Related Rates (Differentiation)?

Question

. A draining conical reservoir Water is flowing at the rate of 50m^3/min
from a shallow concrete conical reservoir (vertex
down) of base radius 45 m and height 6 m.
a. How fast (centimeters per minute) is the water level falling
when the water is 5 m deep?
b. How fast is the radius of the water’s surface changing then?
Answer in centimeters per minute.

Answer 1

(a).

The radius of the cone is 45m, height = 6m and rate change of volume is 50 m³/min.

The volume of the cone is V = (1/3) πr²h.

The relation between r and h: r_cone / h_cone = r_water / h_water.

6/45 = r_water / h_water ⇒ r = 2h/15.

Substitute r = 2h/15 in the volume formula.

V = (1/3) π(2h/15)²h

V = (π/3) (4h³/225)

V = (4π/675) h³,

Differentiate with respect to time t.

dV/dt = (4π/675) 3h² dh/dt.

dV/dt = (4π/225) h² dh/dt.

Substitute dV/dt = 50 m/min, h = 5 in the above equation.

50 = (4π/225) 5² dh/dt.

25 = 2π/9 dh/dt.

dh/dt = 225/2π = 35.7954545 m/min = 3579.545 cm/min.

The rate change of height of the water is 3579.545 cm/min.

 

Answer 2

(b).

The radius of the cone is 45m, height = 6m and rate change of volume is 50 m³/min.

The volume of the cone is V = (1/3) πr²h.

The relation between r and h: r_cone / h_cone = r_water / h_water.

6/45 = r_water / h_water ⇒ h = 15r/2.

If h = 5 then r = 2h/15 = 2(5)/15 = 2/3.

Substitute h = 15r/2 in the volume formula.

V = (1/3) πr²(15r/2)

V = (5π/2) r³.

Differentiate with respect to time t.

dV/dt = (5π/2) 3r² dr/dt.

dV/dt = (15π/2) r² dr/dt.

Substitute dV/dt = 50 m/min, r = 2/3 in the above equation.

50 = (15π/2) (2/3)² dr/dt.

5 = π/3 dr/dt.

dr/dt = 15/π = 4.7727 m/min = 477.273 cm/min.

The rate change of radius of the water is 477.273 cm/min.