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 0.16 inches per second and that r = 38 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 = 280 seconds.

Answer 1

V(t) = 4pi(38+0.16t)^3/3 ; 1,129,910.14 in.3

V(t) = 4pi(38+0.16t)^3 ; 2,377,823.53 in.3

V(t) = 4π(0.16t)2; 25,221.21 in.3

V(t) = 4π(38 + 0.16t)2; 1,325,689.77 in.3

Answer 2

V = 4/3*pi*r^3
Take the derivative with respect to t
dV/dt = 4/3(pi) (3r^2)dr/dt
dr/dt = 0.16
substitute. Then integrate to get V(t)

Answer 3

i don't know how to do that

Answer 4

Are you in pre-calculus?

Answer 5

yes...

Answer 6

We will have to do this without calculus then.
Find r as a function of time t.
r starts at 38 inches and steadily increases at 0.16 inch per second. So in t seconds r would have increased by 0.16t when added to the initial 38 inches will give:
r(t) = 38 + 0.16t

Volume of a sphere = 4/3(pi)r^3. Since radius changes with time, volume will change with time too.

V(t) = 4/3(pi)r^3 = 4/3(pi)(38 + 0.16t)^3

Answer 7

Put t = 280 seconds and find V.