The radius of a cylinder is decreasing at a constant rate of 4 feet per minute, and the volume is decreasing at a rate of 841 cubic feet per minute. At the instant when the height of the cylinder is

9

9 feet and the volume is 513 cubic feet, what is the rate of change of the height? The volume of a cylinder can be found with the equation

V

=

π

r

2

h

.

V=πr

2

h. Round your answer to three decimal places.

The rate of change of height = 19.34 ft/min

Step-by-step explanation:

The Volume of a cylinder (v) = πr²h ... equation 1

Where r = radius of the base, h = height of the cylinder.

Using chain rule,

dv/dt = (dv/dr). (dr/dt) ... equation 2

Where dv/dt = rate of change of the volume, dv/dr = differentiation of the volume of cylinder with respect to the radius, dr/dt = rate of change of the radius.

dv/dt = 841 ft³/min,

dr/dt = 4 ft/min,

dv/dr = differentiation of equation 1 with respect to r = 2πrh

where h = 9 ft. ∴ dv/dr = 2 (3.143*9) r

dv/dr = 56.574r

Substituting these values into equation 2,

841 = 56.57r (4)

841 = 226.296r

226.296r = 841

dividing both sides of the equation by the coefficient of r,

226.296r/226.296 = 841/226.296

r = 3.72 ft.

Also using chain rule for the rate of change of the height,

dv/dt = (dh/dt) (dv/dh) ... equation 3

∴ dh/dt = (dv/dt) / (dv/dh) ... equation 4

Where dv/dt = rate of change of volume = 841 ft³/min,

dv/dh = differentiation of volume with respect to height = πr² (differentiate equation 1 with respect to height.)

dv/dh = 3.143 (3.72) ² = 43.49 ft

applying the values above into equation 4

∴dh/dt = 841/43.49 = 19.34 ft/min

The rate of change of height = 19.34 ft/min