0
$\begingroup$

I've been tasked to provide a proof for the following:

for $$ f(x)=\frac{e^x}{2}+e^{-x} $$ show that arc length of the curve over any interval equals the area under the curve for the same interval.

I tried plugging $f(x)$ into the arc length formula, and the integral I arrived at was different from the integral I used for area under the curve.

I'm not sure whether this is due to a mistake on my part, or if the question itself has a typo.

Any advice/help would be appreciated!

$\endgroup$
  • $\begingroup$ Are you sure the second term isn't also divided by 2? $\endgroup$ – user84413 Apr 7 '16 at 0:29
  • $\begingroup$ ^yeah, I rechecked and it isn't. $\endgroup$ – thevioletsaber Apr 7 '16 at 0:32
  • $\begingroup$ Thanks - I am pretty sure the question does have a typo. $\endgroup$ – user84413 Apr 7 '16 at 0:35
2
$\begingroup$

You have copied the problem down incorrectly. It should read $$y=\cosh x=\frac{e^x+e^{-x}}2$$ Then $$dy=\frac{e^x-e^{-x}}2dx=\sinh x\,dx$$ The element of arc length is given by $$\begin{align}(ds)^2 & =(dx)^2+(dy)^2=\left(1+\frac{e^{2x}-2+e^{-2x}}4\right)(dx)^2\\ & =\left(\frac{e^{2x}+2+e^{-2x}}4\right)(dx)^2=\left(\frac{e^x+e^{-x}}2\right)^2(dx)^2\\ & =\cosh^2x(dx)^2\end{align}$$ So the arc length is $$s=\int ds=\int\cosh x\,dx$$ And this is the same as the area $$A=\int y\,dx=\int\cosh x\,dx$$ Famous theorem for a catenary.

EDIT: OK, so how did I know there was an error in the original problem statement? If $dA=ds$, then $y\,dx=\sqrt{1+(y^{\prime})2}\,dx$. Differentiating, $$y^{\prime}=\frac12\cdot\frac{2y^{\prime}y^{\prime\prime}}{\sqrt{1+(y^{\prime})^2}}=\frac{y^{\prime}y^{\prime\prime}}y$$ Then we can write it as a second-order differential equation: $y^{\prime\prime}=y$ with solution $y=c_1e^x+c_2e^{-x}$. Then $y^{\prime}=c_1e^x-c_2e^{-x}$. Plugging into the original differential equation, $$c_1e^x+c_2e^{-x}=\sqrt{1+c_1^2e^{2x}-2c_1c_2+c_2^2e^{-2x}}$$ Squaring out and simplifying, we see that $2c_1c_2=1-2c_1c_2$ so $c_2=\frac1{4c_1}$. Then $$y=c_1e^x+\frac1{4c_1}e^{-x}=\frac12\left(2c_1e^x+\frac1{2c_1}e^{-x}\right)$$ Since $y>0$, we can take the natural logarithm of $2c_1$ to get $\ln2c_1=-x_0$ and then $$y=\frac{e^{x-x_0}+e^{-(x-x_0)}}2=\cosh(x-x_0)$$ EDIT: I was careless in my proof above. Early on I divided by $y^{\prime}$ without considering what would happen if $y^{\prime}=0$. In fact, in that case $y=1$ is a (singular) solution to the problem as well.

$\endgroup$
  • $\begingroup$ Thank you for the correction; I would like to note, however, that I did not copy the problem down incorrectly. It was in fact a typo in the original question - prntscr.com/ap35jj $\endgroup$ – thevioletsaber Apr 7 '16 at 0:50
  • $\begingroup$ I knew there was a mistake somewhere in there because a catenary is the only curve that satisfies this condition. I'll append the proof to my answer. $\endgroup$ – user5713492 Apr 7 '16 at 0:54
0
$\begingroup$

$f(x)=(\frac{e^x}{2})+(e^{-x})$

$f(x)=\frac{e^x}{2}+\frac{1}{e^{x}}$

$f(x)=\frac{e^{2x}+2}{2e^x}$

Just for clarity's sake. Now I am assuming that you do not need to be so rigorous as to prove the arc length formula itself.

We intend to establish that

$∫_a^b \sqrt{1+[f'(x)]^2} dx$ = $∫_a^b f(x) dx$

$f'(x)=\frac{e^x}{2}-e^{-x}$

$f'(x)^2+1=(\frac{e^{2x}-2}{2e^x})^2+1$

$\sqrt{[f'(x)]^2+1}$=$\sqrt{\frac{e^{4x}+4-2e^{2x}}{2e^x}}$

$∫_a^b\sqrt{f'(x)+1} dx$=$∫_a^b \sqrt{\frac{e^{4x}+4-2e^{2x}}{2e^x}} dx$

And, like you, I am not sure this is true...

Take for example $a=0, b=1$, the relationship does not hold.

$\endgroup$
  • $\begingroup$ I think maybe you forgot to square the derivative. $\endgroup$ – user84413 Apr 7 '16 at 0:37
  • $\begingroup$ @user84413 Great catch. Sorry of the mistake; I should really get to bed now. Will fix immediately $\endgroup$ – KR136 Apr 7 '16 at 0:38
  • $\begingroup$ @user84413 Still doesn't change the result. $\endgroup$ – KR136 Apr 7 '16 at 0:48
0
$\begingroup$

This means $$ f(x) = \cosh(x) - \frac{e^{-x}}{2} $$ and gives the area $$ A = \int\limits_a^b f(x) \, dx = \int\limits_a^b \left( \cosh(x) - \frac{e^{-x}}{2} \right) \, dx$$ while the arc length is $$ S = \int\limits_a^b \sqrt{1 + f'(x)^2} \, dx = \int\limits_a^b \sqrt{1 + \left( \sinh(x) + e^{-x} / 2\right)^2} \, dx $$ at this point it is likely that one needs to apply $\cosh^2 x = 1 + \sinh^2 x$ to achieve identity of $A$ and $S$, but the additional $e^{-x}/2$ term is a hinderance.

Looks like an error in the task.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.