Calculating the Rate of Decrease in Diameter of a Melting Snowball

[hatelove]

A snowball melts in volume @ 1 cm³/min. At what rate is the diameter decreasing when the diameter is 10 cm?

Volume of a sphere: \frac{4}{3}\pi r^{3}
Volume of a sphere in terms of diameter: \frac{\pi d^{3}}{6}

Rate of volume: \frac{d}{d(time)}volume = -1 cm^{3}/min.

Rate of diameter: \frac{d}{d(time)}10 cm = ?

I'm not sure what to do at this point

 

[Reckoner]

Volume of a sphere: \frac{4}{3}\pi r^{3}
Volume of a sphere in terms of diameter: \frac{\pi d^{3}}{6}

I'll use \(s\) for the diameter, to avoid confusion with the derivatives. We have

\[V = \frac{\pi s^3}6.\]

Differentiate both sides with respect to time \(t\):

\[\Rightarrow\frac{dV}{dt} = \frac{\pi s^2}2\cdot\frac{ds}{dt}\]

Now solve for \(ds/dt\) and set \(s = 10\) an \(\frac{dV}{dt} = -1\).

 

[hatelove]

How did you know to differentiate both sides?

I see we are trying to find \frac{ds}{dt} but how did you know that differentiating both sides would result in \frac{ds}{dt} appearing in the end?

 

[Reckoner]

How did you know to differentiate both sides?

I see we are trying to find \frac{ds}{dt} but how did you know that differentiating both sides would result in \frac{ds}{dt} appearing in the end?

Since \(V\) is a function of \(s\) and \(s\) is a function of time \(t\), the chain rule tells us that

\[\frac{dV}{dt} = \frac{dV}{ds}\cdot\frac{ds}{dt}.\]

This allows us to solve for \(\frac{ds}{dt}\) in terms of \(\frac{dV}{dt}\).

You can follow this same procedure in many other related rate problems: find a relationship between the variables, differentiate implicitly using the chain rule, solve for the unknown rate and make the appropriate substitutions.