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 @
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.