SOLUTION If f(x)=5x2 - 1 40 In(x), then 1 f'(x)= 10x 40x 1 + [f 1 + [f'(x)]² p>'(x)]² = - ( 1 10x 40x = 1 + 100x20.5 + 1 1600x2 1 = 100x² + 0.5+ 1600x2 1 10x + 40x 1 √1 + [f'(x)]² 10x + 40x Thus the arc length is given by s(x) 1+ [f'(t)]² dt = 10t+ 1 40t 5(2) In(1) + 40 5x(2) In(x) + - 5 40 I dt For instance, the arc length along the curve from (1, 5) to (3, f(3)) is s(3) 1 + In(3)-5 40 In(3) + 40 (rounded to four decimal places)
SOLUTION If f(x)=5x2 - 1 40 In(x), then 1 f'(x)= 10x 40x 1 + [f 1 + [f'(x)]² p>'(x)]² = - ( 1 10x 40x = 1 + 100x20.5 + 1 1600x2 1 = 100x² + 0.5+ 1600x2 1 10x + 40x 1 √1 + [f'(x)]² 10x + 40x Thus the arc length is given by s(x) 1+ [f'(t)]² dt = 10t+ 1 40t 5(2) In(1) + 40 5x(2) In(x) + - 5 40 I dt For instance, the arc length along the curve from (1, 5) to (3, f(3)) is s(3) 1 + In(3)-5 40 In(3) + 40 (rounded to four decimal places)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
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![SOLUTION
If f(x) = 5x2
-
1
40
In(x), then
f'(x) 10x
=
-
1
40x
1
1 + [f 1 + [f'(x)]² p>'(x)]²
10x
40x
= 1 + 100x20.5 +
1
1600x2
1
=
100x² + 0.5 +
1600x2
1
10x +
40x
1
√1+ [f'(x)]²
10x +
40x
Thus the arc length is given by
s(x)
=
√1 + [f'(t)]² dt
=
10t+
1
=
5t(2)
5 (2)
5x
dt
40t
In(t)
+
40
In(x)
+
- 5
40
For instance, the arc length along the curve from (1, 5) to (3, f(3)) is
s(3) =
1
+ In(3) - 5
40
+
In(3)
40
(rounded to four decimal places)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F442584ec-78ec-4980-82e9-4016ffe9c3e5%2F2978eab9-38d5-42ba-a735-37a6fa0ebd83%2Fisv63nq_processed.png&w=3840&q=75)
Transcribed Image Text:SOLUTION
If f(x) = 5x2
-
1
40
In(x), then
f'(x) 10x
=
-
1
40x
1
1 + [f 1 + [f'(x)]² p>'(x)]²
10x
40x
= 1 + 100x20.5 +
1
1600x2
1
=
100x² + 0.5 +
1600x2
1
10x +
40x
1
√1+ [f'(x)]²
10x +
40x
Thus the arc length is given by
s(x)
=
√1 + [f'(t)]² dt
=
10t+
1
=
5t(2)
5 (2)
5x
dt
40t
In(t)
+
40
In(x)
+
- 5
40
For instance, the arc length along the curve from (1, 5) to (3, f(3)) is
s(3) =
1
+ In(3) - 5
40
+
In(3)
40
(rounded to four decimal places)
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