1. ## Differential Equations

A water trough, 6 metres long, has uniform cross section in the shape of a semicircle with horizontal diameter 2 metres. It contains water to a depth of x metres. THe centre of the circle is at O and the triangle POQ is isosceles with angle POQ = @.

I've managed to get parts. But am stuck..

a. Find the area of the segment that is filled with water.
A = 0.5(@-sin@)

b. Find the volume of water in trough.
V = 6*A = 3(@-sin@)

c. If x metres is the depth of the water in the trough, find
i. x in terms of @
ii. dx/d@
iii. dx/dt in terms of @ and d@/dt

d. Find dV/dt.
dV/dt = 3(1-cos@)*(d@/dt)

e. If d@/dt = 0.05 rad/min, when @ = pie/3, find
i. Rate in m^3 per min at which volume of water is increasing.
ii. Rate in metres per min at which water level is rising.

2. May I ask, why would someone in a calculus or DE class spell $\pi$ as 'pie'?.

This is a math forum, not a baking forum

The depth of the water in the trough is the 'middle ordinate' of the segment.

It is given by $x=R(1-cos(\frac{\theta}{2}))=1-cos(\frac{\theta}{2})$