Derivatives - Examples Of Tangent And Normals

Derivatives - Examples of tangent and normals

Tangent and normal are two important concepts in Calculus
They are used to analyze the behavior of a function at a specific point
Let’s understand them with some examples

Example 1

Consider the function f(x) = x^2 + 2x + 1. Find the equation of the tangent line and the normal line at the point x = 2.

First, we need to find the derivative of the function f(x)
Differentiating f(x), we get f’(x) = 2x + 2
Substituting x = 2, we find f’(2) = 2(2) + 2 = 6
This is the slope of the tangent line

Example 1 (contd.)

The equation of a line can be given by y = mx + c, where m is the slope and c is the y-intercept
We already know the slope (m) is 6
To find the y-intercept (c), we substitute the point (2, f(2)) into the equation
Substituting x = 2, we get y = 2^2 + 2(2) + 1 = 9
Thus, the equation of the tangent line is y = 6x + 9

Example 1 (contd.)

The normal line is perpendicular to the tangent line
The slope of the normal line is the negative reciprocal of the slope of the tangent line
Therefore, the slope of the normal line is -1/6

Example 1 (contd.)

Using the point-slope form of a line, we can find the equation of the normal line
The equation would be of the form y - y1 = m(x - x1)
Substituting the point (2, f(2)), we get y - 9 = -(1/6)(x - 2)
Simplifying, we get y - 9 = -1/6x + 1/3
Rearranging, the equation of the normal line is y = -1/6x + 28/3

Example 2

Consider the function g(x) = 3x^3 - 2x^2 + 5x - 4. Find the equation of the tangent line and the normal line at the point x = 1.

Differentiating g(x), we get g’(x) = 9x^2 - 4x + 5
Substituting x = 1, we find g’(1) = 9(1)^2 - 4(1) + 5 = 10
This is the slope of the tangent line

Example 2 (contd.)

Substituting x = 1 into g(x), we get g(1) = 3(1)^3 - 2(1)^2 + 5(1) - 4 = 2
Thus, the equation of the tangent line is y = 10x - 8

Example 2 (contd.)

The slope of the normal line is the negative reciprocal of the slope of the tangent line
Therefore, the slope of the normal line is -1/10

Example 2 (contd.)

Substituting the point (1, g(1)), we get y - 2 = -(1/10)(x - 1)
Simplifying, we get y - 2 = -1/10x + 1/10
Rearranging, the equation of the normal line is y = -1/10x + 21/10

Slide 11

Tangent and normal lines are used to analyze the behavior of a function at a specific point
The tangent line touches the curve at a specific point and has the same slope as the curve at that point
The normal line is perpendicular to the tangent line and has a slope that is the negative reciprocal of the tangent line’s slope
Both lines provide valuable information about the function’s behavior near a specific point

Slide 12

To find the equation of a tangent line, we need the slope of the curve at the desired point
The slope can be found by taking the derivative of the function and evaluating it at the desired point
The point-slope form of a line (y - y1 = m(x - x1)) can be used to write the equation of the tangent line using the slope and the coordinates of the point

Slide 13

Example 3: Consider the function h(x) = 2x^3 + 4x^2 - 3x + 1. Find the equation of the tangent line at the point x = 0.
The derivative of h(x) is h’(x) = 6x^2 + 8x - 3. Evaluating it at x = 0, we find h’(0) = -3.
The point (0, h(0)) is (0, 1), so the equation of the tangent line is y - 1 = -3(x - 0), which simplifies to y = -3x + 1.

Slide 14

Example 4: Consider the function f(x) = sin(x) + cos(x). Find the equation of the tangent line at the point x = π/4.
The derivative of f(x) is f’(x) = cos(x) - sin(x). Evaluating it at x = π/4, we find f’(π/4) = √2/2 - √2/2 = 0.
The point (π/4, f(π/4)) is (√2/2 + √2/2), so the equation of the tangent line is y - (√2) = 0(x - π/4), which simplifies to y = √2.

Slide 15

The slope of the normal line is the negative reciprocal of the tangent line’s slope
To find the equation of the normal line, we use the point-slope form of a line and substitute the slope and coordinates of the point
The equation of the normal line is in the form y - y1 = m(x - x1), where m is the negative reciprocal of the tangent line’s slope

Slide 16

Example 5: Consider the function g(x) = x^2 + 3x - 2. Find the equation of the normal line at the point x = -2.
The derivative of g(x) is g’(x) = 2x + 3. Evaluating it at x = -2, we find g’(-2) = -1.
The point (-2, g(-2)) is (-2, 0), so the equation of the tangent line is y - 0 = -1(x - (-2)), which simplifies to y = -x + 2.

Slide 17

The slope of the normal line is the negative reciprocal of the tangent line’s slope, which is 1 in this case
Substituting the point (-2, 0) into the point-slope form, we get y - 0 = -1(x - (-2)), which simplifies to y = -x + 2
Therefore, the equation of the normal line is y = -x + 2

Slide 18

Example 6: Consider the function h(x) = e^x + ln(x). Find the equation of the normal line at the point x = 1.
The derivative of h(x) is h’(x) = e^x + 1/x. Evaluating it at x = 1, we find h’(1) = e + 1.
The point (1, h(1)) is (1, e + ln(1)), so the equation of the tangent line is y - (e) = (e + 1)(x - 1), which simplifies to y = (e + 1)x - (e + 1).

Slide 19

The slope of the normal line is the negative reciprocal of the tangent line’s slope, which is -1/(e + 1)
Substituting the point (1, e + ln(1)) into the point-slope form, we get y - (e) = -1/(e + 1)(x - 1), which simplifies to y = -1/(e + 1)x + (e + 1)/(e + 1)
Therefore, the equation of the normal line is y = -1/(e + 1)x + 1

Slide 20

Tangent and normal lines provide valuable information about the behavior of functions at specific points
They can help determine the direction of the curve, the rate of change, and the behavior of the function near the point of tangency
Calculating the slopes and equations of these lines is an important skill in calculus.

Slide 21

Tangent and normal lines play a crucial role in finding the rate of change of a function at a specific point
These lines help us understand the behavior of the function and the slope of the curve
By analyzing the tangent and normal lines, we can make predictions about the function’s behavior near the point of tangency
They are fundamental concepts in differential calculus
Let’s explore their applications through more examples

Slide 22

Example 7: Consider the function f(x) = x^3 - 2x. Find the equation of the tangent line and the normal line at the point x = 3.
Differentiating f(x), we get f’(x) = 3x^2 - 2. At x = 3, f’(3) = 3(3)^2 - 2 = 25.
The coordinates of the point (3, f(3)) are (3, 21). Therefore, the equation of the tangent line is y - 21 = 25(x - 3).

Slide 23

Substituting the point (3, 21) into the point-slope form, we get y - 21 = 25(x - 3), which simplifies to y = 25x - 54.
The slope of the normal line is -1/25, the negative reciprocal of the tangent line’s slope.
Therefore, the equation of the normal line is y - 21 = -1/25(x - 3).

Slide 24

Example 8: Consider the function g(x) = √x + ln(x). Find the equation of the tangent line and the normal line at the point x = 4.
The derivative of g(x) is g’(x) = 1/(2√x) + 1/x.
At x = 4, g’(4) = 1/(2√4) + 1/4 = 1/4 + 1/4 = 1/2.
The coordinates of the point (4, g(4)) are (4, 2√2 + ln(4)). Therefore, the equation of the tangent line is y - (2√2 + ln(4)) = (1/2)(x - 4).

Slide 25

Substituting the point (4, 2√2 + ln(4)) into the point-slope form, we get y - (2√2 + ln(4)) = (1/2)(x - 4), which simplifies to y = (1/2)x - √2 + ln(4).
The slope of the normal line is -2, the negative reciprocal of the tangent line’s slope.
Therefore, the equation of the normal line is y - (2√2 + ln(4)) = -2(x - 4).

Slide 26

Example 9: Consider the function h(x) = e^x - e^(-x). Find the equation of the tangent line and the normal line at the point x = 0.
The derivative of h(x) is h’(x) = e^x + e^(-x).
At x = 0, h’(0) = e^0 + e^(-0) = 2.
The coordinates of the point (0, h(0)) are (0, 0). Therefore, the equation of the tangent line is y - 0 = 2(x - 0).

Slide 27

Substituting the point (0, 0) into the point-slope form, we get y - 0 = 2(x - 0), which simplifies to y = 2x.
The slope of the normal line is -1/2, the negative reciprocal of the tangent line’s slope.
Therefore, the equation of the normal line is y - 0 = -1/2(x - 0), which further simplifies to y = -1/2x.

Slide 28

Example 10: Consider the function f(x) = e^x - e^(-x) + 2x. Find the equation of the tangent line and the normal line at the point x = 1.
The derivative of f(x) is f’(x) = e^x + e^(-x) + 2.
At x = 1, f’(1) = e^1 + e^(-1) + 2 ≈ 5.73.
The coordinates of the point (1, f(1)) are (1, e - e^(-1) + 2). Therefore, the equation of the tangent line is y - (e - e^(-1) + 2) = 5.73(x - 1).

Slide 29

Substituting the point (1, e - e^(-1) + 2) into the point-slope form, we get y - (e - e^(-1) + 2) = 5.73(x - 1), which simplifies to y = 5.73x - 4.73e + 4.73.
The slope of the normal line is approximately -1/5.73, the negative reciprocal of the tangent line’s slope.
Therefore, the equation of the normal line is y - (e - e^(-1) + 2) = -1/5.73(x - 1), which further simplifies to y = -1/5.73x + 1/5.73e + 1/5.73.

Slide 30

Tangent and normal lines provide valuable insight into the behavior of a function at a specific point
By analyzing the slopes and equations of these lines, we can make predictions about the function’s behavior near the point of tangency
Tangent lines have the same slope as the curve at the point of tangency, while normal lines have slopes that are negative reciprocals
Calculating these lines’ equations is an essential skill in calculus and helps us understand the rate of change and behavior of functions.