Rate question of cylinder

Question

The radius and height of a circular cylinder are changing with time in such a way that the volume remains constant at 1 liter (=1000 cubic cm). If, at a certain time, the radius is 4 centimeters and is increasing at the rate of 1/2 cm per second, what is the rate of change of the height

Answer 1

Volume of cylinder V = πr²h, where r is radius of base and h is height. Let differentiate this equation by time. Because V is const V ' = 0. After differentiation of formula for volume we get: 0 = π(2rr'h + r²h'). bFrom here

h' = - 2r'h/r. By conditions r = 4 cm, r' = 1/2 cm/sec; h = V/(πr²) = 1000 cm³/(16π cm²) = 125/(2π) cm.

Becausr h' = - 2·1/2 cm/sec·125/(2π) cm/4 cm = - 125/(8π) cm/sec ≈ - 4.97 cm/sec

Answer: height of cylinder decreasing with rate h' ≈ - 4.97 cm/sec