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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- Jul 15th 2014, 08:55 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? - Jul 16th 2014, 03:59 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... - Jul 18th 2014, 06:59 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?

- Jul 18th 2014, 09:09 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?

- Jul 18th 2014, 10:35 PMJason76Re: Arc Length Problem
Would the answer be evaluated at 1 upper bound and 0 lower bound?

- Jul 18th 2014, 11:07 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. - Jul 19th 2014, 12:40 AMJason76Re: 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, 01:19 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. - Jul 19th 2014, 01:30 AMJason76Re: Arc Length Problem
- Jul 19th 2014, 01:59 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 $\displaystyle \begin{align*} \theta \end{align*}$?

- Jul 19th 2014, 11:19 AMJason76Re: Arc Length Problem
- Jul 19th 2014, 11:22 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.

- Jul 19th 2014, 11:25 AMJason76Re: Arc Length Problem
Wouldn't it be?

- Jul 19th 2014, 09:21 PMProve ItRe: Arc Length Problem
Don't listen to me then...

- Jul 19th 2014, 10:24 PMJason76Re: Arc Length Problem
is not the answer on homework.