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Chapter 1: Problem 24

Determine an appropriate domain of each function. Identify the independent and dependent variables. The volume \(V\) of a balloon of radius \(r\) (in meters) filled with helium is given by the function \(f(r)=\frac{4}{3} \pi r^{3} .\) Assume the balloon can hold up to \(1 \mathrm{m}^{3}\) of helium.

Short Answer

Expert verified
Answer: The domain of the function is \(0 < r \leq \sqrt[3]{\frac{3}{4\pi}}\), where r is the radius of the balloon.

Step by step solution

01

Identify the independent and dependent variables

The independent variable is the input of the function, which, in this case, is the radius \(r\). The dependent variable is the output of the function, which is the volume \(V\). In this function, the volume of the balloon is dependent on its radius.
02

Write down the given function

The function \(f(r) = \frac{4}{3} \pi r^3\) gives us the volume of the balloon as a function of its radius.
03

Determine the maximum volume constraint

The problem states that the balloon can hold up to 1 cubic meter (\(1 m^3\)) of helium. This piece of information helps us in determining the range of values for the radius.
04

Define the constraint inequality for the radius

Since the volume cannot exceed 1 cubic meter, we can write the inequality as: $$ f(r) = \frac{4}{3} \pi r^3 \leq 1 $$
05

Solve the inequality for the radius

To find the range of values for the radius \(r\), we need to solve the inequality for \(r\): $$ \frac{4}{3} \pi r^3 \leq 1 $$ Divide both sides of the inequality by \(\frac{4}{3} \pi\): $$ r^3 \leq \frac{3}{4 \pi} $$ Now take the cube root of both sides of the inequality: $$ r \leq \sqrt[3]{\frac{3}{4 \pi}} $$
06

Define the domain of the function

Since the radius of the balloon must be positive and it cannot exceed \(\sqrt[3]{\frac{3}{4 \pi}}\), the domain of the function is: $$ 0 < r \leq \sqrt[3]{\frac{3}{4\pi}} $$

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