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 0.02 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 = 360 seconds.

Question

anonymous · Answer

a. V(t) = 4π(0.02t)2; 651.44 in.3

b. V(t) = 4π(36 + 0.02t)2; 1,694,397.14 in.3

c. V(t) = ; 4690.37 in.3

 d. V(t) = ; 337,706.83 in.3

These are the answer choices!:)

abb0t · Answer

First of all, do you know the volume of a sphere? $$V = \frac{ 4 }{ 3 } \pi r^3 $$

abb0t · Answer

At the time it is first being launched, time (t) equals zero.

anonymous · Answer

okay i got 195432.1958 of V....

abb0t · Answer

and rate is changing, hence you have (rt) - radius changes with time, which means you're multiplying.
add this change to the radius. it adds up.

abb0t · Answer

I don't know if answer two is correct, because volume of a sphere is r cubed. and u have r squared.

abb0t · Answer

You should get: $$V = \frac{ 4 }{ 3 }\pi (r+∆rt)^3$$

anonymous · Answer

@abb0t oh so it will be the choice b right?

anonymous · Answer

Hmmm.

anonymous · Answer

Basically, I would solve for $r(t)$  using $$
\int_0^{t}0.02du = \int_0^t\frac{dr}{du}du = r(t)-r(0) = r(t) - 36\implies r(t) = \int_0^t0.02dt+36
$$

anonymous · Answer

Clearly $$
V(t) = \frac{4}{3}[r(t)]^3
$$As @abb0t said.