Suppose a tank is made of glass and has the shape of a right-circular cylinder of radius 1 ft. Assume that h(0) = 2 ft corresponds to water filled to the top of the tank, a hole in the bottom is circular with radius in., g = 32 ft/s2, and c = 0.6. Use the differential equation in Problem 12 to find the height h(t) of the water.
Answer:
Height of the water = √(128)/147456 ft
Explanation:
Given
Radius, r = 1 ft
Height, h = 2 ft
Radius of hole = 1/32in
Acceleration of gravity, g = 32ft/s²
c = 0.6
Area of the hold = πr²
A = π(1/32)² ---- Convert to feet
A = π(1/32 * 1/12)²
A = π/147456 ft²
Area of water = πr²
A = π 1²
A = π
The differential equation is;
dh/dt = -A1/A2 √2gh where A1 = Area of the hole and A2 = Area of water
A1 = π/147456, A2 = π
dh/dt = (π/147456)/π √(2*32*2)
dh/dt = 1/147456 * √128
dh/dt = √128/147456 ft
Height of the water = √(128)/147456 ft
