Arc Length Problem

**Jason76** · Jul 15th 2014, 07:55 PM

$y = \sqrt{2 - x^{2}}$ evaluated at upper bound 1 and lower bound 0

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

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

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

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

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

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

$du = 8x$

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

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

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

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

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

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

$[(\dfrac{1}{12})(1 + 4(1)^{2})^{3/2}] - (\dfrac{1}{12})(1 + 4(0)^{2})^{3/2}$ - Is this right?

**Prove It** · Jul 16th 2014, 02:59 AM

Why have you written your derivative in such an odd format?

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

so your arclength is

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

Go from here...

**Jason76** · Jul 18th 2014, 05:59 PM

Sorry for delay in responding.

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

$0\leq x \leq 1$

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

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

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

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

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

Using trig substitution you get to this point:

$a = 2$

$x = 2\sin\theta$

$dx = 3\cos\theta$

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

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

$\int_{0}^{1}\sqrt{2}$

**Prove It** · Jul 18th 2014, 08:09 PM

Originally Posted by Jason76

Sorry for delay in responding.

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

$0\leq x \leq 1$

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

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

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

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

Fixed that step

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

Using trig substitution you get to this point:

$a = 2$

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

$x = 2\sin\theta$

$dx = 3\cos\theta$

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

$\int_{0}^{1}\dfrac{\sqrt{2} (2\cos \theta)}{\sqrt {4 cos^{2}\theta}}$ - 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: $\displaystyle \begin{align*} \int{ \mathrm{d}x} = \int{ 1 \, \mathrm{d}x } \end{align*}$...

**Jason76** · Jul 18th 2014, 09:35 PM

Would the answer be $\sqrt{2} \theta$ evaluated at 1 upper bound and 0 lower bound?

**Prove It** · Jul 18th 2014, 10:07 PM

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.

**Jason76** · Jul 18th 2014, 11:40 PM

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.

**Prove It** · Jul 19th 2014, 12:19 AM

You NEED the trig functions to convert back to a function of x.

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

**Jason76** · Jul 19th 2014, 12:30 AM

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?

**Prove It** · Jul 19th 2014, 12:59 AM

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 $\displaystyle \begin{align*} \theta \end{align*}$?

**Jason76** · Jul 19th 2014, 10:19 AM

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 $\dfrac{\pi}{2}$

Answer would be $\sqrt{2}\sin(1) - \sqrt{2}\sin(0) = \sqrt{2}(\dfrac{\pi}{2}) - \sqrt{2}(0) = \sqrt{2}\dfrac{\pi}{2}$ ??

**Prove It** · Jul 19th 2014, 10:22 AM

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.

**Jason76** · Jul 19th 2014, 10:25 AM

Wouldn't it be?

$\sqrt{2}\sin(1) - \sqrt{2}\sin(0) = \sqrt{2}(\dfrac{\pi}{2}) - \sqrt{2}(0) = \sqrt{2}\dfrac{\pi}{2}$

**Prove It** · Jul 19th 2014, 08:21 PM

Don't listen to me then...

**Jason76** · Jul 19th 2014, 09:24 PM

$\sqrt{2}\dfrac{\pi}{2}$ is not the answer on homework.

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