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Prove that a straight line is the shortest distance between two points in $E_3$. Use the following scheme; let $\alpha: [a,b]\to E_3$ be an arbitrary curve segment from $p = \alpha(a) , q = \alpha(b)$.Let $u = (q — p)/||q — p || $.

(a) If $\sigma$ is a straight-line segment from $ p$ to $ q$ , say $$\sigma (t) = (1 - t)p + tq ,\quad 0\leq t\leq1$$ show that $L(\sigma ) = d(p,q)$.

What I have done $$ L(\sigma)=\int_{0}^{1}||\sigma'(t)||dt=\int_{0}^{1}(p^2+q^2)^{1/2}dt=\sqrt{(p^2+q^2)}(1), $$ $d(p,q)=\sqrt{(b-a)^2+(q-p)^2}$. Where am I doing wrong? It's a problem from O'Neill Elementary Differential Geometry.

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  • $\begingroup$ How did you get your expression for $\vert\vert \sigma ' (t) \vert\vert$? $\endgroup$ – πr8 Jan 8 '16 at 14:38
  • $\begingroup$ You should integrate from 0 to 1. $\endgroup$ – Paul K Jan 8 '16 at 14:41
  • $\begingroup$ integrating the speed of curve from a to b,yes t should be from 0 to 1 $\endgroup$ – Nebo Alex Jan 8 '16 at 14:42
  • $\begingroup$ But he's integrating the special curve $\sigma$ which is only defined on $[0,1]$. $\endgroup$ – Paul K Jan 8 '16 at 14:43
  • $\begingroup$ It should be $\int_{0}^1$ not $\int_{a}^b$ $\endgroup$ – Thomas Andrews Jan 8 '16 at 14:44
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It is correct that $\sigma'(t)=q-p,$ what is wrong is the norm of this vector, that is not $\sqrt{p^2+q^2},$ but $||q-p||=d(p,q).$

In coordinates, if $p=(x_1,y_1,z_1)$ and $q=(x_2,y_2,z_2)$, then $q-p$ is the vector of components $(x_2-x_1,y_2-y_1,z_2-z_1)$ whose norm is $$ \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}=d(p,q). $$

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  • $\begingroup$ i am confused by these $ p=α(a),q=α(b)$ $\endgroup$ – Nebo Alex Jan 8 '16 at 16:40
  • $\begingroup$ @Cielo: to visualize, imagine $t\in[a,b]$ is the time, so $p=\alpha(a)$ is the position at the initial time $a$, it is a point in space $\mathbb{E}_3$, while $q=\alpha(b)$ is the position at the final time $b$. $\endgroup$ – enzotib Jan 8 '16 at 16:43
  • $\begingroup$ oh thanks you are great you gave me good intuition. $\endgroup$ – Nebo Alex Jan 8 '16 at 16:58

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