Derivative as Rate Measure: Definition, Formula, Examples

Understanding Change Measurement

The rate of any variable with respect to time represents its rate of change. It shows how much one quantity changes when time changes by a small amount. This concept forms the basis of differentiation and rate measurement in calculus.

For instance, the following quantities denote the rate of change with respect to time:

$
\Rightarrow \frac{dx}{dt}, \frac{dy}{dt}, \frac{dR}{dt} \text{ (Linear Rate)}
$

$
\Rightarrow \frac{dS}{dt}, \frac{dA}{dt} \text{ (Rate of Change of Area)}
$

$
\Rightarrow \frac{dV}{dt} \text{ (Rate of Change of Volume)}
$

$
\Rightarrow \frac{dV}{V} \times 100 \text{ (Percentage Change in Volume)}
$

The derivative $\frac{ds}{dt}$ represents the rate of change of distance ($s$) with respect to time ($t$). Similarly, if a quantity $y$ varies with another quantity $x$, satisfying the rule $y = f(x)$, then $\frac{dy}{dx}$ or $f'(x)$ represents the rate of change of $y$ with respect to $x$.

At a specific point $x = x_0$, the rate of change is given by $\left.\frac{dy}{dx}\right|_{x = x_0}$ or $f'(x_0)$.

Chain Rule in Rate Measurement

When two variables $x$ and $y$ both vary with respect to another variable $t$, such that $x = f(t)$ and $y = g(t)$, we use the Chain Rule to connect their rates of change.

$
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \quad \text{if } \frac{dx}{dt} \neq 0
$

Hence, the rate of change of $y$ with respect to $x$ can be determined by dividing the rate of change of $y$ with respect to $t$ by that of $x$ with respect to $t$.

This concept is crucial in related rates problems, where two or more changing quantities are connected through a common variable, usually time.

Derivative as a Rate Measure

If two related quantities change over time, their rates of change are also connected. For example, if a balloon is being inflated, both its radius ($r$) and volume ($V$) increase simultaneously. Here, the derivative helps us measure how quickly one quantity changes in relation to another.

If $y$ depends on $x$, then the rate of change of $y$ with respect to $x$ is:

$
\frac{dy}{dx}
$

When the rate of change is with respect to time, it is known as a time rate of change. For instance, the rate of change of displacement ($s$) with respect to time gives velocity ($v$):

$
v = \frac{ds}{dt}
$

This relationship is the foundation for many applications of derivatives in physics, engineering, and economics.

Example: Rate of Change in a Balloon’s Radius

Let’s understand the concept with a real-life problem involving volume and radius change.

A spherical balloon is being filled with air at a constant rate of $2 \text{ cm}^3/\text{sec}$. How fast is the radius increasing when the radius is $3 \text{ cm}$?

The volume of a sphere of radius $r$ is given by:

$
V = \frac{4}{3} \pi r^3
$

As both volume and radius vary with time, we express them as functions of $t$:

$
V(t) = \frac{4}{3} \pi [r(t)]^3
$

Differentiating both sides with respect to time $t$ and applying the chain rule gives:

$
\frac{dV}{dt} = 4 \pi [r(t)]^2 \frac{dr}{dt}
$

Given that $\frac{dV}{dt} = 2 \text{ cm}^3/\text{sec}$, we substitute the values:

$
2 = 4 \pi [r(t)]^2 \frac{dr}{dt}
$

Solving for $\frac{dr}{dt}$:

$
\frac{dr}{dt} = \frac{1}{2\pi [r(t)]^2} \text{ cm/sec}
$

When $r = 3 \text{ cm}$:

$
\frac{dr}{dt} = \frac{1}{18\pi} \text{ cm/sec}
$

Thus, when the radius of the balloon is 3 cm, it increases at a rate of $\frac{1}{18\pi} \text{ cm/sec}$.

Importance of Derivatives in Measuring Change

Derivatives are a fundamental tool for understanding how one quantity changes in response to another. Whether it’s the growth rate of a population, the speed of a car, or the expansion of a balloon, derivatives help us quantify instantaneous change.

Key applications include:

• Measuring velocity and acceleration in motion.

• Calculating growth rates in economics and biology.

• Determining slope of curves in geometry.

Geometrical Interpretation of Derivative

The geometrical meaning of a derivative helps visualize how a function changes at a particular point on its curve. In simple terms, the derivative gives the slope of the tangent line drawn to the curve at that point. This slope represents the instantaneous rate of change of the function - how fast or slow the function’s value changes with respect to $x$.

Derivative as Slope of Tangent

If a function is represented as $y = f(x)$, then the slope of the tangent at any point $x = a$ is given by the value of its derivative:

$m = f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$

This means that the derivative $f'(a)$ provides the instantaneous rate of change of $y$ with respect to $x$ at the point $(a, f(a))$. In graphical terms, if you zoom in infinitely around the point of tangency, the curve appears almost like a straight line - that line’s slope is the derivative.

For example, for $y = x^2$, the derivative is $f'(x) = 2x$. So at $x = 2$, the slope of the tangent is $m = 4$, which indicates that the curve is rising steeply at that point.

The slope of tangent therefore connects algebraic differentiation to geometrical visualization, showing how a mathematical concept translates into a measurable geometric property.

Derivative and Direction of Motion

The sign of the derivative tells us the direction in which the function moves:

• When $f'(x) > 0$, the slope of the tangent is positive — meaning the curve rises as $x$ increases. The function is increasing in this interval.

• When $f'(x) < 0$, the slope is negative — indicating the curve falls as $x$ increases. The function is decreasing in this region.

• When $f'(x) = 0$, the tangent is horizontal, often corresponding to a turning point or extremum (maximum or minimum).

Hence, the derivative not only measures rate of change but also gives insight into the behavior of the curve, helping us understand whether a function is rising, falling, or stationary at a particular point.

Solved Examples Based On Derivative as Rate Measure:

Example 1: If the surface area of a sphere of radius $r$ is increasing uniformly at the rate $8 \mathrm{~cm}^2 / \mathrm{s}$, then the rate of change of its volume is :
1) Constant
2) Proportional to $\sqrt{r}$
3) Proportional to $r^2$
4) Proportional to $r$

Solution

$
\begin{aligned}
& \mathrm{V}=\frac{4}{3} \pi r^3 \Rightarrow \frac{d \mathrm{~V}}{d t}=4 \pi r^2 \cdot \frac{d r}{d t} \\
& \mathrm{~S}=4 \pi r^2 \Rightarrow \frac{d \mathrm{~S}}{d t}=8 \pi r \cdot \frac{d r}{d t} \\
& \Rightarrow 8=8 \pi r \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{1}{\pi r}
\end{aligned}
$
Putting the value of $\frac{d r}{d t}$ in (i), we get $\frac{d \mathrm{~V}}{d t}=4 \pi r^2 \times \frac{1}{\pi r}=4 r$
$\Rightarrow \frac{d \mathrm{~V}}{d t}$ is proportional to $r$.

Hence, the answer is option 4.

Example 2: A spherical balloon is being inflated at the rate of $35 \mathrm{cc} / \mathrm{min}$. The rate of increase in surface area( in $\mathrm{cm}^2 / \mathrm{min}$.) of the balloon when its diameter is 14 cm , is :
1) 10
2) $\sqrt{10}$
3) 100
4) $10 \sqrt{10}$

Solution

Volume of the spherical ballon is, $\mathrm{V}=\frac{4}{3} \pi r^3$

$
\begin{aligned}
& \frac{d \mathrm{~V}}{d t}=\frac{4}{3} \cdot \pi \cdot 3 r^2 \cdot \frac{d r}{d t} \\
& 35=4 \pi r^2 \cdot \frac{d r}{d t} \text { or } \frac{d r}{d t}=\frac{35}{4 \pi r^2}
\end{aligned}
$
Now, surface area of sphere is, $S=4 \pi r^2$

$
\begin{aligned}
& \frac{d S}{d t}=(4 \pi) 2 r \frac{d r}{d t} \\
& \frac{d \mathrm{~S}}{d t}=8 \pi r \times \frac{35}{4 \pi r^2}=\frac{70}{r}
\end{aligned}
$
Now, diameter $=14 \mathrm{~cm}, r=7$

$
\therefore \quad \frac{d \mathrm{~S}}{d t}=10
$
Hence, the answer is option 1.

Example 3: The population $\mathrm{P}=\mathrm{P}(\mathrm{t})$ at time ' $t$ ' of ascertain species follows the differential equation. Then the time at which the population becomes zero:
1) $\log _e 18$
2) $\frac{1}{2} \log _e 18$
3) $\log _e 9$
4) $2 \log _e 18$

Solution

$
\begin{aligned}
& \frac{\mathrm{dP}(\mathrm{t})}{\mathrm{dt}}=\frac{\mathrm{P}(\mathrm{t})-900}{2} \\
& \int_0^t \frac{\mathrm{dP}(\mathrm{t})}{\mathrm{P}(\mathrm{t})-900}=\int_0^{\mathrm{t}} \frac{\mathrm{dt}}{2} \\
& \{\ell \ln |\mathrm{P}(\mathrm{t})-900|\}_0^{\mathrm{t}}=\left\{\frac{\mathrm{t}}{2}\right\}_0^{\mathrm{t}} \\
& \operatorname{\ell n}|\mathrm{P}(\mathrm{t})-900|-\ln |\mathrm{P}(0)-900|=\frac{\mathrm{t}}{2} \\
& \ln |\mathrm{P}(\mathrm{t})-900|-\ln 50=\frac{\mathrm{t}}{2}
\end{aligned}
$
Let at $t=t_1, P(t)=0$ hence

$
\begin{aligned}
& \ln |\mathrm{P}(\mathrm{t})-900|-\ln 50=\frac{\mathrm{t}_1}{2} \\
& \mathrm{t}_1=2 \ell \mathrm{n} 18
\end{aligned}
$
Hence, the answer is the option 4.

Example 4: The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of the balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is:
1) $9$
2) $10$
3) $11$
4) $12$

Solution

Let r: radius

$
\begin{aligned}
& \mathrm{s}=4 \pi \mathrm{r}^2 \\
& \frac{\mathrm{ds}}{\mathrm{dt}}=8 \pi \mathrm{r}\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)=\lambda \text { (constant) } \\
& 4 \pi \mathrm{r}^2=\lambda \mathrm{t}+\mathrm{c} \\
& \text { at } \mathrm{t}=0, \mathrm{r}=3 \quad \Rightarrow \quad \mathrm{c}=36 \pi \\
& \text { at } \mathrm{t}=5, \mathrm{r}=7 \quad \Rightarrow \quad \lambda=32 \pi \\
& \therefore \lambda^2=8 \mathrm{t}+9 \\
& \therefore \text { at } \mathrm{t}=9 \\
& \quad \mathrm{r}^2=81 \\
& \mathrm{r}=9 \mathrm{unit}
\end{aligned}
$
Hence, the answer is the option 1.

Example 5: Water is being filled at the rate of $1 \mathrm{~cm}^3 / \mathrm{sec}_{\text {in }}$ in a right circular conical vessel ( vertex downwards ) of height 35 cm and diameter 14 cm . When the height of the water level is 10 cm , the rate ( in $\mathrm{cm}^2 / \mathrm{sec}$ ) at which the wet conical surface area of the vessel increases is:
1) 5
2) $\frac{\sqrt{21}}{5}$
3) $\frac{\sqrt{26}}{5}$
4) $\frac{\sqrt{26}}{10}$

Solution

Let the volume of the cone be $\mathrm{V} \mathrm{cm}{ }^3$
Given $\frac{\mathrm{dV}}{\mathrm{dt}}=1 \mathrm{~cm}^3 / \mathrm{sec}, \mathrm{h}=35 \mathrm{~cm}, \mathrm{r}=7 \mathrm{~cm}$
i.e $\frac{\mathrm{h}}{\mathrm{r}}=5$

We know for a cone $\mathrm{l}^2=\mathrm{r}^2+\mathrm{h}$
Lateral Surface area,

$
\begin{aligned}
& \mathrm{S}=\pi \mathrm{r} \sqrt{\mathrm{r}^2+\mathrm{h}^2} \\
& \mathrm{~S}=\pi \frac{\mathrm{h}}{5} \sqrt{\frac{\mathrm{h}^2}{25}+\mathrm{h}^2}=\pi \frac{\sqrt{26}}{25} \mathrm{~h}^2 \\
& \mathrm{~V}=\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h} \frac{1}{3} \pi\left(\frac{\mathrm{h}}{5}\right)^2 \mathrm{~h}=\frac{\pi}{75} \mathrm{~h}^3 \\
& \Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\pi}{25} \mathrm{~h}^2 \frac{\mathrm{dh}}{\mathrm{dt}} \Rightarrow \frac{\pi}{25} \mathrm{~h}^2 \frac{\mathrm{dh}}{\mathrm{dt}}=1 \\
& \Rightarrow \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{25}{\pi \mathrm{h}^2}
\end{aligned}
$

$
\begin{aligned}
& \frac{\mathrm{ds}}{\mathrm{dt}}=\frac{\pi \sqrt{26}}{25} \times 2 \mathrm{~h} \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{2 \sqrt{26}}{\mathrm{~h}} \\
& \left(\frac{\mathrm{ds}}{\mathrm{dt}}\right)_{\mathrm{h}=10}=\frac{\sqrt{26}}{5}
\end{aligned}
$
Hence, the answer is the option (3).

List of topics related to Derivative as Rate Measure

This section lists important subtopics and related concepts you’ll study under the idea of derivatives as rate measures, helping you connect calculus with real-world rate-based applications.

Differentiability and Existence of Derivative

Examining differentiability Using Graph of Function

Continuity of Composite Function

Continuity And Differentiability

NCERT Resources

This section compiles all NCERT study resources, including notes, solutions, and exemplar exercises, to strengthen your conceptual clarity and exam preparation.

NCERT Class 12 Maths Notes for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Solutions for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Exemplar Solutions for Chapter 5 - Continuity and Differentiability

Practice Questions based on Derivative as Rate Measure

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Derivative As Rate Measure- Practice Question MCQ

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