The height of a cylinder is decreasing at a constant rate of 8 inches per minute, and the volume is decreasing at a rate of 161 cubic inches per minute. at the instant when the height of the cylinder is 66 inches and the volume is 919 cubic inches, what is the rate of change of the radius? the volume of a cylinder can be found with the equation v=\pi r^2 h. v=πr 2 h. round your answer to three decimal places.

0.056 inches per minute

Explanation:

dh/dt = 8 inches per minute

dV/dt = 161 cubic inch per minute

h = 66 inches

V = 919 cubic inch

dr/dt = ? rate of change of radius

The volume of cylinder is given by

V = πr²h

where, r be the radius of cylinder

Differentiate both sides with respect to t

dV/dt = πr² x dh/dt + πh x 2r dr/dt (1)

When h = 66 inches, V = 919 cubic inches

So, 919 = 3.14 x r² x 66

r = 2.11 inch

Substitute the values in equation (1)

161 = 3.14 x 2.11 x 2.11 x 8 + 3.14 x 66 x 2 x 2.11 x dr/dt

dr/dt = 0.056 inches per minute

found in the nucleus

has mass of one amu