Related rates problems with inflating and deflating balloons
Steps we use to solve a related rates problem
Related Rates are an application of implicit differentiation, and are usually easy to spot.
They ask you to find how quickly one variable is changing when you know how quickly another variable is changing.
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To solve a related rates problem, complete the following steps:
Construct an equation containing all the relevant variables.
Differentiate the entire equation with respect to (time), before plugging in any of the values you know.
Plug in all the values you know, leaving only the one you’re solving for.
Solve for your unknown variable.
Inflating and deflating balloon related rates problems
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Finding rate of change of the radius, given rate of change of volume
Example
How fast is the radius of a spherical balloon increasing when the radius is ???100??? cm, if air is being pumped into it at ???400??? cm???^3???/s?
In this example, we’re asked to find the rate of change of the radius, given the rate of change of the volume.
The formula that relates the volume and radius of a sphere to one another is simply the formula for the volume of a sphere.
???V=\frac{4}{3}\pi r^3???
Before doing anything else, we use implicit differentiation to differentiate both sides with respect to time ???t???.
???\frac{dV}{dt}=\frac{4}{3}(3)\pi r^{3-1}\frac{dr}{dt}???
???\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}???
They ask you to find how quickly one variable is changing when you know how quickly another variable is changing.
Now we plug in everything that we know. Keep in mind that ???dV/dt??? is the rate at which the volume is changing, ???dr/dt??? is the rate at which the radius is changing, and ???r??? is the length of the radius at a specific moment.
Our problem tells us that the rate of change of the volume is ???400???, and that the length of the radius at the specific moment we’re interested in is ???100???.
???400=4\pi (100)^2\frac{dr}{dt}???
Solving for ???dr/dt??? gives
???400=40,000\pi \frac{dr}{dt}???
Therefore, we know that the radius of the balloon is increasing at a rate of ???1/100\pi??? cm per second.
???\frac{1}{100\pi}=\frac{dr}{dt}???
