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

help meeee

anonymous · Answer

@Hoa

anonymous · Answer

@abb0t could you help me?

anonymous · Answer

@abb0t 
V(t) = 4π(0.02t)2; 651.44 in.3

 V(t) = 4π(36 + 0.02t)2; 1,694,397.14 in.3

 V(t) = ; 4690.37 in.3

 V(t) = ; 337,706.83 in.3

these are the options

abb0t · Answer

Volume of a sphere: $$V=\frac{ 4 }{ 3 } (\pi) r^3$$
when balloon is lauched, t=0
r = 36
solve for V

abb0t · Answer

However, for this you're increasing t, therefore, +0.02t 
since it's increasing with time.

anonymous · Answer

@abb0t  for v, i got 1728pi

abb0t · Answer

$$V = \frac{ 4 }{ 3 } \pi (r+ nt)^3$$
where say, n = rate.

abb0t · Answer

since rate implies a change in time.

abb0t · Answer

plug n solve.

anonymous · Answer

@abb0t which one from the answer choices did you get as an answer?

anonymous · Answer

@abb0t i got the last one!