# Arc Length Problem

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• Jul 15th 2014, 07:55 PM
Jason76
Arc Length Problem
$\displaystyle y = \sqrt{2 - x^{2}}$ evaluated at upper bound 1 and lower bound 0

$\displaystyle y'(\sqrt{2 - x^{2}}) = -2x$

$\displaystyle L = \int_{0}^{1} \sqrt{1 + (-2x)^{2}}$

$\displaystyle L = \int_{0}^{1} \sqrt{1 + 4x^{2}}$

$\displaystyle L = \int_{0}^{1} (1 + 4x^{2})^{1/2}$

$\displaystyle L = \int_{0}^{1} (u)^{1/2}$

$\displaystyle u = 1 + 4x^{2}$

$\displaystyle du = 8x$

$\displaystyle \dfrac{1}{8}du = dx$

$\displaystyle L = \int_{0}^{1} (u)^{1/2}$

$\displaystyle L = \dfrac{1}{8} \int_{0}^{1} ((u)^{1/2}$

$\displaystyle (\dfrac{2}{3})(\dfrac{1}{8})(u)^{3/2}$ evaluated at upper bound 1 and lower bound 0

$\displaystyle (\dfrac{1}{12})(u)^{3/2}$ evaluated at upper bound 1 and lower bound 0

$\displaystyle (\dfrac{1}{12})(1 + 4x^{2})^{3/2}$ evaluated at upper bound 1 and lower bound 0

$\displaystyle [(\dfrac{1}{12})(1 + 4(1)^{2})^{3/2}] - (\dfrac{1}{12})(1 + 4(0)^{2})^{3/2}$ - Is this right?
• Jul 16th 2014, 02:59 AM
Prove It
Re: Arc Length Problem
Why have you written your derivative in such an odd format?

\displaystyle \begin{align*} y &= \sqrt{ 2 - x^2} \\ \frac{ \mathrm{d}y }{\mathrm{d}x} &= -\frac{2x}{2 \sqrt{2 - x^2} } \\ &= -\frac{x}{\sqrt{2 - x^2}} \end{align*}

\displaystyle \begin{align*} L &= \int_a^b{ \sqrt{1 + \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 } \, \mathrm{d}x } \\ &= \int_0^1{ \sqrt{ 1 + \left( -\frac{x}{\sqrt{2 - x^2} } \right) ^2 } \, \mathrm{d}x } \end{align*}

Go from here...
• Jul 18th 2014, 05:59 PM
Jason76
Re: Arc Length Problem
Sorry for delay in responding.

$\displaystyle \sqrt{2 - x^{2}}$

$\displaystyle 0\leq x \leq 1$

$\displaystyle y = (2 - x^{2})^{2}$

$\displaystyle y' = -x(2 - x^{2})^{-1/2}$

$\displaystyle \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}}$

$\displaystyle \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x}}$

$\displaystyle \int_{0}^{1} \dfrac{\sqrt{2}}{\sqrt{2 - x^{2}}}$

Using trig substitution you get to this point:

$\displaystyle a = 2$

$\displaystyle x = 2\sin\theta$

$\displaystyle dx = 3\cos\theta$

$\displaystyle 4 - (2\sin\theta)^{2} = 4 - 4 \sin^{2}\theta = 4(1 - \sin^{2}\theta) = 4 cos^{2}\theta$

$\displaystyle \int_{0}^{1}\dfrac{\sqrt{2} (2\cos \theta)}{\sqrt {4 cos^{2}\theta}}$ - What next it seems like all the thetas cancel out, so how can we evaluate it?

$\displaystyle \int_{0}^{1}\sqrt{2}$
• Jul 18th 2014, 08:09 PM
Prove It
Re: Arc Length Problem
Quote:

Originally Posted by Jason76
Sorry for delay in responding.

$\displaystyle \sqrt{2 - x^{2}}$

$\displaystyle 0\leq x \leq 1$

$\displaystyle y = (2 - x^{2})^{2}$

$\displaystyle y' = -x(2 - x^{2})^{-1/2}$

$\displaystyle \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}}$

$\displaystyle \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x^2}}$

Fixed that step

Quote:

$\displaystyle \int_{0}^{1} \dfrac{\sqrt{2}}{\sqrt{2 - x^{2}}}$

Using trig substitution you get to this point:

$\displaystyle a = 2$
NO! Here your \displaystyle \begin{align*} a = \sqrt{2} \end{align*} so the substitution you must make is \displaystyle \begin{align*} x = \sqrt{2}\sin{ \left( \theta \right) } \implies \mathrm{d}x = \sqrt{2}\cos{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}

Quote:

$\displaystyle x = 2\sin\theta$

$\displaystyle dx = 3\cos\theta$

$\displaystyle 4 - (2\sin\theta)^{2} = 4 - 4 \sin^{2}\theta = 4(1 - \sin^{2}\theta) = 4 cos^{2}\theta$

$\displaystyle \int_{0}^{1}\dfrac{\sqrt{2} (2\cos \theta)}{\sqrt {4 cos^{2}\theta}}$ - What next it seems like all the thetas cancel out, so how can we evaluate it?
Even though you are out by a significant factor from the above mistake, you come to the right conclusion that the "functional" part cancels out. So how would you evaluate \displaystyle \begin{align*} \int{ \mathrm{d}x } \end{align*}? Hint: \displaystyle \begin{align*} \int{ \mathrm{d}x} = \int{ 1 \, \mathrm{d}x } \end{align*}...
• Jul 18th 2014, 09:35 PM
Jason76
Re: Arc Length Problem
Would the answer be $\displaystyle \sqrt{2} \theta$ evaluated at 1 upper bound and 0 lower bound?
• Jul 18th 2014, 10:07 PM
Prove It
Re: Arc Length Problem
Well, if you mean evaluated where x is between 0 and 1, then yes.

But you will be better off changes your bounds to \displaystyle \begin{align*} \theta \end{align*} bounds, so that you don't have to convert back to a function of x.
• Jul 18th 2014, 11:40 PM
Jason76
Re: Arc Length Problem
How could I convert back to a function of x without all the trig functions? If I had those, then all I would have to do is make a right triangle.
• Jul 19th 2014, 12:19 AM
Prove It
Re: Arc Length Problem
You NEED the trig functions to convert back to a function of x.

Convert your bounds, it's a much easier process.
• Jul 19th 2014, 12:30 AM
Jason76
Re: Arc Length Problem
Quote:

Originally Posted by Prove It
You NEED the trig functions to convert back to a function of x.

Convert your bounds, it's a much easier process.

how to convert bounds? hint?
• Jul 19th 2014, 12:59 AM
Prove It
Re: Arc Length Problem
You know that \displaystyle \begin{align*} x = \sqrt{2}\sin{ \left( \theta \right) } \end{align*}. If x = 0 then what is \displaystyle \begin{align*} \theta \end{align*}? If x = 1 then what is \displaystyle \begin{align*} \theta \end{align*}?
• Jul 19th 2014, 10:19 AM
Jason76
Re: Arc Length Problem
Quote:

Originally Posted by Prove It
You know that \displaystyle \begin{align*} x = \sqrt{2}\sin{ \left( \theta \right) } \end{align*}. If x = 0 then what is \displaystyle \begin{align*} \theta \end{align*}? If x = 1 then what is \displaystyle \begin{align*} \theta \end{align*}?

right I see. The sin of 0 is 0 and the sin of 1 is ?? Is it $\displaystyle \dfrac{\pi}{2}$

Answer would be $\displaystyle \sqrt{2}\sin(1) - \sqrt{2}\sin(0) = \sqrt{2}(\dfrac{\pi}{2}) - \sqrt{2}(0) = \sqrt{2}\dfrac{\pi}{2}$ ??
• Jul 19th 2014, 10:22 AM
Prove It
Re: Arc Length Problem
NO! You are trying to figure out which ANGLES give those sine values! What angle gives the sine value as 0? (You have correctly identified it as 0). What angle gives \displaystyle \begin{align*} \sqrt{2}\sin{ \left( \theta \right) } = 1 \implies \sin{ \left( \theta \right) } = \frac{1}{\sqrt{2}} \end{align*}. This is another one that you should have committed to memory.
• Jul 19th 2014, 10:25 AM
Jason76
Re: Arc Length Problem
Wouldn't it be?

$\displaystyle \sqrt{2}\sin(1) - \sqrt{2}\sin(0) = \sqrt{2}(\dfrac{\pi}{2}) - \sqrt{2}(0) = \sqrt{2}\dfrac{\pi}{2}$
• Jul 19th 2014, 08:21 PM
Prove It
Re: Arc Length Problem
Don't listen to me then...
• Jul 19th 2014, 09:24 PM
Jason76
Re: Arc Length Problem
$\displaystyle \sqrt{2}\dfrac{\pi}{2}$ is not the answer on homework.
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