Shortest distance between two points

Question

Everyone knows that the shortest distance between two points is a straight line. How can we prove it mathematically?

Answer 1

If $y$ is a $C^2$ function of $x$ and traces a curve between points $(a,c)$ and $(b,d)$, then the length of the curve is $$ \ell = \int_a^b \sqrt{1+(y')^2}\,dx $$ This is the kind of functional that we can apply the calculus of variations to. According the Euler-Lagrange equations (see link), the quantity $\ell$ is minimized when $$ \frac{d}{dx} \frac{\partial}{\partial y'}\sqrt{1+(y')^2} = \frac{\partial}{\partial y}\sqrt{1+(y')^2} $$ Therefore, $$ \frac{d}{dx}\frac{y'}{\sqrt{1+(y')^2}} = 0 \implies \frac{y''}{(1+(y')^2)^{3/2}} = 0 \implies y'' = 0 $$ If $y'' = 0$, the curve is a straight line.

Answer 2

This follows from the Euler-Lagrange equation from the calculus of variations. Let $(a,y(a)),(b,y(b))$ be two points in the plane, where $y$ is a differentiable function on $[a,b]$. The lenght $S$ of this curve between $a \le x \le b$ is given by: $$S =\int_{a}^{b}\sqrt{1+\bigg{(}\frac{dy}{dx}\bigg{)}^{2}}dx$$ So we would like to find out which curve $y$ minimizes $S$. Write $F(y') = \sqrt{1+(y')^{2}}$. We have: $$\frac{\partial}{\partial y} F = 0 \quad \mbox{and} \quad \frac{\partial}{\partial y'}F = \frac{y'}{\sqrt{1+y'}}$$ Thus, using Euler-Lagrange we have: $$\frac{\partial}{\partial y}F - \frac{d}{dx}\bigg{(}\frac{\partial}{\partial y'}F\bigg{)} = 0 \Rightarrow -\frac{d}{dx}\bigg{[}\frac{y'}{\sqrt{1+(y')^{2}}}\bigg{]} = 0 \hspace{1cm} (1)$$ The solution of equation (1) is given by: $$ y = \alpha x + \beta$$ which is exactly the equation of a straight line.