What curve lying above the x-axis has the property that the length of the arc joining any two points on it is proportional to the area under the arc?
Calculus 4. Area under arc.
1
$\begingroup$
calculus
-
0You would prove it by finding the length of the arc joining two general points on the horizontal line, and finding the area under this arc, and noting the simple relation between them. – 2012-09-20
1 Answers
1
Let $y(x)$ be the function, which I assume is "nice" enough. Then $0=\int_a^b dx\,\left[\sqrt{1+y'(x)^2}-\alpha y(x)\right]$ for some constant of proportionality $\alpha$. Since $a$ and $b$ are arbitrary here, we must have that the integrand vanishes identically, $\sqrt{1+y'(x)^2}=\alpha y(x)$ for all $x\in\mathbb{R}$. Rearranging, $\frac{dy}{dx}=\pm\alpha\sqrt{y^2-\frac{1}{\alpha^2}}.$ We have to be careful dividing through by the radical, because there is a solution $y(x)=\frac{1}{\alpha}=\textrm{const}.$ However, there is also a solution $y(x)=\frac{1}{\alpha}\cosh\left(\alpha x+A\right),$ with $A$ a constant of integration. You should doublecheck details of the separation of variables for cases such as $\alpha<0$.