The surface area S (in square meters) of a hot-air balloon is given by S(r)=4(pi)r^{2}, where r is the radius of the balloon (in meters). If the radius is increasing with time t (in seconds) according to the formula r(t)=2/3t^{3}, t≥0, find the surface area of the balloon as a function of the time t.

AdamTaurus
Oct 9, 2017

#1**+2 **

surface area = 4 * π * (radius)^{2}

And they tell us that the

radius = 2/3 t^{3}

And we want an equation that gives the surface area for any value of t ( that is ≥ 0 ) .

So....instead of calling the radius "radius", we want to say the radius is 2/3 t^{3} .

surface area = 4 * π * (2/3 t^{3})^{2} Then simplify this equation.

surface area = 4 * π * 4/9 * t^{6}

surface area = 16 π t^{6} / 9 So we can say...

s(t) = 16 π t^{6} / 9

hectictar
Oct 10, 2017

#1**+2 **

Best Answer

surface area = 4 * π * (radius)^{2}

And they tell us that the

radius = 2/3 t^{3}

And we want an equation that gives the surface area for any value of t ( that is ≥ 0 ) .

So....instead of calling the radius "radius", we want to say the radius is 2/3 t^{3} .

surface area = 4 * π * (2/3 t^{3})^{2} Then simplify this equation.

surface area = 4 * π * 4/9 * t^{6}

surface area = 16 π t^{6} / 9 So we can say...

s(t) = 16 π t^{6} / 9

hectictar
Oct 10, 2017