INSTRUCTORS Raffi Hovasapian John Zhu

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• ## Related Books

 1 answerLast reply by: Si Jia WenWed Aug 21, 2013 1:21 PMPost by Steve Denton on October 11, 2012On 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 answerLast reply by: Steve DentonThu Oct 11, 2012 10:49 PMPost by James Xie on July 5, 2012For the second to last example, I got y = -(7/9)x + 26/9as an answer.

### 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 = ex at x = 0
• y′ = ex
• 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

*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.

### 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 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