Connecting...

This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Calculus AB

Next Lecture

Previous Lecture

Discussion
Answer Engine
Study Guides
Practice Questions
Download Lecture Slides
Table of Contents
Related Books

Lecture Comments (4)

1 answer

Last reply by: Si Jia Wen
Wed Aug 21, 2013 1:21 PM

Post by Steve Denton on October 11, 2012

On example 3, (1,-1) is not even on the curve x^2 - 4x. It should be (1,-3). lol. This reminds me of Microsoft products that are put out there prematurely and the public finds the mistakes instead of the company quality control.

1 answer

Last reply by: Steve Denton
Thu Oct 11, 2012 10:49 PM

Post by James Xie on July 5, 2012

For the second to last example, I got

y = -(7/9)x + 26/9

as an answer.

Answer EngineGet answers to any question!Ask any question related to Calculus AB

Working on the solution...

Tangent & Normal Lines

Finding tangent and normal lines

Take derivative at desired coordinate
Find appropriate slope
Use slope-intercept form:
Or point-slope form:

Tangent & Normal Lines

Find the slope of the tangent line of y = 4x⁵ + 3x + 1 at x = 1

y′ = [d/dx] (4x⁵ + 3x + 1)
y′ = 20 x⁴ + 3
Substitute in x = 1
y′ = 20 (1⁴) + 3 at x = 1

The tangent line of y at x = 1 has a slope of 23

Find the slope of the normal line of y = 4x⁵ + 3x + 1 at x = 1

We already found the slope of the tangent line in the previous problem. The slope of the normal line at the same point is the negative inverse of the slope of the tangent line.

The normal line of y at x = 1 has a slope of −[1/23]

Find the equation of the tangent line to the curve y = 4x⁵ + 3x + 1 at x = 1

We already have the slope, we just need to find the full equation of that tangent line.
We can use point-slope form (y − y₁) = m(x − x₁), but first we need a point on the graph. Let's plug in x = 1 into the original equation since it shares that same point with the tangent.
y = 4(1⁵) + 3(1) + 1
y = 8
y − 8 = 23(x − 1)

y = 23x − 15 is the equation of the tangent line.

Find the equation of the tangent and normal lines to the curve y = 9x at x = 12

y′ = 9
The slope of the tangent line is a constant 9.
Plugging in x = 12 into the original equation.
y = 9(12)
y = 108
Point-slope form for tangent line.
y − 108 = 9(x − 12)
y = 9x + 108 − 108
Point-slope form for normal line
y − 108 = −[1/9] (x − 12)

tangent line y = 9x. normal line y = −[1/9] (x − 12) + 108

Find the equation of the tangent and normal lines to the curve y = √{3x + 1} at x = 1

y′ = [1/2] [1/(√{3x + 1})] [d/dx] (3x + 1)
y′ = [3/(2 √{3x + 1})]
y′ = [3/(2 √{3(1) + 1})] = [3/4] at x = 1
Plugging in to the original equation
y = √{3(1) + 1} = 2
Point-slope form for tangent line
y − 2 = [3/4] (x − 1)
Point-slope form for normal line
y − 2 = − [4/3] (x − 1)

tangent line y = [3/4](x − 1) + 2. normal line y = −[4/3](x − 1) + 2

Find the equation of the tangent line to the curve y = sinx at x = [(π)/2]

y′ = cosx
y′ = cos[(π)/2] = 0 at x = [(π)/2]
Plugging in to the original equation
y = sin[(π)/2] = 1 at x = [(π)/2]
Point-slope form for tangent line
y − 1 = 0 (x − [(π)/2]
Point-slope form for normal line
y − 1 = [(−1)/0] (x − [(π)/2])
Slope here is undefined, which implies a vertical line at x = [(π)/2]

tangent line y = 1. normal line x = [(π)/2]

Find the equation of the tangent line to the curve y = e^x at x = 0

y′ = e^x
y′ = e⁰ = 1 at x = 0
Point-slope form for tangent line
y − 1 = 1 (x − 0)
Point-slope form for normal line
y − 1 = −1 (x − 0)

tangent line y = x + 1. normal line y = −x + 1

Find the equation of the tangent and normal lines to the curve y = 2x at x = 3

y′ = 2
y = 2(3) = 6 at x = 3
Point-slope form for tangent line
y − 6 = 2 (x − 3)
Point-slope form for normal line
y − 6 = [(−1)/2] (x − 3)

tangent line y = 2x. normal line y = −[1/2](x − 3) + 6

Find the equation of the tangent and normal lines to the curve y = lnx + √x at x = 1

y′ = [1/x] + [1/2] [1/(√x)]
y′ = [1/1] + [1/2] [1/(√1)] = [3/2] at x = 1
Plugging in to the original equation
y = ln1 + √1 = 0 + 1 = 1 at x = 1
Point-slope form for tangent line
y − 1 = [3/2] (x − 1)
Point-slope form for normal line
y − 1 = −[2/3] (x − 1)

tangent line y = [3/2] (x − 1) + 1. normal line y = −[2/3] (x − 1) + 1

Find the equation of the tangent and normal lines to the curve y = sin⁻¹ x + cos⁻¹ x at x = 0

y′ = [1/(√{1 − x²})] − [1/(1 − x²)] = 0
Plugging in to the original equation
y = sin⁻¹ 0 + cos⁻¹ 0 = 0 + [(π)/2] = [(π)/2] at x = 0
Point-slope form for tangent line
y − [(π)/2] = 0 (x − 0)
Point-slope form for normal line
y − [(π)/2] = −[1/0](x − 0)
Slope here is undefined, which implies a vertical line at x = 0

tangent line y = [(π)/2]. normal line x = 0

« Previous Question | Next Question »

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Tangent & Normal Lines

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Tangent and Normal Lines

Tangent and Normal Lines: Example 2

Tangent and Normal Lines

Example 3

Tangent and Normal Lines: Example 4

Tangent and Normal Lines: Example 5

Tangent and Normal Lines: Example 6

Tangent and Normal Lines: Example 7

Intro 0:00
Tangent and Normal Lines 0:10

Definition
Example 1

Tangent and Normal Lines: Example 2 2:43
Tangent and Normal Lines 5:21

Example 3

Tangent and Normal Lines: Example 4 9:14
Tangent and Normal Lines: Example 5 12:27
Tangent and Normal Lines: Example 6 15:54
Tangent and Normal Lines: Example 7 19:05

Mathematics: AP Calculus AB

Section 1: Functions
	Definitions & Properties of Functions	11:26
	Graphing	13:58
	Inverse Functions	6:47
	Polynomial Functions	5:04
	Trigonometric Functions	6:45
	Inverse Trigonometric Functions	5:58
	Trigonometric Identities	17:42
	Exponential Functions	5:53
	Logarithmic Functions	7:08
	Rational Functions	15:36
	Conic Sections	14:58
Section 2: Limits and Continuity
	Limit Definition & Properties	7:15
	Solving Limits with Algebra	8:01
	Rational Limit Rules	3:16
	One Sided Limits	9:57
	Special Trigonometric Limits	8:28
	Limits & Continuity	10:14
	Limits: Multiple Choice Practice	6:16
Section 3: Derivatives
	Derivative Definition & Properties	4:11
	Basic Rules of Differentiation	7:07
	Power Rule	7:14
	Trigonometric Rules	7:53
	Product Rule	11:11
	Quotient Rule	16:50
	Chain Rule	19:48
	Higher Order Derivatives	15:00
	Derivatives of Exponential Functions	13:14
	Derivatives of Logarithmic Functions	11:30
	Derivatives of Inverse Trigonometric Functions	16:54
	Implicit Differentiation	16:53
	Multiple Choice Practice: Derivatives	11:07
Section 4: Applications of Derivatives
	Tangent & Normal Lines	22:36
	Position Velocity & Acceleration	18:42
	Related Rates	26:22
	Minimum & Maximum	12:22
	Concavity	11:43
	Rolles Theorem	8:28
	Mean Value Theorem	9:39
	Differentials	12:25
	Applications of Derivatives: Multiple Choice Practice	13:21
	Applications of Derivatives: Free Response Practice	10:22
Section 5: Integrals
	Definition of Integrals	1:08
	Integrals of Power Rule	8:50
	Integrals Basic Rules of Integration	9:43
	Trigonometric Rules of Integrals	8:58
	Chain Rule	13:59
	Integrals of Exponential Functions	12:52
	Integrals of Natural Logarithmic Functions	13:00
	Integrals of Inverse Trigonometric Functions	8:29
	Integrals: Multiple Choice Practice	15:37
Section 6: Applications of Integrals
	Fundamental Theorem of Calculus	15:55
	Area Under A Curve	18:34
	Reimann Sums	10:35
	Trapezoid Rule	12:46
	Mean Value Theorem	11:22
	Second Fundamental Theorem of Calculus	4:44
	Area Between Curves	16:39
	Revolving Solids Washer Disk Methods	21:09
	Revolving Solids Cylindrical Shells Method	26:46
	Revolving Solids Known Cross Sections	27:41
	Differential Equations Eulers Method	17:54
	Differential Equations Slope Fields	16:30
	Application of Integrals: Multiple Choice Practice	14:19
	Application of Integrals: Free Response Practice	9:14
Section 7: Sample AP Test
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 1	17:50
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 2	17:32
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 3	22:14
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 4	19:35
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 5	25:43
	AP Calculus AB Practice Test: Section 2: Free Response Part 1	16:50
	AP Calculus AB Practice Test: Section 2: Free Response Part 2	21:36
	AP Calculus AB Practice Test: Section 2: Free Response Part 3	13:39

Related Books

	Cracking the AP Calculus AB Exam, 2015 Edition Author: Princeton Review ISBN: 804124809 Publisher: Princeton Review Year: 2014 The book includes 3 full length practice tests with detailed explanations, a review of all the key concepts, and targeted strateeies to ace th exam for your highest score.
	Cliffs AP: Calculus AB & BC, 3rd Edition Authors: Dale W Johnson, Kerry J King ISBN: 764586831 Publisher: Cliffs Notes Year: 2001 The book has a topic-by topic breakdown and lots of problem approach suggestions for both free response and multiple choice calculus questions. The book also includes two full length tests.

Related Books

Name	Description	Link
BookRenter.com	BookRenter.com is simply the most reliable online textbook rental service.	Visit BookRenter.com
PhysicsForums.com Homework Help	Physics Forums is a scientific community for students looking for math & science help.	Visit PhysicsForums.com Homework Help

John Zhu

Tangent & Normal Lines

Slide Duration:

Table of Contents

Section 1: Functions

Definitions & Properties of Functions

11m 26s

Intro