lawn sprinkler problem

[Guest]

Hopefully somebody can help with this one. I can't seem to find anybody that knows how to do this one. It is in a chapter on differentiation/ extrema on an interval. The problem is:

a sprinkler is constructed in such a way that dtheta/dt is constant, where theta ranges between 45 degrees & 135 degrees. the distance the water travels horizontally is

x=(v^2sin2theta)/32 where v is the speed of water. find dx/dt & explain why sprinkler does not water evenly. what part of lawn receives most water.

This has a diagram but I can't link it since its not a public website...It shows the water moving at theta = 45 deg (V^2/32), 75 deg (V^2/64), 105 deg (-V^2/64) and 135 deg (-v^2/32).

I have dx/dt=v^2[32(cos2theta)(2)-sin2theta(0)]/(32)^2= v^2cos2theta/16(dtheta/dt). I am not sure where to go from here to solve the question.
Can anybody provide some insight????

Thanks

 

[daon]

What do you mean by the "distance traveled horizontally"?

I imagine this as a graph with the sprinkler at the origin, so do you mean the distance the water lands from the origin at a given theta? Please clarify.

Also, are we to assume that the lawn occupies exactly which parts of the graph receive water, or is the lawn a predefined shape such as a portion of a circle?

edit: Another point that need clarification: You say that d(theta)/dt is constant. But that would mean the "sprinkler head" would have to travel in a full circle, since it never switches to negative (opposite direction). From real-life experience, I know that sprinklers change direction. Let me know.

 

[Guest]

the sprinkler is a oscillating bar sprinkler that oscillates between 45 and 135 degrees. looking at the graph (crosssection), we could say that the lawn is on the x-axis and the oscillating sprinkler is on the y-axis. the horizontal distance travelled is where the water hits the lawn at theta.