Math Problem Statement

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.

Solution

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=43πr3V = \frac{4}{3} \pi r^3

Where:

  • VV is the volume,
  • rr is the radius of the cantaloupe,
  • and π\pi is a constant.

We are given that the radius is increasing at a rate of drdt=0.2cm/week\frac{dr}{dt} = 0.2 \, \text{cm/week}, and at the moment of interest, the radius r=8.6cmr = 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 dVdt\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 tt using the

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Math Problem Analysis

Mathematical Concepts

Calculus
Related Rates
Derivatives

Formulas

V = (4/3)πr^3
dV/dt = 4πr^2(dr/dt)

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12