Arc Length

  • Thread starter chapsticks
  • Start date
  • #1
38
0

Homework Statement



so I did

Homework Equations



s=∫ba√(1+[f'(x)]2dx

The Attempt at a Solution


y=2/3x3/2+7
y'=x1/2 0≤x≤1
=∫6√(1+(x1/2)2)dx
=[2/3(1+x)3/2]7

my answer came out wrong
 

Attachments

  • 29.jpg
    29.jpg
    16.7 KB · Views: 326

Answers and Replies

  • #2
35,125
6,872

Homework Statement



so I did

Homework Equations



s=∫ba√(1+[f'(x)]2dx

The Attempt at a Solution


y=2/3x3/2+7
y'=x1/2 0≤x≤1
Why do you have this inequality?
=∫6√(1+(x1/2)2)dx
=[2/3(1+x)3/2]7

my answer came out wrong
What is that 7 doing?
 
  • #3
38
0
I saw it in an example for the inequality.. The 7 is for the a & b part I don't know how to place the 0 on the bottom.
 
  • #4
35,125
6,872
I saw it in an example for the inequality.. The 7 is for the a & b part I don't know how to place the 0 on the bottom.
Your inequality is saying that the interval is [0, 1]. It's actually [0, 6].

Here's how your integral looks using LaTeX.
[tex]\int_0^6 \sqrt{1+x}~dx[/tex]

Here's how to write that integral in LaTeX.

[ tex]\int_0^6 \sqrt{1+x}~dx[ /tex]

(The extra spaces in the tex and /tex tags prevent the browser from rendering the code inside.)

Here's the antiderivative with limits of integration shown.
[tex]\left. \frac{2}{3}(1 + x)^{3/2}\right|_0^6[/tex]

The LaTeX for that.
[ tex]\left. \frac{2}{3}(1 + x)^{3/2}\right|_0^6[ /tex]
 
  • #5
38
0
I got it correct :))
 

Attachments

  • 29.jpg
    29.jpg
    16.1 KB · Views: 330
  • #6
35,125
6,872
Good to know. That's what I got, too.
 

Related Threads on Arc Length

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
952
  • Last Post
Replies
4
Views
815
  • Last Post
Replies
3
Views
1K
  • Last Post
2
Replies
25
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
7
Views
1K
Top