Arc Length with Parametric Equations — Question 10

Question 10

Problem

Find $a>0$ so that the line path $x=at$, $y=(1-a)t$, $0\le t\le1$, has minimum length.

Question 10 – Solution

Solution

1. Differentiate the line parametrization: $$x'(t)=a, \qquad y'(t)=1-a.$$ The speed is constant, so $$L(a)=\int_0^1\sqrt{a^2+(1-a)^2}\,dt =\sqrt{a^2+(1-a)^2}.$$

2. Because the square-root function is increasing, minimizing $L(a)$ is equivalent to minimizing $$L(a)^2=a^2+(1-a)^2=2a^2-2a+1.$$ Complete the square: $$2a^2-2a+1 =2\left(a-\frac12\right)^2+\frac12.$$

3. The squared term is smallest when $a=1/2$, which satisfies $a>0$. Therefore, $$\boxed{a=\frac12}.$$

4. At this value, the minimum length is $$L_{\min}=\sqrt{\frac12}=\boxed{\frac1{\sqrt2}}.$$ Geometrically, the endpoint $(a,1-a)$ is then $(1/2,1/2)$, the closest point on the line $x+y=1$ to the origin.