The volume of a sphere is increasing at 100 cm^3 / sec

[RobLiefeld]

arg, Calc help needed here. Brain freeze has taken hold

A balloon is being inflated "It's volume increases at a rate of 100 cm^3/s. How fast is the radius of the balloon increasing when the diameter is 50 cm?"

The first step is figuring out the initial equation. I know the formula for the volume of a sphere is 4/3pi r^3, but I can't figure out what the function is Been up too late and I completely blank.

The second is figuring out the rate of increase at a specific point (diameter=50 cm i.e. radius =25cm). We haven't done that in a few weeks and I haven't retained it, so I'm blanking on step 2 before I even get past step 1. Any help would be appreciated.

 

[skeeter]

Re: The volume of a sphere is increasing at....

you were given ... \(\displaystyle \frac{dV}{dt} = 100 \, \frac{cm^3}{s}\) and you are asked to determine \(\displaystyle \frac{dr}{dt}\) when r = 50 cm

volume of a sphere is ... \(\displaystyle V = \frac{4}{3}\pi r^3\)

Here's what you need to do ... take the derivative of the volume equation with respect to time, substitute in your given values, then determine the value of \(\displaystyle \frac{dr}{dt}\).

 

[RobLiefeld]

Re: The volume of a sphere is increasing at....

Ok, but what about the original function? I can usually understand something better once I've seen it. What would I plug into a graphic calculator? What's my y=?

 

[skeeter]

Re: The volume of a sphere is increasing at....

A calculator is not necessary to solve this problem ... do you know how to take the time derivative of the volume formula for a sphere?

 

[Deleted member 4993]

Re: The volume of a sphere is increasing at....

\(\displaystyle V \, = \, \frac{4\pi}{3}r^3\)

\(\displaystyle \frac{d}{dt}V \, = \, \frac{d}{dt}[\frac{4\pi}{3}r^3]\)

\(\displaystyle \frac{dV}{dt} \, = \, \frac{dr}{dt}\cdot \frac{d}{dr}[\frac{4\pi}{3}r^3]\)