A trough is shaped like a half cylinder with length 5 m and

A trough is shaped like a half cylinder with length 5 m and radius 1 m. The trough is full of water when a valve is opened and water flows out of the bottom of the trough at a rate of 6.5m3/hr. (Hint: The area of a sector of a circle of radius r subtended by an angle alpha is r2 alpha / 1.) Complete parts (a) and (b) below. How fast is the water level changing when the water level is 0.5 m from the bottom of the trough? dh / dt = m / hr (Type an exact answer using radicals as needed.) What is the rate of change of the surface area of the water when the water is 0.5 m deep? Determine an equation for the surface area of the water given the graph to the right.

Solution

a. volume of water when level is 0.5 below the top= (2*pi*l*r^2)/3 - r^2 *l*sin(2*60)
here r=2h, where h is the height from the top.
as h=r cos(alpha) here alpha=120 degree=2 pi/3.. and l=5m.
so volume=33.25*h^2
== dv/dt = 33.25*2*h*dh/dt.
n as we know the value of dv/dt.
dh/dt=0.195.

b. area =2*r*sin (alpha)*l

again r=2*h, alpha=60.

da/dt will be calculated using dh/dt.

da/dt=10*1.73