# Rate problem

#### daigo

##### Member
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

##### Member
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 $$\frac{dV}{dt} = -1$$.

• Jameson and daigo

#### daigo

##### Member
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

##### Member
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.

• Jameson and Ackbach