Question

If someone can help me with this that would be greatly appreciated, but if no one can that's okay.

Related Rate Problem:

A hemispherical tank with a radius of 10 m is filled from an inflow pipe at a rate of 3 m^3/min. (Hint: The volume of a cap of thickness h sliced from a sphere of radius r is ∏h^2(3r-h)/3)

a. how fast is the water level rising when the water level is 5 m from the bottom of the tank?

b. what is the rate of change of the surface area of the water when the water is 5 m deep?

Answer 1

When depth is h, volume in cap = V = πh²(30−h)/3 = 10πh² − πh³/3

The rate at which the water level is rising = dh/dt. To find it we differentiate above with respect to t. Note for this purpose both V & h are regarded as functions of t and so we use the chain rule “ df/dt = f'.dh/dt “ to get the t-derivative of any function f of h.

dV/dt = (20πh − πh²).dh/dt

The question gives dV/dt=3 when h=5 so we can use this equation to get dh/dt.

3 = (100π−25π).dh/dt → dh/dt = 1/(25π) = 0.0127 m/min

(b) If we can find a formula connecting surface area S and h then we can differentiate again. This will gives us a link between dS/dt and dh/dt. Since we now know dh/dt, dS/dt will follow.

Draw a diagram of the central vertical cross-section and apply Pythagoras to find the radius R of the surface disc.

10² = R² + (10−h)² → R² = 20h−h²

Surface area of water, S = πR² = π(20h−h²)

Differentiate wrt t : dS/dt = π(20−2h).dh/dt = π*10*{1/(25π)} = 0.4 m²/min

( I used the π version of dh/dt because it was obvious the π would cancel )

Answer 2

got it ?

Answer 3

yeah! i just need to read over it a couple of times lol. thank you so much for your help!