The volume of a sphere is [tex]V(r) = \frac{4}{3} \pi r^3[/tex], and the radius is increasing at a rate of [tex]2 \, \text{mm/s}[/tex]. The function [tex]r(t) = 2t[/tex] gives the radius at time [tex]t[/tex] seconds. Which function gives the volume at time [tex]t[/tex]?

A. [tex](V \circ r)(t)[/tex]
B. [tex](r \circ V)(t)[/tex]
C. [tex](r+V)(t)[/tex]
D. [tex](V \cdot r)(t)[/tex]



Answer :

To determine which function gives the volume of the sphere at time \( t \), let's break down the problem step by step:

1. Given Information:
- The volume of a sphere is given by the function \( V(r) = \frac{4}{3} \pi r^3 \).
- The radius is increasing at a rate of \( 2 \) mm per second.
- The radius as a function of time is \( r(t) = 2t \).

2. Finding the Volume as a Function of Time:
- To find the volume at time \( t \), we need to express the volume function in terms of time \( t \).
- We know \( r(t) = 2t \).

3. Using the Composite Function:
- The volume \( V \) depends on \( r \), and \( r \) depends on \( t \).
- We need to find \( V(r(t)) \), which is the composite function \((V \circ r)(t)\).
- Substitute \( r(t) = 2t \) into \( V(r) \):
[tex]\[ V(2t) = \frac{4}{3} \pi (2t)^3 \][/tex]
- Simplify the expression:
[tex]\[ V(2t) = \frac{4}{3} \pi \cdot 8 t^3 = \frac{32}{3} \pi t^3 \][/tex]

4. Conclusion:
- The volume of the sphere as a function of time \( t \) is \( V(t) = \frac{32}{3} \pi t^3 \).

Given the explanation and the steps we followed to determine the correct function, we see that the volume at time \( t \) is represented by the composite function \((V \circ r)(t)\).

Thus, the correct choice is:

A. [tex]\((V \circ r)(t)\)[/tex]