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

Calc?

Answer 2

Yeah

Answer 3

this is a related rates problem. you have volume which depends on radius but radius is some other function of time (perhaps). so when you take the derivative of volume, you have to apply the chain rule.
\[V=\frac{ 4 }{ 3 }\pi r ^{3} \text{, } \frac{ dV }{ dt }=4\pi r ^{2}\frac{ dr }{ dt }\]

does this make sense so far?

Answer 4

Kind of what is the dV/dt but the V=4 /3 pi r ^3 is the formula

Answer 5

oops... i mis-read.

they just want you to put in the radius function into the volume formula.

can you tell me what the radius should be as a function of time? let's call that r(t).

Answer 6

So it would be 36(0.07)

Answer 7

not exactly... where does t (time) fit into it?

Answer 8

what would it be when t is 0?
when t is 1?
etc...

Answer 9

when t is 0 it would be 0,07 right?

Answer 10

Suppose that the radius r increases at the rate of 0.07 inches per second and that r = 36 inches at time t = 0.

so when t is 0, r = ?

Answer 11

36 ?

Answer 12

yep.
then what about when t is 1?

Answer 13

Would it be 1* 0.07 then multiply it by 36 ?

Answer 14

nope... let's read
Suppose that the radius r increases at the rate of 0.07 inches per second and that r = 36 inches at time t = 0.

so it's increasing by .07 inches per second.

36 + .07t where t is in seconds.
so after 1 second it would be 36+.07*1 = 36+.07 = 36.07 inches

make sense?

Answer 15

Ok yeah now it does.

Answer 16

so r(t) = 36 + .07t

then\[V \left( t \right)=\frac{ 4 }{ 3 }\pi \left[ r \left( t \right) \right]^{3}\]

does that make sense?

Answer 17

Yeah

Answer 18

just plug in what r(t) is...

Answer 19

Ok thanks could you help with an other problem

Answer 20

yep

Answer 21

you want to post here or as a new problem?

Answer 22

Ok let me post it as new problem