Question

A high-altitude spherical weather balloon expands as it rises, due to the drop in atmospheric pressure. Suppose that the radius r increases at the rate of 0.07 inches per second, and that r = 36 inches at time t = 0. Determine the equation that models the volume V of the balloon at time t, and find the volume when t = 400 seconds.

Answer 1

Joshua missed that both questions have to do with Volume, not radius

Volume of a sphere = (4/3)πr³

At time 0, the radius of the sphere is 36 inches & each second after 0 the radius increases by 0.07 inches, so at any time t our radius is 36 + 0.07t

Therefore at any time t our volume is (4/3)π(36 + 0.07t)³ so your equation is

V = (4/3)π(36 + 0.07t)³

At t = 400, this becomes V = (4/3)π(36 + 0.07t)³ = (4/3)π(36 + 28)³ (4/3)π(64)³ = 1098066 cubic inches = 635.4 cubic feet

Answer 2

1098066.2 cubic inches

Explanation:

When time(t) = 0, radius of spherical weather balloon(r) = 36 inches

Since, the radius increases at 0.07 inches per second. Hence, the radius of the balloon after t=0 will be, r(t) = 36+0.07t

Now the volume of sphere (V) = \frac{4}{3}\pi r^3

V(t) = \frac{4}{3}\pi (36+0.07t)^3

We need to calculate volume when t = 400 seconds

V(400) = \frac{4}{3}\pi (36+0.07*(400))^3

= 4/3*pi*(64)^3

= 1098066.2 cubic inches