Derivative as Rate Measure - Practice Questions & MCQ

Derivative as Rate Measure

If two related quantities are changing over time, the rates at which the quantities change are related. For example, if a balloon is being filled with air, both the radius of the balloon and the volume of the balloon are increasing.

If a variable quantity y depends on and varies with a quantity x, then the rate of change of y concerning x is $\frac{d y}{d x}$.
A rate of change concerning time is simply called the rate of change.
For example, the rate of change of displacement (s) of an object w.r.t. time is velocity (v).

$
v=\frac{d s}{d t}
$
Illustration:

Consider the balloon example again, a spherical balloon is being filled with air at a constant rate of $2 \mathrm{~cm}^3 / \mathrm{sec}$. How fast is the radius increasing when the radius is 3 cm?
The volume of a sphere of radius r centimetres is,

$
V=\frac{4}{3} \pi r^3 \mathrm{~cm}^3
$
Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, $t$ seconds after beginning to fill the balloon with air, the volume of air in the balloon is

$
V(t)=\frac{4}{3} \pi[r(t)]^3 \mathrm{~cm}^3
$
Differentiating both sides of this equation concerning time and applying the chain rule, we see that the rate of change in the volume is related to the rate of change in the radius by the equation.

$
\begin{aligned}
\frac{d}{d t}(V(t)) & =\frac{4}{3} \pi \frac{d}{d t}\left([r(t)]^3\right) \mathrm{cm}^3 \\
V^{\prime}(t) & =4 \pi[r(t)]^2 r^{\prime}(t)
\end{aligned}
$
The balloon is filled with air at a constant rate of $2 \mathrm{~cm}^3 / \mathrm{sec}$. So, $\mathrm{V}^{\prime}(\mathrm{t})=2 \mathrm{~cm}^3 / \mathrm{sec}$

$
\begin{aligned}
& \therefore & 2 \mathrm{~cm}^3 / \mathrm{sec} & =\left(4 \pi[r(t)]^2 \mathrm{~cm}^2\right) \cdot r^{\prime}(t) \\
& \Rightarrow & r^{\prime}(t) & =\frac{1}{2 \pi[r(t)]^2} \mathrm{~cm} / \mathrm{sec}
\end{aligned}
$
When the radius $\mathrm{r}=3 \mathrm{~cm}$

$
\Rightarrow \quad r^{\prime}(t)=\frac{1}{18 \pi} \mathrm{~cm} / \mathrm{sec}
$