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This is a question from a mathematical contest.

Let $y=y(x)$ be the extremal of the functional $$ I[y(x)] = \int_{x_1}^{x_2} \sqrt{ 1+ \left( \frac{dy}{dx} \right)^2 }dx $$ Subject to the condition that the left end of the extremal moves along $y=x^2$, while the rihgt end moves along $x-y=5$. Then:

  1. Shortest distance between the parabola and the straight line is $\left( \frac{19 \sqrt{2}}{8} \right) $.
  2. Slope of the extermal at $(x,y)$ is $(-3/2)$.
  3. Point $(3/4,0)$ lies on the extremal.
  4. Extremal is orthogonal to the curve $y= x/2$.

My problem:

Since the integrand is independent of $x,y$, we have: $y= ax+b$, where $a,b$ are constants depending on the boundary conditions. Now, since the end of the extremals lie on the given curves so $y(0)=0$, and $y(5,0)$. But this is clearly not satisfied by $y=ax+b$.

So, where is the wrong step? and how to solve this problem.

Any help is highly appreciated.

Kashif
  • 710
  • How do you find the slope of the extremal at $(x,y)$ is $-\frac{3}{2}$? Because the shortest distance is a line, and this line must be perpendicular to the other line which has a slope of $1$, the slope of the shortest path will be $-1$. – Merosity Nov 06 '21 at 08:03

0 Answers0