The height of a cylinder is increasing at a constant rate of 6 feet per second, and the volume is decreasing at a rate of 171 cubic feet per second. at the instant when the height of the cylinder is 6 feet and the volume is 1254 cubic feet, what is the rate of change of the radius? the volume of a cylinder can be found with the equation v = tr²h. round your answer to three decimal places.

The rate of change of the radius with time is a partial derivative of the rate

of change of volume with time.

At the instant the height of the cylinder is 6 feet is the radius is decreasing

at a rate of approximately 4.634 ft./s.

Reasons:

The given parameters are;

Rate of increase of the height of the cylinder, = 6 ft./s

Rate of decrease of the volume of the cylinder, = 171 ft.³/s

Height of the cylinder, h = 6 ft.

Volume of the cylinder, V = 1254 ft.³

The formula for the volume of the cylinder, V = π·r²·h

Solution:

By partial differentiation, we have;

Therefore;

The value of the radius, r, in the given condition is therefore;

Which gives;

Therefore, we get;

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Step-by-step explanation:

This is related rates in calculus, as far as I can tell from the info you've provided.  It's nice to see students are still trying to conquer its vagueness!

Start with the formula for volume of a cylinder:

and then find its derivative with respect to time:

From that formula it looks like we need a value for dV/dt, r, dh/dt, and h; dr/dt is our unknown.  We have that

dV/dt = -171

dh/dt = 6

h = 6 and

V = 1254

We need a value for r.  We can find it using the last 2 values listed above in the volume formula:

and

and

and

r = 8.156394

NOW we have everything we need to fill in the derivative for the volume:

which simplifies to

and

so

ft/sec

if i put something wrong your not gonna like it so rather not say nothing.

570, the answer is 570, hope it