Problem 96 - College Algebra with Corequisite Support

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

(a) \( r(V(t)) = \sqrt[3]{\frac{30 + 60t}{4\pi}} \); (b) \( t = \frac{2000\pi - 15}{30} \).

Step 1: Understand Given Functions

We are given two functions, \( r(V) = \sqrt[3]{\frac{3V}{4\pi}} \) and \( V(t) = 10 + 20t \). The function \( r(V) \) gives the radius of a balloon with volume \( V \), and \( V(t) \) gives the volume as a function of time \( t \).

Step 2: Compose Functions

To find \( r(V(t)) \), substitute \( V(t) = 10 + 20t \) into \( r(V) = \sqrt[3]{\frac{3V}{4\pi}} \). This yields:\[r(V(t)) = \sqrt[3]{\frac{3(10 + 20t)}{4\pi}}.\]

Step 3: Simplify Expression

Simplify the composed function's expression:\[r(V(t)) = \sqrt[3]{\frac{30 + 60t}{4\pi}}.\] This is the composite function that expresses the radius as a function of time.

Step 4: Set Up Equation for Part (b)

To find when the radius is 10 inches, set \( r(V(t)) = 10 \). Thus, solve:\[\10 = \sqrt[3]{\frac{30 + 60t}{4\pi}}.\] Cube both sides to eliminate the cube root.

Step 5: Solve for t

Cubing both sides of the equation gives:\[1000 = \frac{30 + 60t}{4\pi}.\] Multiply both sides by \( 4\pi \) to clear the fraction:\[1000 \times 4\pi = 30 + 60t.\]Then, simplify and solve for \( t \):\[4000\pi = 30 + 60t.\]\[60t = 4000\pi - 30.\]\[t = \frac{4000\pi - 30}{60}.\]

Step 6: Calculate Exact Time

Calculate \( t \) using the expression derived:\[t = \frac{4000\pi - 30}{60}.\]Simplify:\[t = \frac{2000\pi - 15}{30}.\] Thus, this is the exact formula for the time when the radius is 10 inches.

Key Concepts

Radius of a Sphere

The radius of a sphere is an important concept in geometry. It is the distance from the center of the sphere to any point on its surface. The formula for the radius in terms of the sphere's volume is derived from the formula for volume of a sphere, which is: \[ V = \frac{4}{3}\pi r^3 \] To express the radius as a function of the volume, you rearrange the formula to solve for \( r \). The rearranged formula is: \[ r(V) = \sqrt[3]{\frac{3V}{4\pi}} \] This function tells us how the radius grows as the volume changes. It's essential to understand that the function uses a cube root, indicating that the radius increases at a different rate compared to how volume increases. Knowing how to handle cube roots is crucial to working with these types of equations.

Volume as a Function of Time

When studying dynamic systems, such as a balloon being inflated, it's often useful to express measurements, like volume, as a function of time. In our problem, the volume of the balloon is given by the function: \[ V(t) = 10 + 20t \] This function reveals that the initial volume is 10 cubic inches, and the volume increases by 20 cubic inches per second. Understanding such a function allows us to predict the growth of the volume over time. Calculating volume as a function of time can be very useful, especially in fields like engineering and physics, where knowing how quickly a system changes can inform design and safety considerations.

Solving Equations with Roots

Solving equations with roots, such as cube roots, can initially seem complex but is manageable with clear steps. In this case, the task is to find when the radius of the balloon reaches 10 inches, which requires solving: \[ 10 = \sqrt[3]{\frac{30 + 60t}{4\pi}} \] To eliminate the cube root, we cube both sides: \[ 1000 = \frac{30 + 60t}{4\pi} \] This step transforms the equation into a linear form, which is much easier to manipulate. By multiplying both sides by \( 4\pi \), we clear the fraction: \[ 1000 \times 4\pi = 30 + 60t \] We then solve for \( t \) by isolating it on one side of the equation: \[ 60t = 4000\pi - 30 \] Finally, solve for \( t \) by dividing through by 60: \[ t = \frac{4000\pi - 30}{60} \] Clear steps like these simplify the process and help develop a robust understanding of handling roots in equations.