Explanation Given: y = ln ( f ( x ) ) Formula Used: Assume that f ' ( x ) exists and is continuous over [ a , b ] . Then the arc length s of y = f ( x ) over [ a , b ] is equal to s = ∫ a b 1 + [ f ' ( x ) ] 2 d x Proof: The arc length of y = ln ( f ( x ) ) over a ≤ x ≤ b is: s = ∫ a b 1 + [ y ' ] 2 d x ... ( 1 ) Consider y = ln ( f ( x ) ) Therefore, d y d x = f ' ( x ) f ( x ) ( d d x

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

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Chapter 9.1, Problem 28E

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To prove:

The arc length of $y = \ln (f (x))$ for $a \leq x \leq b$ is $\int_{a}^{b} \frac{\sqrt{f {(x)}^{2} + f^{'} {(x)}^{2}}}{f (x)} d x$

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Chapter 9.1, Problem 27E

Chapter 9.1, Problem 29E

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Chapter 9 Solutions

Calculus - Standalone book

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To prove: The arc length of y = ln ( f ( x ) ) for a ≤ x ≤ b is ∫ a b f ( x ) 2 + f ' ( x ) 2 f ( x ) d x

Solution Summary: The author proves that the arc length of y = ln is a b f ( x ).

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To prove:

The arc length of $y = \ln (f (x))$ for $a \leq x \leq b$ is $\int_{a}^{b} \frac{\sqrt{f {(x)}^{2} + f^{'} {(x)}^{2}}}{f (x)} d x$

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Chapter 9 Solutions

To prove: The arc length of y = ln ( f ( x ) ) for a ≤ x ≤ b is ∫ a b f ( x ) 2 + f ' ( x ) 2 f ( x ) d x

Solution Summary: The author proves that the arc length of y = ln is a b f ( x ).

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To prove: The arc length of y=ln(f(x)) for a≤x≤b is ∫abf(x)2+f'(x)2f(x)dx

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Chapter 9 Solutions

To prove:

The arc length of $y = \ln (f (x))$ for $a \leq x \leq b$ is $\int_{a}^{b} \frac{\sqrt{f {(x)}^{2} + f^{'} {(x)}^{2}}}{f (x)} d x$