Irodov 1.17: Minimizing Time, Not Distance – Amit & Deepali Sinha Foundation

Problem Irodov 1.17

A car starts at point $A$ on a straight highway and must reach point $B$ , which sits in the open field at perpendicular distance $l$ from the highway; the foot of that perpendicular on the highway is the point $D$ . The car travels at speed $v$ on the highway and at the slower speed $v/\eta$ once it leaves the highway to cross the field, with $\eta>1$ a given constant. The car drives along the highway from $A$ , turns off at some point (call it $C$ , between $A$ and $D$ ), and then drives in a straight line across the field to $B$ . Find the distance $CD$ that makes the total travel time from $A$ to $B$ as short as possible.

The highway runs through A, C and D; B sits in the field at perpendicular distance l below D (dashed). The car drives from A to the turn-off point C, then cuts straight across the field to B. — The highway runs through $A$ , $C$ and $D$ ; $B$ sits in the field at perpendicular distance $l$ below $D$ (dashed). The car drives from $A$ to the turn-off point $C$ , then cuts straight across the field to $B$ .

Given

$l$
$\eta$ (field speed $=$ highway speed $/\eta$ )

Find

CD

, the turn-off distance from

D

, that minimizes the travel time

Solution

Step 1Write the travel time as a function of the turn-off point

Let $v$ be the highway speed and let $x=CD$ be the turn-off distance sought. The highway leg is $AD-x$ at speed $v$ , and the field leg is the slant distance $\sqrt{x^2+l^2}$ at speed $v/\eta$ .

t(x) = \frac{AD-x}{v} + \frac{\eta\sqrt{x^2+l^2}}{v}

Step 2Minimize over the turn-off distance

Since $AD/v$ is a constant, differentiating $t(x)$ with respect to $x$ and setting the result to zero locates the minimum.

\begin{align*}\frac{dt}{dx} &= \frac{1}{v}\left(-1+\frac{\eta x}{\sqrt{x^2+l^2}}\right) \\ 0 &= -1+\frac{\eta x}{\sqrt{x^2+l^2}}\end{align*}

\frac{\eta x}{\sqrt{x^2+l^2}} = 1

Step 3Solve for the turn-off distance

Squaring clears the square root, and the remaining equation is linear in $x^2$ .

\begin{align*}\eta^2 x^2 &= x^2+l^2 \\ x^2(\eta^2-1) &= l^2 \\ x^2 &= \frac{l^2}{\eta^2-1}\end{align*}

CD = \dfrac{l}{\sqrt{\eta^2-1}}

(1)

Check

The result needs $\eta>1$ , i.e. the field really is slower than the highway, or the square root is imaginary and no interior minimum exists, consistent with $t(x)$ then being minimized by not using the highway at all. As $\eta\to1^+$ (field and highway equally fast), $CD\to\infty$ : the car should drive straight at $B$ from wherever it starts, never favoring the highway, which matches the formula blowing up.

Remark

The stationary condition $\eta x/\sqrt{x^2+l^2}=1$ says $\sin\theta = 1/\eta$ for the angle $\theta$ the field leg $CB$ makes with the perpendicular $DB$ , the same law a light ray obeys at an interface between a fast and a slow medium (Snell’s law, with $\eta$ playing the role of the relative refractive index): least-time paths bend toward the normal on entering the slower medium, whether the traveler is light or a car.