At the instant when the radius of a cone is 3 inches, the volume of the cone is increasing at the rate of 28.27433388 cubic inches per minute. If the height is always 3 times the radius, find the rate of change of the radius at that instant?

For a cone, V = (1/3)(pi)(r^2)(h).

But h=3r, so V = (pi)r^3.

Differentiate with respect to time.

dV/dt = (3pi)(r^2)(dr/dt)

dr/dt = (3pi)(r^2)/(dV/dt)

To find the rate of change of the radius at a given instant, we can use related rates.

First, let's set up the given information in terms of variables. Let's denote the radius as r (in inches), the height as h (in inches), and the volume as V (in cubic inches).

Given:
- Radius (r) = 3 inches
- Volume (V) is increasing at a rate of dV/dt = 28.27433388 cubic inches per minute
- Height (h) is always 3 times the radius, so h = 3r

We know the formula for the volume of a cone is V = (1/3) * π * r^2 * h.

Differentiating both sides of this equation with respect to time (t), we get dV/dt = (1/3) * π * 2r * (dh/dt).

Now, let's plug in the given values and re-arrange the equation to solve for dh/dt (rate of change of the height):

28.27433388 = (1/3) * π * 2(3) * (dh/dt)

Simplifying:

28.27433388 = (2/3) * π * 6 * (dh/dt)

28.27433388 = 4π * (dh/dt)

To find dh/dt (rate of change of the height), divide both sides by 4π:

28.27433388 / (4π) = dh/dt

Now, we can evaluate this expression to find the rate of change of the height.

Note: Since the problem asks for the rate of change of the radius, we need to use the relationship between the radius and the height (h = 3r) to find dr/dt.

Differentiating both sides of the equation h = 3r with respect to time (t), we get dh/dt = 3(dr/dt).

Now, substituting the value we found for dh/dt into this equation, we have:

28.27433388 / (4π) = 3(dr/dt)

Rearranging the equation to solve for dr/dt (rate of change of the radius):

dr/dt = (28.27433388 / (4π)) / 3

Evaluating this expression will give us the rate of change of the radius at the instant when the radius of the cone is 3 inches.