Question

Inflating a Balloon The volume \(V=(4 / 3) \pi r^{3}\) of a spherical balloon changes with the radius (a) At what rate does the volume change with respect to the radius when \(r=2 \mathrm{ft} ?\) (b) By approximately how much does the volume increase when the radius changes from 2 to 2.2 \(\mathrm{ft}\) ?

Short answer

a) The volume of the balloon is increasing at a rate of \(16\pi ft^{3}/ft\) when the radius is 2 feet. b) The volume of the balloon approximately increases by \(3.2\pi ft^{3}\) when the radius changes from 2 to 2.2 feet.

Step-by-step solution

Calculate the derivative of the volume equation

Start by deriving the volume formula with respect to the radius \( r \): \( \frac{dV}{dr} = \frac{d}{dr} \left( \frac{4}{3} \pi r^{3} \right) \). Using the power rule, you get \( \frac{dV}{dr} = 4\pi r^{2} \). The derivative \( \frac{dV}{dr} \) gives us the rate at which the volume changes with respect to the radius.

Evaluate the derivative at \(r = 2 ft\)

Substitute \( r = 2 ft \) into the derivative to find the rate of change of the volume at this radius: \( \frac{dV}{dr} = 4\pi (2)^{2} = 16\pi ft^{3}/ft \). This means that when the radius is 2 feet, the volume of the balloon is increasing at a rate of \(16\pi ft^{3}/ft\).

Approximate the change in volume

Use the expression for \( \frac{dV}{dr} \) to approximate the volume change when the radius changes from 2 to 2.2 feet. The change in volume \( \Delta V \approx \frac{dV}{dr} \Delta r \), where \( \Delta r = 2.2 - 2 = 0.2 ft \). Substituting, \( \Delta V \approx 16\pi (0.2) = 3.2\pi ft^{3} \). This is the approximate change in volume when the radius changes from 2 to 2.2 feet.

Key concepts

Derivative of Volume

Understanding the derivative of volume is essential when dealing with changing three-dimensional spaces, such as inflation or deflation of a balloon. The derivative of a volume equation with respect to a certain variable—often radius or time—indicates how the volume changes as that variable changes.

As seen in the exercise, the derivative of the volume of a sphere with respect to its radius is calculated using the initial volume formula \[ V = \left( \frac{4}{3} \pi r^{3} \right) \]. When we seek the rate at which the volume changes as the radius changes, we apply calculus, specifically differentiation. The solution uses the power rule to derive this rate, resulting in the formula \[ \frac{dV}{dr} = 4\pi r^{2} \]. This resulting rate of change tells us how much volume is added or lost for each unit of change in the radius.

Power Rule Derivation

The power rule is a fundamental tool in calculus used to derive functions of the form \( x^{n} \), where \( n \) is any real number. It states that the derivative of \( x^{n} \) is \( nx^{n-1} \). In the context of our balloon example, we apply the power rule to the volume equation, which is a function of the radius raised to the third power.

Applying the power rule to our function \( r^{3} \), we multiply the exponent 3 by the coefficient \( \frac{4}{3} \pi \) and reduce the exponent by 1, resulting in the derivative \( 4\pi r^{2} \). The power rule simplifies the process of finding rates of change and is a critical concept in understanding the behavior of polynomial functions as they change.

Spherical Volume Change

When asking how the volume of a sphere changes as its radius alters, we're looking into the practical application of derivatives in a real-world scenario. As the radius increases or decreases, the volume does not change linearly but rather according to the cube of the radius—an insight that is crucial when dealing with spherical objects.

In our exercise, we explored this concept by finding how much the volume increases when the radius goes from 2 ft to 2.2 ft. To do this, we used the rate of volume change with respect to the radius, a derivative we've found earlier. With \( \Delta r = 0.2 ft \), we estimated that the volume change \( \Delta V \) is approximately \( 3.2\pi ft^{3} \). This estimation hinges on the assumption that the rate of change is constant over the small interval of radius change, allowing for a linear approximation of a non-linear rate of volume increase.