Question

The volume V (in cubic meters) of the hot-air balloon described in problem 65 is given by $V(r) = \frac{4}{3} \pi r^{3}$. If the radius r is the same function as t as in problem 65, find the volume V as a function of the time t. Use $r(t) = \frac{2}{3} t^{2}$ to solve.

Answer 1

Answer:

$V(t)=\dfrac{32}{81}\pi t^6$

Step-by-step explanation:

You are given two functions

$V(r)=\dfrac{4}{3}\pi r^3\\ \\r(t)=\dfrac{2}{3}t^2$

You have to find the volume V as a function of the time t. Substitute the expression of r into the function V(r) to get V(t):

$V(t)\\ \\=\dfrac{4}{3}\pi\cdot \left(\dfrac{2}{3}t^2\right)^3\\ \\=\dfrac{4}{3}\pi \cdot \dfrac{2^3}{3^3}(t^2)^3\\ \\=\dfrac{4}{3}\pi\cdot \dfrac{8}{27}t^6\\ \\=\dfrac{32}{81}\pi t^6$