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Arc length problem.

  • Thread starter negatifzeo
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  • #1
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Homework Statement


Find the arclength of the function [tex] y=x^2 [/tex] when x is between 0 and 10.

Homework Equations


Arclength here is [tex] \int_{0}^{10} \sqrt{1+(2x)^2} dx [/tex]
(It's intended to be the integral from 0 to 10 of the quare root of 1+(2x)^2. My latex skills suck.)

The Attempt at a Solution


Trig substitution. 2x= tan (theta)

Integrate secant theta
Which is ln(sec(theta)+tan(theta))

Substituting theta back in for arcsin(2x)
ln(sec(arctan(2x))+tan(arctan(2x))) evaluated from 10 to 0. Solving for arctan(2x) my final answer is
ln(sqrt(1+4x^2)+2x) evaluated from 0 to 10. I know this is wrong as it is much to short of an arc length but I don't know where I went wrong. Any clues?






The Attempt at a Solution

 
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Answers and Replies

  • #2
Tom Mattson
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Homework Statement


Find the arclength of the function [tex] y=x^2 [/tex] when x is between 0 and 10.

Homework Equations


Arclength here is [tex] \int_{0}^{10} \sqrt{1+(2x)^2} dx [/tex]
(It's intended to be the integral from 0 to 10 of the quare root of 1+(2x)^2. My latex skills suck.)
I've edited your post to fix your TeX.

The Attempt at a Solution


Trig substitution. 2x= tan (theta)
That's fine.

Integrate secant theta
Which is ln(sec(theta)+tan(theta))
That's not fine. Why would you integrate [itex]\sec (\theta )[/itex]?
 
  • #3
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I've edited your post to fix your TeX.



That's fine.



That's not fine. Why would you integrate [itex]\sec (\theta )[/itex]?
Well, when I use trig substitution here I change 2x to tan(theta). This changes the integrand to sqrt(1+tan^2(theta)). The trig identity says that 1 + tangent squared equals secant squared, and since it is the quare root of that it just becomes secant.
 
  • #4
Tom Mattson
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But you've forgotten about the [itex]dx[/itex].
 
  • #5
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Oh, duh. Wait, why am I having a hard time remembering how to get the dx here? Does dx here equal 1/2*sec^2(theta) d(theta)?
 
  • #6
Tom Mattson
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Yes, that's right.
 

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