Answer:
The rate at which the height of the water tank is changing is approximately 0.4244 ft/hour
Step-by-step explanation:
The given parameters are;
The diameter of the cone = 10 feet
The height of the cone = 15 feet
The rate at which water is leaking from the tank, (dV/dt) = 12 ft³/h
The volume of water in the tank = 27·π cubic feet
The volume V of a right circular cone with radius r and height h = 1/3×π×r²×h
The rate of change of the volume, dV, with time dt is given as follows;
The radius of the cone when the volume of the water in the tank is 27·π cubic feet is given as follows;
1/3×π×r²×h = 27·π ft³
The ratio of the height to the radius of the cone is h/r = 15/5 = 3
h/r = 3
∴ r =h/3
The volume of the cone, V = 1/3×π×r²×h = 1/3×π×(h/3)²×h = 1/27×π×h³ = h³/27×π
dV/dt = dV/dh × dh/dt
Which gives;
12 = d(h³/27×π)/dh × dh/dt = π×h²/9 × dh/dt
dh/dt = 12/(π×h²/9)
At 27·π ft³ = 1/27×π×h³ ft³, we have;
27 = 1/27×h³
27² = h³
h = ∛27² = 9
∴ dh/dt = 12/(π×h²/9) = 12/(π×9²/9) = 12/(π×9) = 4/(3·π) ≈ 0.4244 ft/hour
The rate at which the height of the water tank is changing ≈ 0.4244 ft/hour.
