172k views
2 votes
Real world compositionsWhat is the volume of a spherical balloon after 11 seconds if the radius of the balloon is increasing at 1.3 cm/sec? Round to the nearest tenth of a centimeter solve using composite functions

User Jaygeek
by
3.3k points

1 Answer

3 votes

In this case we have two functions. The first one is the volume of the balloon and the other is the value of the radius. Both functions depend on the variable time.

Finding the functions to solve the problem, we have:


\begin{gathered} V(t)=(4)/(3)\pi r^3\text{ }^{}\text{ (First function)} \\ r(t)=1.3\cdot t\text{(Second function. Let us suppose that the initial radius is equal to zero)} \end{gathered}

Let us replace the second function in the first one to get a composite function. Doing so, we have:


\begin{gathered} V(t)=(4)/(3)\pi(1.3\cdot t)^3 \\ V(t)=(4)/(3)\pi(2.197\cdot t^3)\text{ (Raising the expression within parentheses to the power of 3)} \\ V(t)=2.93\pi\cdot t^3(^{}\text{ Multiplying constant terms)} \end{gathered}

Now, we can replace t=11 to find the value of the volume.


\begin{gathered} V(11)=2.93\pi\cdot(11)^3\text{ } \\ V(11)=2.93\pi(1331)\text{ (Raising 11 to the power of 3)} \\ V(11)=12248.88\text{ (Multiplying)} \end{gathered}

The answer is 12248.9 cm3 (Rounding to the nearest tenth of a centimeter).

User Pstryk
by
3.2k points