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 \( P(t)=\frac{4}{3} t \). Write a formula for the surface area \( N(t) \) (in square meters) of the balloon after \( t \) seconds. It is not necessary to simplify. \( N(t)=\square \)

To find the surface area \( N(t) \) of the balloon after \( t \) seconds, we can substitute the radius function \( P(t) \) into the surface area formula \( S(r) \). We start with: \[ S(r) = 4 \pi r^2 \] Now, substituting \( r \) with \( P(t) = \frac{4}{3}t \): \[ N(t) = 4 \pi \left( \frac{4}{3} t \right)^2 \] Thus, we have: \[ N(t) = 4 \pi \left( \frac{16}{9} t^2 \right) \] So, the formula for the surface area of the balloon after \( t \) seconds is: \[ N(t) = \frac{64}{9} \pi t^2 \] This represents how the surface area of the balloon grows as it inflates over time!