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!

  • $\begingroup$ 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

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.

  • $\begingroup$ 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




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

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




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

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

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

  • $\begingroup$ 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

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.


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