position, i.e. velocity of the point where the water hits the ground.

when $\bigg|\dfrac{dx}{dt}\bigg| \to 0$ at the endpoints of the oscillation, the amount of water hitting the ground is increased because the water flow being constant has more time to hit the ground ... as $\bigg|\dfrac{dx}{dt}\bigg|$ reaches a maximum at $\theta=\dfrac{\pi}{2}$, the position where the water hits the ground moves fastest, less time for the constant water flow to hit the position close to the sprinkler.

dx/dt is a minimum at $\theta = \dfrac{\pi}{4}$ and $\theta = \dfrac{3\pi}{4}$ ... note the attached graph. Red indicates velocity to the left from , blue to the right.

Of course, all of this is based on the position model presented in this problem. There may well be oscillating sprinklers that are designed differently. Check out the link for a very detailed analysis on oscillating sprinklers.

Design of an Oscillating Sprinkler.pdf

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### a lawn sprinker

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