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Rate of Change, Tangents, Normals & Approximations

Master Rate of Change, Tangents, Normals and Approximations for JEE Main, JEE Advanced & Board Exams. Covers related rates, equations of tangent and normal, orthogonal curves, angle between curves, linear approximation and error analysis with solved examples.

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The derivative $f^{'} (x)$ at a point is the instantaneous rate of change of $f (x)$ — and this single idea unlocks a powerful toolkit for analysing curves. Based on the 2026 syllabus updates, this toolkit is strictly segmented: measuring how quickly one quantity changes relative to another (rates of change) remains a core part of JEE Main and Advanced. However, finding the lines that touch a curve at a point (tangents and normals) is now strictly JEE Advanced Only, and using the derivative to estimate values without exact calculation (approximations) has been removed from JEE entirely, making it CBSE Boards Only.

1. Rate of Change of Quantities JEE Main & Advanced

If $y = f (x)$ , the derivative $\frac{d y}{d x}$ represents the rate of change of $y$ with respect to $x$ . More generally, if both $x$ and $y$ are functions of time $t$ :

$\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$

This is simply the chain rule applied to related rates problems.

If $\frac{d y}{d x} > 0$ : $y$ increases as $x$ increases.
If $\frac{d y}{d x} < 0$ : $y$ decreases as $x$ increases.
If $\frac{d y}{d x} = 0$ : $y$ is momentarily stationary with respect to $x$ .

Related Rates — Standard Results

Quantity	Formula	Rate of Change
Area of circle (radius $r$ )	$A = π r^{2}$	$\frac{d A}{d t} = 2 π r \frac{d r}{d t}$
Volume of sphere (radius $r$ )	$V = \frac{4}{3} π r^{3}$	$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
Surface area of sphere	$S = 4 π r^{2}$	$\frac{d S}{d t} = 8 π r \frac{d r}{d t}$
Volume of cylinder (fixed $h$ )	$V = π r^{2} h$	$\frac{d V}{d t} = 2 π r h \frac{d r}{d t}$
Volume of cube (side $a$ )	$V = a^{3}$	$\frac{d V}{d t} = 3 a^{2} \frac{d a}{d t}$

Worked Example

If the radius of a sphere increases at $2$ cm/s, the rate of increase of its volume when $r = 5$ cm is:

$\frac{d V}{d t} = 4 π r^{2} \cdot \frac{d r}{d t} = 4 π (25) (2) = 200 π$ cm³/s.

2. Tangents and Normals JEE Advanced Only

For a curve $y = f (x)$ , at the point $P (x_{1}, y_{1})$ :

Slope of tangent: $m_{T} = f^{'} (x_{1}) = {\frac{d y}{d x} |}_{(x_{1}, y_{1})}$
Slope of normal: $m_{N} = - \frac{1}{m_{T}} = - \frac{1}{f^{'} (x_{1})}$ (since tangent $⊥$ normal)

Equations of Tangent and Normal

Line	Equation at $(x_{1}, y_{1})$	Special Case
Tangent	$y - y_{1} = f^{'} (x_{1}) (x - x_{1})$	If $f^{'} (x_{1}) = 0$ : horizontal tangent ( $y = y_{1}$ )
Normal	$y - y_{1} = - \frac{1}{f^{'} (x_{1})} (x - x_{1})$	If $f^{'} (x_{1}) = 0$ : vertical normal ( $x = x_{1}$ )

Special Scenarios

Vertical tangent: When $f^{'} (x_{1}) \to \infty$ — tangent is $x = x_{1}$ , normal is $y = y_{1}$ .
Tangent parallel to $x$ -axis: $f^{'} (x_{1}) = 0$ .
Tangent parallel to $y$ -axis: $f^{'} (x_{1})$ is undefined (vertical tangent).
Tangent passing through origin: $y_{1} = f^{'} (x_{1}) \cdot x_{1}$ .
Tangent at parametric curve $x = g (t)$ , $y = h (t)$ : $\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{h^{'} (t)}{g^{'} (t)}$ .

Angle Between Two Curves

The angle $ϕ$ between two curves at their point of intersection, where slopes are $m_{1}$ and $m_{2}$ :

$\tan ϕ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$

If $m_{1} m_{2} = - 1$ : curves are orthogonal (perpendicular).
If $m_{1} = m_{2}$ : curves are tangential (touch at that point).

Length of Tangent, Normal, Subtangent and Subnormal

At point $(x_{1}, y_{1})$ with slope $m = f^{'} (x_{1})$ :

Quantity	Formula
Length of tangent	$\frac{y_{1} \sqrt{1 + m^{2}}}{\| m \|}$
Length of normal	$y_{1} \sqrt{1 + m^{2}}$
Length of subtangent	$\| \frac{y_{1}}{m} \|$
Length of subnormal	$\| y_{1} m \|$

3. Approximations Using Derivatives CBSE Boards Only

If $δ x$ is a small change in $x$ , the corresponding approximate change in $y = f (x)$ is:

$δ y \approx f^{'} (x) \cdot δ x = \frac{d y}{d x} \cdot δ x$

This is called the linear approximation (tangent line approximation). The exact change is $Δ y = f (x + Δ x) - f (x)$ ; the differential $d y = f^{'} (x) d x$ approximates it.

Error Analysis

Absolute error: $Δ y \approx f^{'} (x) Δ x$
Relative error: $\frac{Δ y}{y} \approx \frac{f^{'} (x)}{f (x)} Δ x$
Percentage error: $\frac{Δ y}{y} \times 100$

Standard Approximation Formulae (for small $h$ )

Function	Approximation (small $h$ )
$(1 + h)^{n}$	$\approx 1 + n h$
$\sqrt{1 + h}$	$\approx 1 + \frac{h}{2}$
$e^{h}$	$\approx 1 + h$
$\ln (1 + h)$	$\approx h$
$\sin h$ (in radians)	$\approx h$

Worked Example

Approximate $\sqrt{26}$ : Let $f (x) = \sqrt{x}$ , $x = 25$ , $δ x = 1$ .
$f^{'} (x) = \frac{1}{2 \sqrt{x}}$ , so $f^{'} (25) = \frac{1}{10}$ .
$\sqrt{26} \approx \sqrt{25} + \frac{1}{10} (1) = 5.1$ .

Tangent & Normal — Step-by-Step Approach

Step	Action
1	Find $f^{'} (x)$ — differentiate the curve equation.
2	Substitute the given point $(x_{1}, y_{1})$ to get slope $m = f^{'} (x_{1})$ .
3 — Tangent	Use $y - y_{1} = m (x - x_{1})$ . If $m = 0$ : tangent is $y = y_{1}$ . If $m$ undefined: tangent is $x = x_{1}$ .
4 — Normal	Use $y - y_{1} = - \frac{1}{m} (x - x_{1})$ . If $m = 0$ : normal is $x = x_{1}$ . If $m$ undefined: normal is $y = y_{1}$ .
5	Verify that the point actually lies on the curve — substitute back.

JEE/NEET Power Tips — Avoid These Common Mistakes

Normal slope is $- 1 / f^{'} (x_{1})$ , not $- f^{'} (x_{1})$ . A very common reciprocal error in board exams — the normal slope is the negative reciprocal of the tangent slope, not the negative of the tangent slope.
When the tangent is horizontal ( $m = 0$ ), the normal is vertical ( $x = x_{1}$ ). When the tangent is vertical ( $m \to \infty$ ), the normal is horizontal ( $y = y_{1}$ ). These two special cases are directly tested.
For parametric curves, always use $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ . Never differentiate $x$ and $y$ separately and then divide naively — use the chain rule ratio.
In approximations, $δ x$ must be small. The linear approximation works well only when the change is small. The approximation $δ y \approx f^{'} (x) δ x$ uses only the first term — for larger changes, higher order terms matter.
Percentage error in $x^{n}$ is $n$ times the percentage error in $x$ . This follows directly from differentiating $\ln y = n \ln x$ , giving $\frac{δ y}{y} = n \frac{δ x}{x}$ . A frequently tested board and JEE result.
Orthogonal curves satisfy $m_{1} m_{2} = - 1$ — this is a direct consequence of the formula for angle between curves when $ϕ = 90 °$ , giving $\tan ϕ \to \infty$ , i.e., $1 + m_{1} m_{2} = 0$ .

Practice Questions (JEE / Board Level)

Q1: The slope of the tangent to the curve $y = x^{3} - x + 1$ at the point where $x = 2$ is:

A) 5
B) 11
C) 7
D) 3

Answer: B) 11.

Explanation:

First, find the derivative of the curve to get the slope equation:
$\frac{d y}{d x} = 3 x^{2} - 1$

Substitute $x = 2$ into the derivative:
Slope $= 3 (2)^{2} - 1 = 3 (4) - 1 = 12 - 1 = 11$ .

Q2: The equation of the normal to the curve $y = x^{2} - 4 x + 6$ at the point $(2, 2)$ is:

A) $y = 2$
B) $x = 2$
C) $y = x$
D) $x + y = 4$

Answer: B) $x = 2$ .

Explanation:

Find the derivative to get the slope of the tangent:
$f^{'} (x) = 2 x - 4$

Evaluate the derivative at the given point where $x = 2$ :
$f^{'} (2) = 2 (2) - 4 = 0$

Since the tangent slope is 0, the tangent is a horizontal line ( $y = 2$ ).
The normal is perpendicular to the tangent, so it must be a vertical line passing through $x = 2$ . Therefore, the equation of the normal is $x = 2$ .

Q3: Using differentials, the approximate value of $\sqrt[3]{0.007}$ is:

A) 0.19
B) 0.21
C) 0.18
D) 0.20

Answer: A) 0.19.

Explanation:

Let $f (x) = x^{1 / 3}$ . Choose a nearby perfect cube, $x = 0.008$ , and let $δ x = - 0.001$ .

Find the derivative:
$f^{'} (x) = \frac{1}{3} x^{- 2 / 3}$

Evaluate the function and its derivative at $x = 0.008$ :
$f (0.008) = (0.008)^{1 / 3} = 0.2$
$f^{'} (0.008) = \frac{1}{3 (0.008)^{2 / 3}} = \frac{1}{3 (0.04)} = \frac{1}{0.12} = \frac{25}{3}$

Calculate the approximate change ( $δ y$ ):
$δ y \approx f^{'} (x) \cdot δ x$
$δ y \approx \frac{25}{3} \times (- 0.001) \approx - 0.00833$

Final approximate value:
$\sqrt[3]{0.007} \approx 0.2 - 0.00833 \approx 0.19167 \approx 0.19$ .

Q4: The radius of a sphere is increasing at 0.5 cm/s. The rate of increase of its surface area when the radius is 4 cm is:

A) $8 π {cm}^{2} /s$
B) $16 π {cm}^{2} /s$
C) $4 π {cm}^{2} /s$
D) $32 π {cm}^{2} /s$

Answer: B) $16 π {cm}^{2} /s$ .

Explanation:

The surface area of a sphere is given by $S = 4 π r^{2}$ .

Differentiate both sides with respect to time ( $t$ ):
$\frac{d S}{d t} = 8 π r \cdot \frac{d r}{d t}$

Substitute the given values ( $r = 4$ and $d r / d t = 0.5$ ):
$\frac{d S}{d t} = 8 π (4) (0.5) = 16 π {cm}^{2} /s$ .

Q5: If $y = x^{4} - 10$ and if $x$ changes from 2 to 1.99, the approximate change in $y$ is:

A) -0.32
B) 0.32
C) -3.2
D) 3.2

Answer: A) -0.32.

Explanation:

Find the derivative of $y$ with respect to $x$ :
$\frac{d y}{d x} = 4 x^{3}$

Evaluate the derivative at the initial value $x = 2$ :
$\frac{d y}{d x} = 4 (2)^{3} = 4 (8) = 32$

Calculate the change in $x$ ( $δ x$ ):
$δ x = 1.99 - 2 = - 0.01$

Use the differential formula $δ y \approx \frac{d y}{d x} \cdot δ x$ :
$δ y \approx 32 \times (- 0.01) = - 0.32$ .

Q6: The curves $y = a e^{x}$ and $y = b e^{- x}$ intersect orthogonally when:

A) $a = b$
B) $a = - b$
C) $a + b = 0$
D) $a b = 1$

Answer: D) $a b = 1$ .

Explanation:

First, find the slopes of both curves by differentiating them:
$m_{1} = \frac{d}{d x} (a e^{x}) = a e^{x}$
$m_{2} = \frac{d}{d x} (b e^{- x}) = - b e^{- x}$

For two curves to intersect orthogonally (at a 90-degree angle), the product of their slopes at the point of intersection must be $- 1$ :
$m_{1} \cdot m_{2} = - 1$
$(a e^{x}) (- b e^{- x}) = - 1$

Since $e^{x} \cdot e^{- x} = e^{0} = 1$ , the equation simplifies to:
$- a b = - 1 ⟹ a b = 1$ .

Q7 (JEE type): The percentage error in calculating the volume of a cubical box if an error of 1% is made in measuring the side is:

A) 1%
B) 2%
C) 3%
D) 4%

Answer: C) 3%.

Explanation:

The volume of a cube with side $a$ is $V = a^{3}$ .

Take the derivative to relate the error in volume to the error in the side length:
$d V = 3 a^{2} \cdot d a$

To find the relative (or percentage) error, divide by $V = a^{3}$ :
$\frac{d V}{V} = \frac{3 a^{2} \cdot d a}{a^{3}} = 3 (\frac{d a}{a})$

This shows that the percentage error in the volume is 3 times the percentage error in the side length.
Percentage error in $V = 3 \times 1 % = 3 %$ .

Related Study Material

More Mathematics Notes More Calculus Notes All Application Of Derivatives Topics

Practice Questions

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Rate of Change, Tangents, Normals & Approximations

1. Rate of Change of Quantities JEE Main & Advanced

Related Rates — Standard Results

Worked Example

2. Tangents and Normals JEE Advanced Only

Equations of Tangent and Normal

Special Scenarios

Angle Between Two Curves

Length of Tangent, Normal, Subtangent and Subnormal

3. Approximations Using Derivatives CBSE Boards Only

Error Analysis

Standard Approximation Formulae (for small $h$ )

Worked Example

Tangent & Normal — Step-by-Step Approach

Practice Questions (JEE / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

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Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Rate of Change, Tangents, Normals & Approximations

1. Rate of Change of Quantities JEE Main & Advanced

Related Rates — Standard Results

Worked Example

2. Tangents and Normals JEE Advanced Only

Equations of Tangent and Normal

Special Scenarios

Angle Between Two Curves

Length of Tangent, Normal, Subtangent and Subnormal

3. Approximations Using Derivatives CBSE Boards Only

Error Analysis

Standard Approximation Formulae (for small h)

Worked Example

Tangent & Normal — Step-by-Step Approach

Practice Questions (JEE / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

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We Value Your Privacy

Standard Approximation Formulae (for small $h$ )