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CalculusCalculus (MindTap Course List)8th Edition

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Chapter 2.3, Problem 56E
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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

Solutions

Chapter
Section
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
Problem 46E
Problem 47E
Problem 48E
Problem 49E
Problem 50E
Problem 51E
Problem 52E
Problem 53E
Problem 54E
Problem 55E
Problem 56E
Problem 57E
Problem 58E
Problem 59E
Problem 60E
Problem 61E
Problem 62E
Problem 63E
Problem 64E
Problem 65E
Problem 66E
Problem 67E
Problem 68E
Problem 69E
Problem 70E
Problem 71E
Problem 72E
Problem 73E
Problem 74E
Problem 75E
Problem 76E
Problem 77E
Problem 78E
Problem 79E
Problem 80E
Problem 81E
Problem 82E
Problem 83E
Problem 84E
Problem 85E
Problem 86E
Problem 87E
Problem 88E
Problem 89E
Problem 90E
Problem 91E
Problem 92E
Problem 93E
Problem 94E
Problem 95E
Problem 96E
Problem 97E
Problem 98E
Problem 99E
Problem 100E
Problem 101E
Problem 102E
Problem 103E
Problem 104E
Problem 105E
Problem 106E
Problem 107E
Problem 108E
Problem 109E
Problem 110E
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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem
1 views

Find equations of the tangent line and normal line to the curve at the given point.

y 2 = x 3 ,   ( 1 , 1 )

To determine

To find:

Equation of tangent line and normal line to the curve y2= x3 at (1, 1)

Explanation

Concept:

Differentiate the given function to find the slope of tangent line and then find its negative reciprocal to find slope of normal line. Then, using the given point and these two slopes, find the equation of tangent line and normal line.

Formula used:

Power rule:  ddxxn=nxn-1

Slope point formula:  y-y1=m  x- x1

Given:

We have y2=x3

Calculation:

Solving for y: y= ± x3

This can be rewritten as y= ±x32

According to the given point (1, 1), the required equation of the curve is only for the positive part of the function.

So, y=x32

Then y'=32x12

Substituting x = 1, y'=32112=32

So this is the slope of tangent line at x = 1

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