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Rate problem

daigo

Member
Jun 27, 2012
60
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: [tex]\frac{4}{3}\pi r^{3}[/tex]
Volume of a sphere in terms of diameter: [tex]\frac{\pi d^{3}}{6}[/tex]

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

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

I'm not sure what to do at this point
 

Reckoner

Member
Jun 16, 2012
45
Volume of a sphere: [tex]\frac{4}{3}\pi r^{3}[/tex]
Volume of a sphere in terms of diameter: [tex]\frac{\pi d^{3}}{6}[/tex]
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 \(\frac{dV}{dt} = -1\).
 

daigo

Member
Jun 27, 2012
60
How did you know to differentiate both sides?

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

Reckoner

Member
Jun 16, 2012
45
How did you know to differentiate both sides?

I see we are trying to find [tex]\frac{ds}{dt}[/tex] but how did you know that differentiating both sides would result in [tex]\frac{ds}{dt}[/tex] 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.