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Chapter 2: Problem 47

The volume of a cantaloupe is given by \(V=\frac{4}{3} \pi r^{3}\) The radius is growing at the rate of \(0.7 \mathrm{cm} /\) week, at a time when the radius is \(7.5 \mathrm{cm} .\) How fast is the volume changing at that moment?

Short Answer

Expert verified
\( 157.5 \pi \text{ cm}^3 \text{ per week} \)

Step by step solution

01

Understand the given formula

The volume of a cantaloupe is given by the formula: \[ V = \frac{4}{3} \pi r^{3} \]
02

Identify the rate of change

The radius is changing at a rate of 0.7 cm/week. This is written as: \[ \frac{dr}{dt} = 0.7 \text{ cm/week} \]
03

Differentiate the volume formula with respect to time

To find the rate of change of the volume, differentiate the volume formula with respect to time (t): \[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^{3} \right) \]
04

Apply the chain rule

Using the chain rule, the differentiation is:\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^{2} \cdot \frac{dr}{dt} \]Simplifying, we get:\[ \frac{dV}{dt} = 4 \pi r^{2} \cdot \frac{dr}{dt} \]
05

Substitute the given values

Substitute \( r = 7.5 \text{ cm} \) and \( \frac{dr}{dt} = 0.7 \text{ cm/week} \):\[ \frac{dV}{dt} = 4 \pi (7.5)^{2} (0.7) \]
06

Calculate the rate of change of the volume

Perform the calculation:\[ \frac{dV}{dt} = 4 \pi (56.25) (0.7) \]\[ \frac{dV}{dt} = 4 \pi (39.375) \]\[ \frac{dV}{dt} = 157.5 \pi \text{ cm}^3 \text{ per week} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume of a Sphere
The volume of a sphere, such as a cantaloupe, can be found using the formula: \[ V = \frac{4}{3} \ \ \pi r^3 \]. This formula stems from geometric principles and helps calculate the space inside a spherical object.
The variables here are:
  • The constant \( \ \pi\ \approx\ \ 3.14159\): It represents the ratio of the circumference of any circle to its diameter.

  • \( \ r \): The radius of the sphere, which is the distance from the center to any point on its surface.
Derivatives
Derivatives represent how a function changes as its input changes. In our cantaloupe problem, we want to find how the volume changes as the radius changes. Mathematically, this is the derivative of volume, \( V \), with respect to the radius, \( r \).
For the volume formula, the derivative \( \frac{dV}{dr} \) tells us how fast the volume is changing at any radius.
  • To find \( dV \) (the change in volume): Differentiate \( \frac{4}{3}\ \ \pi r^3 \) with respect to \( r \).
  • \( \frac{dV}{dr} = 4 \ \pi r^2 \): This gives us the rate of change of volume concerning the radius.
Chain Rule
The chain rule is a fundamental concept in calculus used for differentiating composed functions. In our case, we want to find \( \ \frac{dV}{dt} \), the rate of volume change over time.
Given that \( r \) is a function of \( t \) (time), we can use the chain rule:
  • Write \( \ \ \frac{dV}{dt} \) as \( \frac{dV}{dr}\ \cdot \frac{dr}{dt} \)

From our volume formula's derivative, we already have \( \frac{dV}{dr} = 4 \ \pi r^2 \). We also know \( \ \ \frac{dr}{dt} = 0.7 \text{ cm/week} \).
So, \( \frac{dV}{dt} = 4 \ \pi r^2 \cdot \ 0.7 \).
Differentiation with respect to Time
Differentiation with respect to time involves understanding how a quantity changes over time.
In our example with the cantaloupe:
  • The radius is growing at a rate of 0.7 cm per week, meaning \( \frac{dr}{dt} = 0.7 \text{ cm/week} \).

To find how fast the volume is changing at a specific time when \( r \) is 7.5 cm, we:
  • First differentiate the volume formula with respect to time using the chain rule.
  • Next, substitute the given values into our differentiated formula: \( r = 7.5 \ \text{ cm} \ and \聽 \ \ \frac{dr}{dt} = 0.7 \text{ cm/week} \).
  • Finally, calculate \( \ \frac{dV}{dt} = 4 \ \pi (7.5)^2 (0.7) = 157.5 \ \pi \text{ cm}^3 \text{ per week}\).

This final value represents the rate at which the volume of the cantaloupe is changing with time when the radius is 7.5 cm.

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