Problem 96 - College Algebra

Question

The radius \(r\), in inches, of a spherical balloon is related to the volume, \(V\), by \(r(V)=\sqrt[3]{\frac{3 V}{4 \pi}} .\) Air is pumped into the balloon, so the volume after \(t\) seconds is given by \(V(t)=10+20 t\). a. Find the composite function \(r(V(t))\). b. Find the exact time when the radius reaches 10 inches.

Step-by-Step Solution

Answer

The radius reaches 10 inches when \( t = \frac{\frac{4000\pi}{3} - 10}{20} \).

Step 1: Find the composite function

To find the composite function \( r(V(t)) \), substitute \( V(t) \) into the formula for \( r \). The function \( V(t) = 10 + 20t \) gives the volume at time \( t \). Substitute this into \( r(V) \): \[ r(V(t)) = \sqrt[3]{\frac{3(10 + 20t)}{4 \pi}}. \] This is the expression for the radius in terms of time \( t \).

Step 2: Solve for time when radius is 10 inches

We need to set \( r(V(t)) \) equal to 10 and solve for \( t \). This gives us: \[ \sqrt[3]{\frac{3(10 + 20t)}{4 \pi}} = 10. \] Cube both sides to eliminate the cube root: \[ \frac{3(10 + 20t)}{4 \pi} = 1000. \] Multiply both sides by \( 4\pi \) to clear the fraction: \[ 3(10 + 20t) = 4000\pi. \] Divide everything by 3: \[ 10 + 20t = \frac{4000\pi}{3}. \] Finally, solve for \( t \) by subtracting 10 and then dividing by 20: \[ 20t = \frac{4000\pi}{3} - 10, \] \[ t = \frac{\frac{4000\pi}{3} - 10}{20}. \] Calculating further gives us the exact value of \( t \).

Key Concepts

Understanding Spherical Geometry

Spherical geometry deals with shapes like spheres, unlike flat surfaces which are studied in Euclidean geometry. A sphere is a perfectly round 3D object, and its properties differ from flat surfaces. We primarily deal with concepts like surface area and volume in spherical geometry. A sphere’s surface is all points equidistant from the center in three-dimensional space. To solve problems involving spherical objects, like a balloon, we need equation-based models, much like the formula used here:

• The radius \( r \) of a sphere given the volume \( V \) is \( r = \sqrt[3]{\frac{3V}{4\pi}} \).

This formula allows us to understand how expanding or contracting a sphere changes its dimensions, which is crucial in applications across physics and engineering.

Basics of Volume Calculation

Volume is the measure of the space that a 3D object occupies. For a sphere, the volume\( V \) can be calculated using:

• \( V = \frac{4}{3} \pi r^3 \)

This formula relates the radius of a sphere directly to its volume.When given a time-dependent volume, as in this exercise with \( V(t) = 10 + 20t \), it intertwines the concept of time with geometric dimensions. Understanding how to calculate the volume is vital for solving complex real-world applications, such as determining how much space is needed for air in balloons or determining the capacity of spherical tanks in industries.

Mastering Solving Equations for Composite Functions

In mathematics, composite functions involve substituting one function into another. Here, the composite function \( r(V(t)) \) means substituting the volume function \( V(t) \) into the radius function \( r(V) \). As shown, this leads to:

• \( r(V(t)) = \sqrt[3]{\frac{3(10 + 20t)}{4 \pi}} \)

To solve for problems like finding when the radius reaches a specific size, setting the composite function equal to a desired value and solving for \( t \) becomes critical. The steps involve:

• Setting the equation \( \sqrt[3]{\frac{3(10 + 20t)}{4 \pi}} = 10 \)
• Cubing both sides to eliminate the cube root

Then transforming it by clearing the fractions and solving for \( t \), you can accurately determine when a particular condition—for example, a radius of 10 inches—is met. Through each algebraic step, understanding equation-manipulation is essential for solving such practical problems efficiently.