Related Rates Question (filling a balloon)

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hi.. this is my first time on this site, so im assuming i can just ask a question? Here is my problem

Sally is filling a balloon, which is increasing at 0.2 cm/second. The balloon's radius is 4 cm. In 1 minute, how fast is the volume decreasing?

We have done problems like this, but we never had time there. Moreover, we were always told radius increasing at 0.2 cm/second, not just balloon is....

Typically we have done
dv/dt=dv/dr dr/dt
dv/dt= 4 pi (4)^2 (0.2 cm/second)

But im assuming i just cant ignore that 1 minute thing? Or can I? And is the wording of the dr/dt okay? im so confused... sorry! can anyone help me?
 
I'm a little confused too. Sally is filling the balloon which is increasing at 0.2/sec. What is increasing?. The volume?. Then it should say 0.2cm^3/sec.

If she is filling the balloon, why is the volume decreasing?.

Maybe I am misunderstanding, but it seems like a poorly posed question you were given.
 
she meant that the balloon is increasing in size. but i was wondering, if thats the whole length of the balloon? so then the radius would be increasing at half of that length? but she probably just meant radius... and what do i do about the time? or can i ignore it?
 
Sally is filling a balloon, which is increasing at 0.2 cm/second. The balloon's radius is 4 cm. In 1 minute, how fast is the volume decreasing?
Click to expand...

well, it is intuitively obvious to the most casual observer that there is a mistake here ... the volume must be increasing if the balloon is being filled.

given dr/dt = 0.2 cm/sec, r = 4 cm

in 1 minute, the radius r will be 4 + 60(.2) = 16 cm

now ... we also make the assumption that the balloon is spherical.

V = (4/3)pi*r^3

take the derivative w/r to time ...

dV/dt = 4pi*r^2*(dr/dt)

you have r and dr/dt ... calculate dV/dt.
 
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