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

RobLiefeld

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Nov 11, 2008
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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.
 
Re: The volume of a sphere is increasing at....

you were given ... dVdt=100cm3s\displaystyle \frac{dV}{dt} = 100 \, \frac{cm^3}{s} and you are asked to determine drdt\displaystyle \frac{dr}{dt} when r = 50 cm

volume of a sphere is ... V=43πr3\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 drdt\displaystyle \frac{dr}{dt}.
 
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=?
 
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?
 
Re: The volume of a sphere is increasing at....

V=4π3r3\displaystyle V \, = \, \frac{4\pi}{3}r^3

ddtV=ddt[4π3r3]\displaystyle \frac{d}{dt}V \, = \, \frac{d}{dt}[\frac{4\pi}{3}r^3]

dVdt=drdtddr[4π3r3]\displaystyle \frac{dV}{dt} \, = \, \frac{dr}{dt}\cdot \frac{d}{dr}[\frac{4\pi}{3}r^3]
 
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