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Arc length integral for the curve

$$e^{2x+2y}=x-y$$

So I tried isolating $y$ or $x$ and got stuck with a $0 = 0$ at the end of it.

I tried to also use partial derivatives. Am I going in the right direction?

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    nope it has a equal sign. – 2017-02-14
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    Did you try with parametrize it!, Ex. $x=r\cos^2\theta$ and $y=r\sin^2\theta$ so it simplifies to $e^{2r}=r\cos(\pi-\theta)$ – 2017-02-14
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    I did, Im not sure about x=r(cos(theta))^2. Im only aware of trying parametrizations with x=rcos. I ended up with e^(2r(cos + sin))=rcos - rsin – 2017-02-14
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    @MyGlasses: I am curious, how would you pursue once you have $e^{2r}=r\cos(\pi-\theta)$? – 2017-02-14
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    @Kuifje I've done part of it but could not finish. Do it now if you want. – 2017-02-14

3 Answers 3

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Hint: Rotate your axes by $\pi/4$

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    *Note:* That's a comment, not an answer. – 2017-02-14
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    @user90369, firstly I don't have 50 points for a comment; secondly it solves the problem except for dotting the i's and crossing the t's. Do you want me to dot the i's and cross the t's? – 2017-02-14
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    Please write you solution down. :-) – 2017-02-14
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    It turns into y=exp(x) – 2017-02-14
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    *I meant:* A full answer for the OP because it seems not to be obviously. This will help the OP in an understandable way. – 2017-02-14
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    No, he said 'Am I going the right way', not 'please do my homework for me' – 2017-02-14
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    It's the same (explanation and doing the homework), if he isn't going the right way. :-) – 2017-02-14
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    Very clever! But I think you should detail too. – 2017-02-14
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    You can turn it into a definite integral, that a graphics calculator can handle. – 2017-02-15
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$\displaystyle 2(x-y)e^{2(x-y)}=2e^{4x}\enspace$ => $\displaystyle \enspace x-y=\frac{1}{2}W(2e^{4x})$

where $W(x)$ is the Lambert W-function

We get $\enspace\displaystyle \frac{dy}{dx}=\frac{1-W(2e^{4x})}{1+W(2e^{4x})}$ .

For the arc length integral follows:

$\displaystyle \int\sqrt{(dx)^2+(dy)^2}=\int\sqrt{1+(\frac{1-W(2e^{4x})}{1+W(2e^{4x})})^2}dx=\sqrt{2}\int\frac{\sqrt{1+W(2e^{4x})^2}}{1+W(2e^{4x})}dx$

$\displaystyle =\frac{\sqrt{2}}{4}( \sqrt{1+W(2e^{4x})^2}-\ln(1+\sqrt{1+W(2e^{4x})^2})+\ln(W(2e^{4x})))+C$

$\displaystyle =\frac{\sqrt{2}}{4}( \sqrt{1+(2x-2y)^2}-\ln(1+\sqrt{1+(2x-2y)^2})+\ln(2x-2y))+C$

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For $e^{2x+2y}= x-y$ let variable changing $u=2x+2y$ and $v=x-y$ the curve is $e^u=v$ so Jacobin is $J=-2$ and the arc length integral for this curve from $(a,A)$ to $(b,B)$ is \begin{eqnarray} \ell &=& \int_a^b\sqrt{1+y'^2}dx\\ &=& \int_{\alpha}^{\beta}\sqrt{1+v'^2}\,|J|\,du\\ &=& \int_{\alpha}^{\beta}2\sqrt{1+e^{2u}}du\\ &=& \left(2\sqrt{1+e^{2u}}-2arctanh\sqrt{1+e^{2u}}\right)\Big|_{\alpha}^{\beta}\\ &=& \left(2\sqrt{1+e^{2x+2y}}-2arctanh\sqrt{1+e^{2x+2y}}\right)\Big|_a^b \end{eqnarray}