ivorycoyote6Lv1
19 Nov 2021
Problem 42
Page 308
Section 4.6: Optimization Problems
Chapter 4: Applications of Differentiation
Textbook ExpertVerified Tutor
19 Nov 2021
Given information
1. The equation of parabola is 
.
2. The tangent to the parabola at some point in the first quadrant.
Figure
Step-by-step explanation
Step 1.
Parametric equation of the parabola as shown in figure above:
Slope-intercept form of straight line is
where 
is the gradient and 
is the 
.
Now slope of the tangent at any point on the parabola is given by
Substituting (1) and (3) in (2) we have
Therefore the equation of the tangent is
Putting 
and 
in (4) subsequently we have the 
and as:
and 
Single Variable Calculus: Early Transcendentals
4th Edition, 2018
Stewart
ISBN: 9781337687805
Solutions
Chapter
Section
- Problem 1a
- Problem 1b
- Problem 2
- Problem 3
- Problem 4
- Problem 5
- Problem 6
- Problem 7
- Problem 8
- Problem 9a
- Problem 9b
- Problem 9c
- Problem 10a
- Problem 10b
- Problem 10c
- Problem 10d
- Problem 10e
- Problem 10f
- Problem 11
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- Problem 13a
- Problem 13b
- Problem 14
- Problem 15
- Problem 16
- Problem 17
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- Problem 19
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- Problem 24
- Problem 25
- Problem 26
- Problem 27
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- Problem 30
- Problem 31
- Problem 32a
- Problem 32b
- Problem 33a
- Problem 33b
- Problem 33c
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- Problem 36
- Problem 37
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- Problem 41
- Problem 42
- Problem 43a
- Problem 43b
- Problem 44a
- Problem 44b
- Problem 45a
- Problem 45b
- Problem 46a
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- Problem 47a
- Problem 47b
- Problem 47c
- Problem 48
- Problem 49
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- Problem 52
- Problem 53
- Problem 54
- Problem 55
- Problem 56
- Problem 57
- Problem 58
- Problem 59a
- Problem 59b
- Problem 59c
- Problem 59d
- Problem 60a
- Problem 60b
- Problem 60c
- Problem 61a
- Problem 61b
- Problem 61c
- Problem 62a
- Problem 62b
- Problem 62c
- Problem 62d