evaluated at upper bound 1 and lower bound 0

evaluated at upper bound 1 and lower bound 0

evaluated at upper bound 1 and lower bound 0

evaluated at upper bound 1 and lower bound 0

- Is this right?

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- July 15th 2014, 07:55 PMJason76Arc Length Problem
evaluated at upper bound 1 and lower bound 0

evaluated at upper bound 1 and lower bound 0

evaluated at upper bound 1 and lower bound 0

evaluated at upper bound 1 and lower bound 0

- Is this right? - July 16th 2014, 02:59 AMProve ItRe: Arc Length Problem
Why have you written your derivative in such an odd format?

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

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... - July 18th 2014, 05:59 PMJason76Re: Arc Length Problem
Sorry for delay in responding.

Using trig substitution you get to this point:

- What next it seems like all the thetas cancel out, so how can we evaluate it?

- July 18th 2014, 08:09 PMProve ItRe: Arc Length Problem
Fixed that step

Quote:

Using trig substitution you get to this point:

Quote:

- What next it seems like all the thetas cancel out, so how can we evaluate it?

- July 18th 2014, 09:35 PMJason76Re: Arc Length Problem
Would the answer be evaluated at 1 upper bound and 0 lower bound?

- July 18th 2014, 10:07 PMProve ItRe: 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. - July 18th 2014, 11:40 PMJason76Re: 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.

- July 19th 2014, 12:19 AMProve ItRe: 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. - July 19th 2014, 12:30 AMJason76Re: Arc Length Problem
- July 19th 2014, 12:59 AMProve ItRe: 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 $\displaystyle \begin{align*} \theta \end{align*}$?

- July 19th 2014, 10:19 AMJason76Re: Arc Length Problem
- July 19th 2014, 10:22 AMProve ItRe: 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.

- July 19th 2014, 10:25 AMJason76Re: Arc Length Problem
Wouldn't it be?

- July 19th 2014, 08:21 PMProve ItRe: Arc Length Problem
Don't listen to me then...

- July 19th 2014, 09:24 PMJason76Re: Arc Length Problem
is not the answer on homework.