1. Arc Length Problem

$y = \sqrt{2 - x^{2}}$ 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?

2. 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...

3. Re: Arc Length Problem

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?

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

4. Re: Arc Length Problem

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*}...

5. Re: Arc Length Problem

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

6. 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.

7. 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.

8. 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.

9. Re: Arc Length Problem

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?

10. 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*}?

11. Re: Arc Length Problem

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

12. 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.

13. Re: Arc Length Problem

Wouldn't it be?

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

14. Re: Arc Length Problem

Don't listen to me then...

15. Re: Arc Length Problem

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

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