Explanation Given: y = 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: It is given that f ' ( x ) ≤ 1 [ f ' ( x ) ] 2 ≤ 1 1 + [ f ' ( x ) ] 2 ≤ 1 + 1 1 + [ f ' ( x ) ] 2 ≤ 2 So, the arc length of the graph s = ∫ a b 1 + [ f ' ( x ) ] 2 d x ≤ 2 d x ≤ 2 ( b − a ) Therefore, the arc le...

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ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 9.2, Problem 33E

To determine

To prove:

The arc length of $y = f (x)$ over $[a, b]$ is at most $\sqrt{2} (b - a)$ if $0 \leq f^{'} (x) \leq 1$ for all $x$ . Also, the arc length of $f (x) = x$ is $\sqrt{2} (b - a)$ .

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Chapter 9.2, Problem 32E

Chapter 9.2, Problem 34E

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Chapter 9 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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To prove: The arc length of y = f ( x ) over [ a , b ] is at most 2 ( b − a ) if 0 ≤ f ' ( x ) ≤ 1 for all x . Also, the arc length of f ( x ) = x is 2 ( b − a ) .

Solution Summary: The author proves that the arc length of y = f (x) over [a, b] is at most 2 (b a).

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To prove:

The arc length of $y = f (x)$ over $[a, b]$ is at most $\sqrt{2} (b - a)$ if $0 \leq f^{'} (x) \leq 1$ for all $x$ . Also, the arc length of $f (x) = x$ is $\sqrt{2} (b - a)$ .

Want to see the full answer?

Chapter 9 Solutions

To prove: The arc length of y = f ( x ) over [ a , b ] is at most 2 ( b − a ) if 0 ≤ f ' ( x ) ≤ 1 for all x . Also, the arc length of f ( x ) = x is 2 ( b − a ) .

Solution Summary: The author proves that the arc length of y = f (x) over [a, b] is at most 2 (b a).

Concept explainers

Videos

To prove: The arc length of y=f(x) over [a,b] is at most 2(b−a) if 0≤f'(x)≤1 for all x. Also, the arc length of f(x)=x is 2(b−a).

Want to see the full answer?

Chapter 9 Solutions

To prove:

The arc length of $y = f (x)$ over $[a, b]$ is at most $\sqrt{2} (b - a)$ if $0 \leq f^{'} (x) \leq 1$ for all $x$ . Also, the arc length of $f (x) = x$ is $\sqrt{2} (b - a)$ .