Arc length

  • #1

Homework Statement


y = 5 + 8x^3/2 from 0 to 1


Homework Equations





The Attempt at a Solution


I have tried it a few times Keep getting variations of 1+ 12[tex]\sqrt{12}[/tex]. I would like some to give me a step by step how to work it. Something is killing me on this I am lost.
 

Answers and Replies

  • #2
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Show us what you've tried and we can go from there.
 
  • #3
Arc length is the Integral of sqrt(1+(f '(x))^2) so Derivative of the equation is 12x^1/2. Then I believe you are supposed to square that. Thats the same as (sqrt(12x))^2 right? so I tired it like that I also did 12x^1/2 * 12x^1/2= 144x so then I know you finish of the integral. I used U substitution to do that.
 
  • #4
28
0
Roughly: [tex]\int{\sqrt{1+144x}}\:dx = \frac{1}{216}(1 + 144x)^{3/2}[/tex]
 
  • #5
34,935
6,698
Roughly: [tex]\int{\sqrt{1+144x}}\:dx = \frac{1}{216}(1 + 144x)^{3/2}[/tex]
You're going to have to back up, since your integrand is wrong. Show us what you did to get the part under the radical.j

Edit: Never mind. I misread the exponent as 3, rather than 3/2.
 
Last edited:
  • #6
28
0
I'm not the original poster dude.

[tex]s = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx.[/tex]

[tex]f(x) = 5 + 8x^{\frac{3}{2}}[/tex]

[tex]f'(x) = 12x^{\frac{1}{2}}[/tex]

[tex][f'(x)]^{2} = 144x[/tex]

[tex]s = \int_{0}^{1} \sqrt { 1 + 144x }\, dx = {\frac{1}{216}(1 + 144x)^{3/2}}\left|^{1}_{0}[/tex]
 
  • #7
ok i have got to the that point is that answer right? then you just substitute the 1 in and thats the answer? I just want to know that I did it right my online homework is picky about the answer format.
Thanks to all
 
  • #8
28
0
Mugen, you should be able to check your work (or the posted work) for errors.

Yes, substituting 1 and 0 in is the correct procedure for evaluating this definite integral. Be more confident!
 
  • #9
34,935
6,698
I'm not the original poster dude.

[tex]s = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx.[/tex]

[tex]f(x) = 5 + 8x^{\frac{3}{2}}[/tex]

[tex]f'(x) = 12x^{\frac{1}{2}}[/tex]

[tex][f'(x)]^{2} = 144x[/tex]

[tex]s = \int_{0}^{1} \sqrt { 1 + 144x }\, dx = {\frac{1}{216}(1 + 144x)^{3/2}}\left|^{1}_{0}[/tex]

Looks fine. I misread the exponent in the original problem as x^3 rather than x^(3/2).
 
  • #10
OK I just put it in and it said it was wrong so I dont care anymore. I think its right as do all of you so Im going with the homework having a typo in the answer.
Thanks again
 
  • #11
34,935
6,698
Your answer should be [tex]\frac{1}{216}(145^{3/2} - 1) [/tex]
which can also be written as
[tex]\frac{1}{216}(145\sqrt{145} - 1)[/tex]
 

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