Finding the Rate of Change of a Spherical Balloon's Diameter

[skivail]

i need to be able to solve a problem like this for a test on monday. please solve, showing work, so i can trace your steps to teach myself.

a spherical balloon is inflated so that its volume is increasing at the rate of 6ft^3/minute. how fast is the diameter of the balloon increasing when the radius is 1 feet? reminder: volume of a sphere= 4/3 (pi)r^3

 

[Tide]

Just differentiate with respect to time. The left side will be dV/dt and the right side will contain a factor of dr/dt which is rate at which the RADIUS changes. Be sure to rewrite it in terms of the DIAMETER.

 

[ms. confused]

-Substitute (D(t)/2)^3 for r in V=4/3 (pi)r^3.
-When simplified you get pi/6(D(t))^3.
-Differentiate and you get V'= pi/6[3(D(t))^2] [D'(t)] (implicit diff.)
-plug in the numbers--remember any variable with the ' means it's a rate, a unit over time
- I got 3.8197 ft/min. but if you have an answer key go with that.

I really hope this helps! Good luck on your test!