The volume of a cantaloupe with radius r is given by .V =

-r. The radius is

growing at a at a rate of 0.2 cm/week, at a time when the radius is 8.6 cm. How fast is the volume changing at that moment? Round your answer to 2 decimal places.

It seems like the volume equation for the cantaloupe is incomplete in your query. However, I assume the volume is modeled as a sphere, so the volume formula would likely be:

\[
V = \frac{4}{3} \pi r^3
\]

Where:
- \( V \) is the volume,
- \( r \) is the radius of the cantaloupe,
- and \( \pi \) is a constant.

We are given that the radius is increasing at a rate of \( \frac{dr}{dt} = 0.2 \, \text{cm/week} \), and at the moment of interest, the radius \( r = 8.6 \, \text{cm} \).

To find how fast the volume is changing at that moment, we need to compute the rate of change of the volume \( \frac{dV}{dt} \). We can do this using related rates.

### Step 1: Differentiate the volume equation

We begin by differentiating the volume formula with respect to time \( t \) using the

The volume of a cantaloupe with radius r is given by V = (4/3)πr^3. The radius is growing at a rate of 0.2 cm/week, at a time when the radius is 8.6 cm. How fast is the volume changing at that moment? Round your answer to 2 decimal places.

Learn how to calculate the rate at which the volume of a sphere changes when its radius grows at a constant rate. This problem applies calculus concepts, including related rates and the chain rule, with a step-by-step solution. Suitable for Grades 11-12.

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