I'm trying to prove that the shortest distance between two points in the Euclidean plane is a straight line:

Here is what I've achieved so far; but I've got lost right at the end if anyone could help?

Let $\varphi: [a,b] \rightarrow R$ such that $\varphi(a)=\varphi(b)=0$ and consider $L_t = \text{length}(f + t\varphi )$. If $f$ minimises the length then $\dfrac{d}{dt}L_t |_{t=0} = 0$.

Let us begin by defining the arc length of a curve : \begin{equation} \text{Arc length of a curve} = \int_{0}^1 \sqrt{1+ (f'(x))^2} dx \end{equation}

Let $L$ be the arc length of the graph of $f$, such that: \begin{equation} L_t= \int_b^a \sqrt{1+(f'+t\varphi')^2}\,dx \end{equation} Then we evaluate: \begin{align} \dfrac{d}{dt}L_t\bigg|_{t=0} &= \int_a^b \dfrac{2(f'+t\varphi')\varphi'}{2\sqrt{1+(f'+t\varphi')^2}} \bigg|_{t=0} \\ &= \int_a^b \dfrac{f'\varphi'}{\sqrt{1+(f')^2}} \end{align} Using integration by parts: \begin{equation} = -\int_a^b \varphi \, \dfrac{d}{dx}\bigg(\dfrac{f'}{\sqrt{1+(f')^2}}\bigg)=0 \quad \text{for any} \, \varphi \end{equation} We can now see that: \begin{align} \dfrac{d}{dx}\bigg(\dfrac{f'}{\sqrt{1+(f')^2}}\bigg) &=0 \\ \Rightarrow \dfrac{f'}{\sqrt{1+(f')^2}} &= \text{constant} \\ f' &= \text{constant} \,(\sqrt{1+(f')^2}) \\ (f')^2 &= \text{constant} \, (1+(f')^2) \end{align}



$$\dfrac{f'}{\sqrt{1+(f')^2}} = \text{constant}$$

square both sides and take the reciprocal, to get

$$\dfrac{1+(f')^2}{f'^2} = \text{constant}$$


$$\dfrac{1}{f'^2} = \text{constant}$$

$$f'^2 = \text{constant}$$

$$f' = \text{constant}$$

  • $\begingroup$ So then would it just be that if f' = constant, then f=constant*x + constant(other) ?? $\endgroup$ – Sarah Jayne Mar 20 '14 at 19:10
  • $\begingroup$ If the slope of $f$ is constant, then the graph of $f$ is a straight line. $\endgroup$ – TonyK Mar 20 '14 at 19:18
  • $\begingroup$ Great, thank you so much! $\endgroup$ – Sarah Jayne Mar 20 '14 at 19:20

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