Question

Water runs into a sunken concrete hemispherical bowl of radius 5 m at the rate of 0.2 m^3/sec. How fast is the water level in the bowl rising when the water is 4 m deep?

Answer 1

Rate at which the water is filled into the hemisphere = 0.2 m^3/sec = dV/dt

Radius of the hemisphere = r = 5 m

You are asked to find dh/dt at h = 4 m  

dh/dt = rate of increase of height.

dV/dt = 0.2  (by integrating,we get)

V = ∫ 0.2 dt  

V = 0.2t + c  (c is the constnat)

When V = 0, t = 0; therefore c = 0  

V = 0.2t  

Volume of hemisphere with respect to height is give as

V = π(3rh² - h³) / 3  

Equating these expressions for V:  

0.2t = π(3rh² - h³) / 3  

t = (π / 3 * 0.2)(3rh² - h³)  

Substituting r we get,  

t = (π / 0.6)(15h² - h³)  

dt/dh = (π / 0.6)(30h - 3h²)  (Differentiating with height, we get)

When h = 4  

dt/dh = (π / 0.6)(120 - 48)  

dt/dh = (π / 0.6)(72)  

dt/dh = 120π  

dh/dt = 1 / 120π m/sec