The radius of a right circular cylinder is increasing at the rate of 7 in./s, while the height is decreasing at the rate of 3 in./s. At what rate is the volume of the cylinder changing when the radius is 17 in. And the height is 5 in.?

1014.73 in³/s

Step-by-step explanation:

We know the volume of a cylinder, V = πr²h where r = radius of cylinder and h = height of cylinder.

Now, to find the rate of change of volume, we differentiate it with respect to time. So

dV/dt = d(πr²h)/dt

= πhdr²/dr × dr/dt + πr²dh/dh × dh/dt

= 2πrh × dr/dt + πr² × dh/dt

when r = 17 in, dr/dt = rate of change of radius = + 7 in/s, h = 5 in and dh/dt = rate of change of height = - 3in/s (negative since it is decreasing).

So, dV/dt = 2πrh × dr/dt + πr² × dh/dt

= 2π × 17 in × 5 in × (+ 7 in/s) + π(17 in)² × (- 3 in/s)

= 1190π - 867π

= 323π in³/s

= 1014.73 in³/s

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!

step-by-step explanation: