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

a is the best answer your welcome

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west