The surface ares S (in square meters) of a hot air ballon is given by s(r)=4 pi r^2 where r is radius of balloon (in meters). If the radius r is increasing with time (in seconds) according to the formula r(t) = 2/3t^3, t is greater than or equal to 0, find the surface area S of the balloon as a function of the time t.
Ok, your first function give S(r) or surface area as a function of the radius. You second function r(t)=(2/3)t^3 gives the radius as a function of time. If you compose the function to get
S(r(t)) = 4*pi*((2/3)t^3)^2
you have the surface area as a function of time alone. You can write it as
S(t) = something you can simplify.
To find the surface area S of the balloon as a function of time t, we need to substitute the expression for the radius r(t) = (2/3)t^3 into the formula for the surface area S(r) = 4πr^2.
Starting with the equation r(t) = (2/3)t^3, we substitute r(t) into S(r):
S(r(t)) = 4π( (2/3)t^3 )^2
To simplify this expression, we square the term inside the parentheses:
S(r(t)) = 4π(4/9)t^6
We can now simplify further by multiplying 4π(4/9) to get a constant term:
S(r(t)) = (16/9)πt^6
Therefore, the surface area S of the balloon as a function of time t is given by:
S(t) = (16/9)πt^6