Question

The surface area of a balloon can be represented by the function S(r) = 4πr2, where r is the radius of the balloon. If r increases with time, t, and is represented by the function r(t) = 12t2, What is the surface area of the balloon expressed as a function of time?

A) S(t)=576π
B) S(t)=576t^4
C) S(t)=4πt^2
D) S(t)=12πt^4

Answer 1

The surface area of the balloon as a function of time, given the functions S(r) = 4πr^2 and r(t) = 12t^2, is S(t) = 576πt^4, which is option B.

Step-by-step explanation:

The surface area of a balloon as a function of radius is given by the formula S(r) = 4πr2. Since the radius increases with time and is represented by the function r(t) = 12t2, we can substitute this expression for r into the surface area formula to find the surface area as a function of time.

Substituting r(t) into S(r), we get:

S(t) = 4π(12t2)2

Expanding and simplifying this, we get:

S(t) = 4π(144t4)

S(t) = 576πt4

Therefore, the surface area of the balloon expressed as a function of time is S(t) = 576πt4, which corresponds to option B.