Asked by Anonymous
You are letting the air out of a hot air balloon at a steady rate. The volume of the balloon is decreasing at the rate of 10pi m^3/min when the diameter is 45 meters. At what rate is the radius of the balloon changing? I need help! How do I solve this problem? Thank you so much for the help!
Answers
Answered by
Reiny
V = (4/3)pi(r^3)
so dV/dt = 4pi(r^2)dr/dt
so when diameter is 45 m
r = 22.5, dV/dt = -10pi
plug that in to solve for dr/dt
so dV/dt = 4pi(r^2)dr/dt
so when diameter is 45 m
r = 22.5, dV/dt = -10pi
plug that in to solve for dr/dt
Answered by
Anonymous
Thanks, that's helpful. Would the answer be .20m/min?
Answered by
Reiny
I had
-10pi = 4pi(22.5)^2 dr/dt
dr/dt = -10/(4(22.5)^2)
= - 0.00494 m/min
or -.494 cm/min
seems like a small amount, but remember that balloon is huge, so a small change in the radius is going to result in a large change in the volume.
BTW, the negative indicates that the radius is decreasing.
-10pi = 4pi(22.5)^2 dr/dt
dr/dt = -10/(4(22.5)^2)
= - 0.00494 m/min
or -.494 cm/min
seems like a small amount, but remember that balloon is huge, so a small change in the radius is going to result in a large change in the volume.
BTW, the negative indicates that the radius is decreasing.
Answered by
Anonymous
Thanks, I forgot to divide the 4. The negative makes sense.
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