Questions: Graphs and Functions
Word problem involving composition of two functions
The surface area S(r) (in square meters) of a spherical balloon with radius r meters is given by S(r)=4 π r^2.
The radius P(t) (in meters) after t seconds is given by P(t)=8/3 t.
Write a formula for the surface area N(t) (in square meters) of the balloon after t seconds.
It is not necessary to simplify.
N(t)=
Transcript text: Graphs and Functions
Word problem involving composition of two functions
The surface area $S(r)$ (in square meters) of a spherical balloon with radius $r$ meters is given by $S(r)=4 \pi r^{2}$.
The radius $P(t)$ (in meters) after $t$ seconds is given by $P(t)=\frac{8}{3} t$.
Write a formula for the surface area $N(t)$ (in square meters) of the balloon after $t$ seconds.
It is not ne ${ }^{\text {ary to simplify. }}$
\[
N(t)=
\]
Solution
Solution Steps
To find the surface area of the balloon as a function of time, we need to compose the two given functions. The surface area \( S(r) \) is a function of the radius \( r \), and the radius \( P(t) \) is a function of time \( t \). Therefore, the surface area as a function of time \( N(t) \) can be found by substituting \( P(t) \) into \( S(r) \).
Step 1: Define the Functions
The surface area \( S(r) \) of a spherical balloon with radius \( r \) is given by the formula:
\[ S(r) = 4 \pi r^2 \]
The radius \( P(t) \) of the balloon as a function of time \( t \) is given by:
\[ P(t) = \frac{8}{3} t \]
Step 2: Compose the Functions
To find the surface area as a function of time, substitute \( P(t) \) into \( S(r) \):
\[ N(t) = S(P(t)) = 4 \pi \left(\frac{8}{3} t\right)^2 \]