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!
Best Answer
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.