Applications of Differentiation — JEE Advanced Physics

Key Concepts

5.4 Approximating Function Values Without a Calculator

Suppose you already know the exact value of a function $f(x)$ at some point $x_0$ , and you need the value of $f(x)$ at a nearby point $x_0+\Delta x$ , where $\Delta x$ is small. Rather than recomputing $f$ from scratch at the new point, the derivative lets us estimate it directly — this is the idea of linear approximation (also called differential approximation).

Recall that the derivative $f'(x_0)$ is the slope of the tangent line to $y=f(x)$ at $x_0$ . Near $x_0$ , the curve and its tangent line are almost indistinguishable, so instead of following the curve itself to $x_0+\Delta x$ , we can follow the tangent line, which is much easier to evaluate. Moving a small distance $\Delta x$ along that tangent line changes the height by approximately $\Delta x \cdot f'(x_0)$ . Adding this change to the starting value $f(x_0)$ gives the approximation:

$f(x_0+\Delta x) \approx f(x_0) + \Delta x\,\frac{d(f(x))}{dx}\bigg|_{x=x_0}$

The right-hand side is simply the sum of the known value of the function at $x_0$ and the estimated change in the function due to the small step $\Delta x$ , where that change is found by multiplying the rate of change of $f(x)$ at $x_0$ by $\Delta x$ itself.

Accuracy of the Approximation

The value obtained for $f(x_0+\Delta x)$ from this formula is only approximate, not exact. Its accuracy depends on two things: (i) how linear the function $f(x)$ actually is near $x_0$ — the less it curves, the better the straight-line estimate; and (ii) how small $\Delta x$ is — the smaller the step, the closer the tangent line stays to the true curve. Two fully worked examples applying this formula — estimating $(2.01)^3$ and $(8.01)^{2/3}$ without directly computing either cube or cube root — are given as Examples 29 and 30 in the Examples section.

Using linear approximation with $f(x) = x^2$ , $f(5) = 25$ and $f'(x) = 2x$ , estimate $(5.02)^2$ .

In the linear approximation $f(x_0+\Delta x) \approx f(x_0) + \Delta x\, f'(x_0)$ , at which point is the derivative evaluated?

Examples

Worked Examples: Applications of the Derivative

The examples below work through every application introduced in this chapter, in the same order: slope of the tangent (including implicit differentiation), the increasing/decreasing test, maxima and minima, and linear approximation.

Example 24: Slope of the Tangent to y = x²

Find the slope of the tangent drawn at $x=\dfrac{1}{2}$ on the curve $y=x^2$ .

Solution: The derivative of $y=x^2$ is

$\frac{dy}{dx} = 2x$

The slope of the tangent at $x=\dfrac{1}{2}$ is

$\tan\theta = \left.\frac{dy}{dx}\right|_{x=\frac{1}{2}} = 2\left(\frac{1}{2}\right) = 1$

Since $\tan\theta = 1$ , the tangent drawn at $x=\dfrac{1}{2}$ is inclined at $45^\circ$ to the x-axis.

The parabola y = x^2 with its tangent at x = 1/2, inclined at 45 degrees to the x-axis.

Example 25: Slope of a Circle via Implicit Differentiation

Find the slope of the circle $x^2+y^2=25$ at the point $(3,-4)$ .

Solution: The circle is not the graph of a single function of $x$ — it is the combined graphs of two differentiable functions, $y_1=\sqrt{25-x^2}$ (the upper semicircle) and $y_2=-\sqrt{25-x^2}$ (the lower semicircle). The point $(3,-4)$ lies on the graph of $y_2$ , so one way to proceed is to differentiate $y_2$ explicitly with the chain rule:

$\left.\frac{dy_2}{dx}\right|_{x=3} = \left.-\frac{-2x}{2\sqrt{25-x^2}}\right|_{x=3} = -\frac{-6}{2\sqrt{25-9}} = \frac{6}{8} = \frac{3}{4}$

But the same result is reached far more easily by differentiating the equation of the circle implicitly with respect to $x$ , treating $y$ as an (unspecified) function of $x$ and applying the chain rule to the $y^2$ term:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$

$2x + 2y\frac{dy}{dx} = 0$

$\frac{dy}{dx} = -\frac{x}{y}$

The slope at $(3,-4)$ is then

$\left.\frac{dy}{dx}\right|_{(3,-4)} = -\frac{3}{-4} = \frac{3}{4}$

which matches the explicit calculation, without ever needing to solve for $y_1$ or $y_2$ separately.

The circle x^2 + y^2 = 25 with upper branch y1 and lower branch y2, and the tangent of slope 3/4 at (3, -4).

Example 26: Testing Whether sin x Increases or Decreases

Check whether $\sin x$ increases or decreases at $x=\dfrac{\pi}{3}$ .

Solution: Let $y=\sin x$ . Then

$\frac{dy}{dx} = \cos x$

The value of $\cos x$ at $x=\dfrac{\pi}{3}$ is $\cos\dfrac{\pi}{3} = \dfrac{1}{2}$ , which is positive. Since $\dfrac{dy}{dx}>0$ at $x=\dfrac{\pi}{3}$ , the function $\sin x$ is increasing at that point.

Example 27: Classifying the Extrema of Two Simple Functions

For what value of $x$ is $f(x)$ minimum or maximum in each of the following cases? (i) $f(x)=x^2$ (ii) $f(x)=-x^2+2x$

Solution: A maximum or minimum can only occur where $\dfrac{d(f(x))}{dx}=0$ . The sign of the second derivative there then decides which one it is:

$\text{Maximum if } \frac{d}{dx}\!\left(\frac{d(f(x))}{dx}\right) < 0, \qquad \text{Minimum if } \frac{d}{dx}\!\left(\frac{d(f(x))}{dx}\right) > 0$

(i) $\dfrac{d(f(x))}{dx}=2x$ , which is zero at $x=0$ . Its derivative is $\dfrac{d(2x)}{dx}=2$ , which is positive. Hence $f(x)=x^2$ has a minimum at $x=0$ .

(ii) $\dfrac{d(f(x))}{dx}=-2x+2$ , which is zero at $x=1$ . Its derivative is $\dfrac{d(-2x+2)}{dx}=-2$ , which is negative. Hence $f(x)=-x^2+2x$ has a maximum at $x=1$ .

Example 28: Maximum Height of a Projectile

The height of a particle ( $h$ , in metres) above the ground, in terms of time ( $t$ , in seconds), is given by

$h = (20\,\text{m/s})\,t + \frac{1}{2}(-10\,\text{m/s}^2)\,t^2$

Find the instant at which the particle attains its maximum height, and find that maximum height.

Solution: If $h$ is maximum at $t=t_0$ , the rate of change of $h$ at that instant must be zero:

$\frac{dh}{dt} = (20\,\text{m/s}) - (10\,\text{m/s}^2)\,t = 0$

which gives $t = 2\,\text{s}$ . To confirm this is a maximum rather than a minimum, check the second derivative:

$\frac{d}{dt}\!\left(\frac{dh}{dt}\right) = \frac{d}{dt}(20-10t) = -10\,\text{m/s}^2$

Since this is negative, $t=2\,\text{s}$ is indeed the instant of maximum height. Substituting back, the maximum height is

$h_{max} = (20\,\text{m/s})(2\,\text{s}) + \frac{1}{2}(-10\,\text{m/s}^2)(2\,\text{s})^2 = 40\,\text{m} - 20\,\text{m} = 20\,\text{m}$

Example 29: Linear Approximation of (2.01)³

Find the value of $(2.01)^3$ using linear approximation.

Solution: Let $f(x)=x^3$ , and take $x_0=2$ since it is close to $2.01$ and easy to evaluate exactly. Then $f(2)=8$ , and

$\frac{d(f(x))}{dx} = 3x^2$

Using the linear approximation formula with $\Delta x = 0.01$ :

$f(2.01) \approx f(2) + (0.01)\times \left.3x^2\right|_{x=2} = (2)^3 + (0.01)\times 3(2)^2$

$f(2.01) \approx 8 + (0.01)(12) = 8 + 0.12 = 8.12$

So $(2.01)^3 \approx 8.12$ (the exact value is $8.120601$ , confirming the approximation is excellent for such a small $\Delta x$ ).

Example 30: Linear Approximation of (8.01)^(2/3)

Find the value of $(8.01)^{2/3}$ using linear approximation.

Solution: Let $f(x)=x^{2/3}$ , and take $x_0=8$ since $8^{2/3}=4$ is easy to evaluate exactly. Then $f(8)=4$ , and

$\frac{d(f(x))}{dx} = \frac{2}{3}x^{-1/3}$

Using the linear approximation formula with $\Delta x = 0.01$ :

$f(8.01) \approx f(8) + (0.01)\times \left.\frac{2}{3}x^{-1/3}\right|_{x=8} = (8)^{2/3} + (0.01)\times\frac{2}{3}(8)^{-1/3}$

$f(8.01) \approx 4 + (0.01)\left(\frac{2}{3}\times\frac{1}{2}\right) = 4 + (0.01)(0.333) \approx 4 + 0.0033 = 4.0033$

So $(8.01)^{2/3} \approx 4.0033$ .

A particle's height is $h = (30\ \text{m/s})\,t + \tfrac{1}{2}(-10\ \text{m/s}^2)\,t^2$ . What is its maximum height?

For the circle x² + y² = 25, implicit differentiation gives dy/dx = -x/y. What is the slope of the tangent at the point (0, 5)?

Instantaneous Velocity and Acceleration

Kinematics and the Derivative

The most direct and powerful application of differentiation in physics is in the study of motion, or kinematics. As we've learned, the derivative measures the instantaneous rate of change. This is precisely what we need to define the velocity and acceleration of an object at a single moment in time.

Instantaneous Velocity

We know that average velocity is the change in position divided by the change in time ( $\frac{\Delta x}{\Delta t}$ ). To find the velocity at a single instant, we need to make that time interval, $\Delta t$ , infinitesimally small. This is the exact definition of the derivative.

The instantaneous velocity, $v(t)$ , is the derivative of the position function, $x(t)$ , with respect to time.

$v(t) = \frac{dx}{dt} = \lim_{\Delta t \to 0} \frac{x(t + \Delta t) - x(t)}{\Delta t}$

Geometrically, this means the velocity at any time $t$ is the slope of the tangent line on a position-time graph at that specific time.

A tangent drawn to the position-time curve at the instant t0. Its slope is the instantaneous velocity v(t0).

Instantaneous Acceleration

Similarly, acceleration is the rate at which velocity changes. The instantaneous acceleration, $a(t)$ , is the derivative of the velocity function, $v(t)$ , with respect to time.

$a(t) = \frac{dv}{dt}$

Since velocity is already the derivative of position, acceleration is the second derivative of the position function.

$a(t) = \frac{d}{dt}\left(\frac{dx}{dt}\right) = \frac{d^2x}{dt^2}$

Geometrically, the acceleration at any time $t$ is the slope of the tangent line on a velocity-time graph at that time.

Example: Analyzing Motion

The position of a particle is given by $x(t) = 2t^3 - 9t^2 + 12t$ . Find its velocity and acceleration at $t=2$ s.

1. Find the velocity function, v(t):
Differentiate $x(t)$ with respect to $t$ .
$v(t) = \frac{dx}{dt} = 6t^2 - 18t + 12$

2. Find the acceleration function, a(t):
Differentiate $v(t)$ with respect to $t$ .
$a(t) = \frac{dv}{dt} = 12t - 18$

3. Evaluate at t=2 s:
$v(2) = 6(2)^2 - 18(2) + 12 = 24 - 36 + 12 = 0$ m/s.
$a(2) = 12(2) - 18 = 24 - 18 = 6$ m/s².

Interpretation: At exactly 2 seconds, the particle is momentarily at rest ( $v=0$ ), but its velocity is about to increase in the positive direction because its acceleration is positive.

If an object's velocity is negative and its acceleration is also negative, what is happening to the object's motion?

Introduction to Applications of Differentiation

From a Rate of Change to a Toolkit

So far, differentiation has been introduced as an abstract idea: the derivative $\frac{dy}{dx}$ is the instantaneous rate of change of $y$ with respect to $x$ , and geometrically it is the slope of the tangent line to the curve $y=f(x)$ . That single idea, it turns out, is powerful enough to answer four very different practical questions that show up constantly in physics — from projectile motion to circuit design. This chapter builds a toolkit around it.

Over the next few sections, we will use the derivative to:

Measure rate of change — decide whether a quantity is increasing, decreasing, or momentarily unchanging, just by checking the sign of its derivative.
Find the slope of a tangent — determine the exact direction of a curve at any point, including curves that are not simple functions $y=f(x)$ , using implicit differentiation.
Locate maxima and minima — find the largest or smallest value a quantity can take, such as the maximum height of a projectile, by finding where the derivative vanishes.
Approximate function values — estimate $f(x_0+\Delta x)$ for a small change $\Delta x$ without recomputing the function from scratch, using a linear (first-order) approximation.

Each of these applications is explored in its own section below, with fully worked examples collected at the end of the chapter.

5.1 Measuring Rate of Change

If a variable $y$ depends on $x$ , then $\dfrac{dy}{dx}$ represents the rate of change of $y$ as $x$ increases. This single number tells us everything about the local behaviour of the function:

If $y$ increases with increasing $x$ , then $\dfrac{dy}{dx}$ is positive.
If $y$ decreases with increasing $x$ , then $\dfrac{dy}{dx}$ is negative.
If $y$ is momentarily unchanged as $x$ increases, then $\dfrac{dy}{dx}$ is zero.

This fact turns differentiation into a test: to find out whether a function is increasing or decreasing at a particular point, we don't need to plot the curve — we just evaluate the derivative there and read off its sign. This is the same sign rule you have already met in the geometric picture of the derivative (an upward-sloping tangent means the function is increasing), stated here as a standalone tool that we will use repeatedly for the rest of the chapter.

A fully worked check of this test — deciding whether $\sin x$ is increasing or decreasing at $x=\frac{\pi}{3}$ — is given as Example 26 in the Examples section.

A quantity $y$ depends on $x$ and $\dfrac{dy}{dx} < 0$ at $x = x_0$ . What can you conclude about $y$ near $x_0$ ?

Finding Maxima and Minima

Optimization: Finding the Best Value

One of the most powerful applications of calculus is in optimization—finding the largest or smallest value that a function can take. This has endless applications, from maximizing the area of a field to minimizing the cost of a manufacturing process. The key to finding these 'extreme' values lies in the derivative.

Critical Points: Where Extrema Can Occur

Local maximums ('peaks') and minimums ('valleys') of a function have a special property: the tangent line at these points is horizontal, meaning its slope is zero. This gives us a crucial clue.

A critical point of a function is a point where the derivative is either zero or undefined. These are the only candidates for where a local maximum or minimum can occur.

$f'(x) = 0 \quad \text{or} \quad f'(x) \text{ is undefined}$

At the local maximum and the local minimum the tangent is horizontal, so f'(x) = 0 at both points.

How to Find Critical Points:

Find the derivative of the function, $f'(x)$ .
Set the derivative equal to zero and solve for $x$ .
Determine if there are any points where the derivative is undefined.

The Second Derivative Test: A Quick Check

Once you have a critical point $c$ (where $f'(c)=0$ ), the Second Derivative Test is often the quickest way to classify it. It uses the concavity of the function at that point.

If $f''(c) > 0$ (concave up), the point is a local minimum.
If $f''(c) < 0$ (concave down), the point is a local maximum.
If $f''(c) = 0$ , the test is inconclusive. You must use the First Derivative Test.

Example: Finding the Minimum of a Function

Find the local extrema of $f(x) = x^3 - 3x^2 + 1$ .

1. Find Critical Points:
First, find the derivative: $f'(x) = 3x^2 - 6x$ .
Set it to zero: $3x^2 - 6x = 0 \implies 3x(x - 2) = 0$ .
The critical points are $x=0$ and $x=2$ .

2. Use the Second Derivative Test:
Find the second derivative: $f''(x) = 6x - 6$ .
Check each critical point:
At $x=0$ : $f''(0) = 6(0) - 6 = -6$ . Since $f''(0) < 0$ , this is a local maximum.
At $x=2$ : $f''(2) = 6(2) - 6 = 6$ . Since $f''(2) > 0$ , this is a local minimum.

You have found a critical point of a function where the first derivative is zero. The second derivative at this point is also zero. What can you conclude?

Where the Second Derivative Test Comes From

The Second Derivative Test is not an arbitrary rule — it falls directly out of the increasing/decreasing behaviour of a smooth function near a peak or a valley. Consider a continuous, smooth function $f(x)$ that is maximum at $x=x_0$ . Just before $x_0$ , the function must still be climbing towards that peak, and just after $x_0$ it must be falling away from it, with the value momentarily unchanged exactly at $x_0$ :

$f(x)$ increases just before $x=x_0$ ,
$f(x)$ decreases just after $x=x_0$ , and
$f(x)$ is momentarily constant at $x=x_0$ .

In terms of the derivative, this means $\dfrac{dy}{dx}$ goes from a positive value, to zero, to a negative value as $x$ increases through $x_0$ . In other words, $\dfrac{dy}{dx}$ is itself continuously decreasing at $x_0$ — which is exactly what a negative second derivative means. This is precisely the pair of conditions used above:

$\left.\frac{dy}{dx}\right|_{x=x_0} = 0 \qquad \text{and} \qquad \left.\frac{d}{dx}\!\left(\frac{dy}{dx}\right)\right|_{x=x_0} < 0$

A curve with a maximum at x = x0: the tangent slope dy/dx is positive before x0, zero at x0, and negative after x0.

The argument runs in exactly the opposite direction for a minimum at $x=x_0$ . Just before $x_0$ the function must still be falling towards the valley, and just after $x_0$ it must be climbing back out, with the value momentarily unchanged at $x_0$ itself:

$f(x)$ decreases just before $x=x_0$ ,
$f(x)$ increases just after $x=x_0$ , and
$f(x)$ is momentarily constant at $x=x_0$ .

Now $\dfrac{dy}{dx}$ goes from a negative value, to zero, to a positive value — it is itself continuously increasing at $x_0$ , which is exactly what a positive second derivative means:

$\left.\frac{dy}{dx}\right|_{x=x_0} = 0 \qquad \text{and} \qquad \left.\frac{d}{dx}\!\left(\frac{dy}{dx}\right)\right|_{x=x_0} > 0$

A curve with a minimum at x = x0: the tangent slope dy/dx is negative before x0, zero at x0, and positive after x0.

This is the same statement as the Second Derivative Test above, just derived from first principles rather than quoted as a rule. Two fully worked applications of this idea — classifying the extrema of simple polynomials, and finding the maximum height of a projectile — are given as Examples 27 and 28 in the Examples section.

Solving Related Rates Problems

When Everything Changes Together

In many real-world situations, several quantities are changing over time, but they are all related to each other. For example, as a spherical balloon is inflated, its radius, volume, and surface area are all increasing, and their rates of change are related. Related rates problems involve finding the rate of change of one quantity by using the known rate of change of another.

The key tool for solving these problems is to find an equation that connects the variables, and then use implicit differentiation with respect to time, often involving the chain rule.

A Step-by-Step Procedure

Identify Variables and Rates: Read the problem carefully. Identify all the quantities that are changing and the rates of change. Assign variables to them. Determine which rate you know and which rate you need to find.
Draw a Diagram: A sketch of the situation is almost always helpful. Label it with your variables.
Find a Relating Equation: Find an equation that connects the variables. This often comes from geometry (Pythagorean theorem, area/volume formulas, similar triangles).
Differentiate Implicitly: Differentiate both sides of your equation with respect to time ( $\frac{d}{dt}$ ). Remember to use the chain rule for every variable, as they are all functions of time. For example, the derivative of $r^2$ with respect to $t$ is $2r \frac{dr}{dt}$ .
Substitute and Solve: Substitute all the known values for the variables and their rates into the differentiated equation. Solve for the unknown rate.

Example: The Sliding Ladder

Problem: A 10-foot ladder is leaning against a vertical wall. The bottom of the ladder is sliding away from the wall at a rate of 2 ft/s. How fast is the top of the ladder sliding down the wall when the bottom is 6 feet from the wall?

The sliding ladder of fixed length L: the foot moves away from the wall at dx/dt while the top drops at dy/dt.

1. Variables & Rates:
Let $x$ be the distance from the wall to the bottom of the ladder.
Let $y$ be the height of the top of the ladder.
Known rate: $\frac{dx}{dt} = 2$ ft/s.
Find: $\frac{dy}{dt}$ when $x=6$ ft.

2. Equation:
By the Pythagorean theorem: $x^2 + y^2 = 10^2$ .

3. Differentiate:
Differentiate with respect to time $t$ :
$\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(100)$
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$

4. Substitute and Solve:
We need to find $y$ when $x=6$ . From the original equation: $6^2 + y^2 = 100 \implies y^2 = 64 \implies y=8$ .
Now substitute everything into the differentiated equation:
$2(6)(2) + 2(8)\frac{dy}{dt} = 0$
$24 + 16\frac{dy}{dt} = 0$
$\frac{dy}{dt} = -\frac{24}{16} = -1.5$ ft/s.
(The negative sign indicates that the height is decreasing, i.e., the ladder is sliding down the wall.)

The volume of a cube is increasing at a rate of 12 cm³/s. What is the primary equation you would use to find the rate at which the side length (s) is changing?

Equations of Tangent and Normal Lines

Geometric Applications of the Derivative

We've established that the derivative of a function at a point gives the slope of the tangent line at that point. This is the fundamental link between calculus and geometry. Using this, we can find the exact equations for two very important lines: the tangent line and the normal line.

5.2 Determination of the Slope of a Tangent

Consider a function $y=f(x)$ whose graph is some curve. For the curve $y=f(x)$ , a tangent drawn at $x=x_0$ has a slope equal to the value of $\dfrac{dy}{dx}$ evaluated at $x=x_0$ :

$\tan\theta = \left.\frac{dy}{dx}\right|_{x=x_0}$

where $\theta$ is the angle the tangent line makes with the positive x-axis. This is exactly the geometric meaning of the derivative applied at one specific point — plug in the x-coordinate of interest, and the derivative hands back the tangent's slope directly.

A tangent drawn to the curve y = f(x) at the point x = x0, making angle theta with the positive x-axis.

Two fully worked examples of this idea — including a curve that is not a simple function y = f(x), handled using implicit differentiation — are given as Examples 24 and 25 in the Examples section.

The Tangent and Normal Lines

The Tangent Line is a straight line that 'just touches' the curve at a single point and has the same direction as the curve at that point.
The Normal Line is a straight line that is perpendicular to the tangent line at that same point.

The tangent touches the curve at P; the normal crosses it there at right angles.

Finding Their Slopes

Let the slope of the tangent line be $m_t$ and the slope of the normal line be $m_n$ .

Slope of the Tangent: This is simply the value of the derivative at the point of tangency, $x_0$ .
$m_t = f'(x_0)$
Slope of the Normal: Since the normal is perpendicular to the tangent, its slope is the negative reciprocal of the tangent's slope.
$m_n = -\frac{1}{m_t} = -\frac{1}{f'(x_0)}$

Finding the Equations

Once we know the slope of a line and a point it passes through, $(x_0, y_0)$ , we can find its equation using the point-slope form: $y - y_0 = m(x - x_0)$ .

Step-by-Step Procedure:

Find the Point: You'll be given an x-value, $x_0$ . Find the corresponding y-value by calculating $y_0 = f(x_0)$ .
Find the Derivative: Calculate the derivative function, $f'(x)$ .
Find the Slopes: Evaluate the derivative at $x_0$ to get the tangent slope, $m_t = f'(x_0)$ . Then find the normal slope, $m_n = -1/m_t$ .
Write the Equations: Plug the point $(x_0, y_0)$ and the slopes into the point-slope formula.
Tangent Line: $y - y_0 = m_t(x - x_0)$
Normal Line: $y - y_0 = m_n(x - x_0)$

Example

Find the equations of the tangent and normal lines to the curve $y = x^2$ at the point $x=2$ .

1. Point: At $x_0=2$ , $y_0 = (2)^2 = 4$ . The point is (2, 4).

2. Derivative: $f'(x) = 2x$ .

3. Slopes:
Slope of tangent: $m_t = f'(2) = 2(2) = 4$ .
Slope of normal: $m_n = -1/4$ .

4. Equations:
Tangent: $y - 4 = 4(x - 2) \implies y = 4x - 4$ .
Normal: $y - 4 = -\frac{1}{4}(x - 2) \implies y = -\frac{1}{4}x + \frac{9}{2}$ .

What is the slope of the normal line to the curve $f(x) = x^3$ at the point $x=1$ ?