Question

A ball is being inflated at a rate that can be modeled by the function V(t) = 10t, where t is the number of seconds spent inflating the ball. The radius at a given volume can be modeled by the function .

Which composite function can be used to determine the radius of the ball depending on the number of seconds spent inflating the ball?

Answer 1

The radius function,
`\(r(t)\)`, determined from the volume function
`\(V(t) = 10t\)`yields
`\(r(t) = \sqrt[3]{(30t)/(4\pi)}\)`. This function calculates the radius of the ball based on seconds spent inflating it.

To find the radius function
`\(r(t)\)` depending on the number of seconds
`\(t\)` spent inflating the ball, we'll follow these steps.

Given:
`\(V(t) = 10t\)` (the volume function).

We know the formula for the volume of a sphere is
`\(V = (4)/(3) \pi r^3\)`, and we aim to express
`\(r\)` as a function of
`\(t\)`.

Start with the formula for the volume of a sphere:

`\[ V = (4)/(3) \pi r^3 \]`

Rearrange the formula to solve for
`\(r\)`:

`\[ r^3 = (3V)/(4\pi) \]`

Now, replace
`\(V\)` with the given function
`\(V(t) = 10t\)`:

`\[ r^3 = (3 \cdot 10t)/(4\pi) \]`

Simplify:

`\[ r^3 = (30t)/(4\pi) \]`

Isolate
`\(r\)` by taking the cube root of both sides:

`\[ r = \sqrt[3]{(30t)/(4\pi)} \]`

Thus, the step-by-step process leads us to the expression for
`\(r\)` as a function of
`\(t\)`:

`\[ r(t) = \sqrt[3]{(30t)/(4\pi)} \]`