ok I really need help with a related rates problem.

Here it is: Draining a half cylinder tank. A trough is shaped like a half cylinder with length 5 m and radius 1 m. the tank is full of water when a valve is opened and water flows out of the bottom of the tank at a rate of 1.5m^3/hr . Hint: the area of a sector of a circle of radius r subtended by an angle theta is r^2/2 times theta.
a: How fast is the water level changing when the water level is 0.5 m from the bottom of the trough?

Question

ok I really need help with a related rates problem.

Here it is: Draining a half cylinder tank. A trough is shaped like a half cylinder with length 5 m and radius 1 m. the tank is full of water when a valve is opened and water flows out of the bottom of the tank at a rate of 1.5m^3/hr . Hint: the area of a sector of a circle of radius r subtended by an angle theta is r^2/2 times theta.
a: How fast is the water level changing when the water level is 0.5 m from the bottom of the trough?

chrisplusian · Answer

thank you for your help. Two quick questions. #1: Why did you write "sin(theta/2)=(od/oc)=(1-h)/1=1-h????? It is under the part that says "How is theta related to H"

chrisplusian · Answer

The second part of the question would be in relation to the same situation but this time it asks what is the rate of change of the surface area of the water when the water is .5 m deep?

anonymous · Answer

Read the following attached file. Actually you are right, 
it is $$ \sin(\theta/2) = \frac{OD}{OC}
$$