The tangent line shows the curve’s exact direction at a given point, while the normal line is perpendicular to the tangent at that same point. They help us to analyze curves and their properties in geometry.
Tangent Line
A tangent line is a straight line that just touches a curve at one point and has the same slope as the curve at that point. It gives us an idea of the curve′s direction at that instant.
Equation
The equation of the tangent line to a function y = f(x) at a given point (x0, y0) is:
y – y0 = m(x – x0)
Here,
m = f′(x0) is the slope of the tangent line at x0
(x0, y0) is the point of tangency, where y0 = f(x0)
Note: The tangent line at any point on a function y = f(x) has the same slope as the function at that point.
Finding the Equation
Now, let us find the equation of the tangent line to the function y = x2 at x = 2
Step 1: Finding the y-coordinate
For y = x2, we substitute x = 2 into the equation to find the y-coordinate of the tangency point.
At x = 2, y0 = (2)2 = 4
Thus, the point of tangency is (2, 4)
Step 2: Calculating the Derivative
The derivative of y = x2 is:
f′(x) = ${\dfrac{d}{dx}\left( x^{2}\right)}$ = 2x
This represents the slope of the tangent line at any point x.
Step 3: Evaluating the Slope at x0 = 2
Now, substituting x = 2 into f′(x), we get
m = 2(2) = 4
Thus, the slope of the tangent line is 4.
Step 4: Using the Tangent Line Formula
Now, using the point-slope equation, we get
y – y0 = m(x – x0)
⇒ y – 4 = 4(x – 2)
⇒ y – 4 = 4x – 8
⇒ y – 4 + 4 = 4x – 8 + 4
⇒ y = 4x – 4
Thus, the equation of the tangent line is:
y = 4x – 4
This means that at x = 2, the curve y = x2 has a tangent line with the equation y = 4x – 4, which moves in the same direction as the curve at that point.
Normal Line
A normal line to a curve at a given point is a straight line that is perpendicular to the tangent line at that point. It shares the same point of tangency but extends in a direction that is perpendicular to the tangent.
Equation
The equation of the normal line to a function y = f(x) at a given point (x0, y0) is:
y – y0 = mn(x – x0)
Here,
mn = ${-\dfrac{1}{m}}$ is the negative reciprocal of the tangent slope
(x0, y0) is the point of tangency
Finding the Equation
Now, let us find the equation of the normal line to the function y = x2 at x = 2
Step 1: Finding the Tangent Slope
As we already calculated, the tangent slope is m = 4
Thus, the normal slope is mn = ${-\dfrac{1}{4}}$
Step 2: Using the Normal Line formula
Now, using the point-slope equation, we get
y – y0 = m(x – x0)
⇒ ${y-4=-\dfrac{1}{4}\left( x-2\right)}$
⇒ ${y-4+4=-\dfrac{1}{4}x+\dfrac{1}{2}+4}$
⇒ ${y=-\dfrac{1}{4}x+\dfrac{1}{2}+4}$
⇒ ${y=-\dfrac{1}{4}x+\dfrac{9}{2}}$
Thus, the equation of the normal line is ${y=-\dfrac{1}{4}x+\dfrac{9}{2}}$
Tangent Line vs Normal Line
Basis
Tangent
Normal
Definition
A line that touches the curve at a single point and follows its direction.
A line that is perpendicular to the tangent line at the point of contact.
Slope
Equal to the derivative of the function at the given point.
Negative reciprocal of the tangent line’s slope.
Equation
y – y0 = m(x – x0), where m = f′(x0)
y – y0 = mn(x – x0), where mn = ${-\dfrac{1}{m}}$
Graphical Representation
Touches the curve at exactly one point.
Passes through the same point but is perpendicular to the tangent.
Uses
Used in motion analysis, derivatives, and approximation of curves.
Used in physics (reflection, forces) and normal force calculations.
Solved Examples
Find the tangent and normal line equations for y = x3 at x = 1
Solution:
Given, y = x3 The y-coordinate at x = 1 is: y0 = (1)3 = 1 Thus, the point of tangency is (1, 1) To find the slope of the tangent line, we differentiate y = x3 ${\dfrac{d}{dx}\left( x^{3}\right) =3x^{2}}$ Now, substituting x = 1, m = 3(1)2 = 3 Thus, the tangent line has a slope of 3. As we know, the equation of a tangent line is: y – y0 = m(x – x0) Substituting (x0, y0) = (1, 1) and m = 3, ⇒ y – 1 = 3(x – 1) ⇒ y – 1 = 3x – 3 ⇒ y = 3x – 3 + 1 ⇒ y = 3x – 2 Thus, the tangent line equation is: y = 3x – 2 Since the normal line is perpendicular to the tangent, its slope is the negative reciprocal of 3: mn = ${-\dfrac{1}{3}}$ As we know, the equation of a normal line is: y – y0 = mn(x – x0) Substituting (x0, y0) = (1, 1) and mn = ${-\dfrac{1}{3}}$, ⇒ ${y-1=-\dfrac{1}{3}\left( x-1\right)}$ ⇒ ${y=-\dfrac{1}{3}x+\dfrac{1}{3}+1}$ ⇒ ${y=-\dfrac{1}{3}x+\dfrac{4}{3}}$ Thus, the normal line equation is: ${y=-\dfrac{1}{3}x+\dfrac{4}{3}}$
Determine the tangent and normal lines to y = sin x at x = ${\dfrac{\pi }{2}}$
Solution:
Given, y = sin x The y-coordinate at x = ${\dfrac{\pi }{2}}$ is: y0 = ${\sin \left( \dfrac{\pi }{2}\right)}$ = 1 Thus, the point of tangency is ${\left( \dfrac{\pi }{2},1\right)}$ The derivative of y = sin x is: ${\dfrac{d}{dx}\left( \sin x\right) =\cos x}$ Now, substituting x = ${\dfrac{\pi }{2}}$, m = ${\cos \left( \dfrac{\pi }{2}\right)}$ = 0 Thus, the tangent line has a slope of 0, meaning the tangent line is horizontal: y = 1 Since the tangent is horizontal, the normal line must be vertical, meaning its equation is: x = ${\dfrac{\pi }{2}}$
Find the equation of the tangent line to y = ex at x = 0
Solution:
Given, y = ex The y-coordinate at x = 0 is: y0 = e0 = 1 Thus, the point of tangency is (0, 1) The derivative of y = ex is: ${\dfrac{d}{dx}\left( e^{x}\right) =e^{x}}$ Now, substituting x = 0, m = e0 = 1 Thus, the slope of the tangent line is 1. As we know, the equation of a tangent line is: y – y0 = m(x – x0) Substituting (x0, y0) = (0, 1) and m = 1, ⇒ y – 1 = 1(x – 0) ⇒ y – 1 = x ⇒ y = x + 1 Thus, the tangent line equation is: y = x + 1
