How to find the minimum length of $f(x)$

Question

Let $f(0)=0, f(1)=t $ and $$ \int_0^1 f(x) dx = 1 $$ How find the minimum length of $f(x)$ in $[0, 1]$ ?

For example when t=2, the minimum length of f(x) is the segment of OA

(O is the origin and A=(1, 2))

But how find the minimum length when t=3 ?

Answer 1

Following Gil Strang's calculus of variations notes online, you want to use a Lagrangian of the form $$ \mathscr{L}(y,y',\lambda) = \int_0^1 \sqrt{1+(y')^2}\,dx + \lambda \left(\int_0^1 y\,dx - 1\right) = \int_0^1\left(\sqrt{1+y'^2}+\lambda y\right)\,dx -\lambda $$ Let the term in parentheses be $F(y,y',\lambda)$. If $\dfrac{\delta \mathscr{L}}{\delta y} =0$, the Euler-Lagrange equations are $$ 0 = \frac{\partial F}{\partial y} -\frac{d}{dx}\frac{\partial F}{\partial y'} = \lambda - \frac{d}{dx}\frac{y'}{\sqrt{1+(y')^2}} $$ So we can integrate once to get: $$ \frac{y'}{\sqrt{1+y'^2}} = \lambda x + c $$ for some constant $c$. Solving, $$ y' = \frac{\lambda x + c}{\sqrt{1-(\lambda x + c)^2}} $$ Let $u = \lambda x + c$. Then $$ y = \int \frac{\lambda x + c}{\sqrt{1-(\lambda x + c)^2}}\,dx = \frac{1}{\lambda}\int \frac{u}{\sqrt{1-u^2}}\,du = -\frac{1}{\lambda} \sqrt{1-u^2} + d $$ Or, \begin{align*} (\lambda y - \lambda d)^2 &= 1-(\lambda x + c)^2\\ \implies \left(x + \frac{c}{\lambda}\right)^2 + (y-d)^2 &= \frac{1}{\lambda^2} \end{align*} The curve is a circular arc. We can find $c$, $d$, and $\lambda$ by placing $(0,0)$ and $(1,t)$ on the curve, and solving $\int_0^1 y\,dx = 1$.

Answer 2

When you did $t=2$ the shortest distance was a straight line. If we increase $t$ then the shortest distance is likely to be an arc of a circle.

This is better justified with we rotate the graph so that $(1,t)$ is now on the x-axis at $(\sqrt{1+t^2},0)$.

The area can be found by considering the triangle take away the sector. The radius can be adjusted until the area enclosed is equal to $1$.