Example 32: Man 2 m tall walks away from a 6 m lamp post at 5 km/h: how fast his shadow grows

Basics you need

Solution

1. Draw the situation and name the lengths

   Let AB be the lamp post with the lamp at B, and let MN be the man at time $t$⁠, standing at M with his head at N. The ray of light from B that just passes N meets the ground at S, so MS is his shadow. Let

   $$\text{AM} = l \ \text{metres}, \qquad \text{MS} = s \ \text{metres}.$$
   Both $l$ and $s$ change as he walks, so both are functions of time. We know how fast $l$ changes and want how fast $s$ changes, so we need an equation linking $l$ and $s$⁠.
   The shadow is the part of the ground that the man blocks from the lamp: from his feet M to the point S where the light grazing his head lands.

2. Similar triangles: AS in terms of $s$

   The post and the man are both vertical, so the angles at A and at M are right angles. Triangles MSN and ASB also share the angle at S. Hence $\triangle\text{MSN} \sim \triangle\text{ASB}$⁠, and

   $$\frac{\text{MS}}{\text{AS}} = \frac{\text{MN}}{\text{AB}} = \frac{2}{6} = \frac{1}{3}, \qquad\text{so}\qquad \text{AS} = 3s.$$
   Similarity is the only link between the two triangles, and it involves the two heights we know (2 m and 6 m). The ratio $\frac{1}{3}$ says the whole distance from the post to the shadow's tip is always three times the shadow.
   Similar triangles (AA): two equal angles make the triangles similar, so corresponding sides are proportional. MS corresponds to AS (bases) and MN to AB (heights).

3. A relation between $l$ and $s$ at every instant

   $$\text{AM} = \text{AS} - \text{MS} = 3s - s = 2s, \qquad\text{so}\qquad l = 2s.$$
   This equation does not depend on where the man is: Step 2 holds at every position. So it is an identity in time, $l(t) = 2s(t)$ for all $t$⁠, and identities in time may be differentiated.
   M lies between A and S, so AS = AM + MS.

4. Differentiate with respect to time

   $$\frac{dl}{dt} = 2\,\frac{ds}{dt}.$$
   Differentiating both sides of $l = 2s$ with respect to $t$ converts a relation between lengths into a relation between their rates. The constant 2 stays as a factor.
   Derivative of a constant multiple: $\frac{d}{dt}(2s) = 2\frac{ds}{dt}$⁠. Basics: related rates

5. Substitute the walking speed

   $\frac{dl}{dt} = 5$ km/h, so

   $$5 = 2\,\frac{ds}{dt}, \qquad \frac{ds}{dt} = \frac{5}{2}\ \text{km/h}.$$

   The rate is positive: the shadow lengthens at $\frac{5}{2}$ km/h.

   The heights were in metres and the speed in km/h, but this causes no trouble: $l = 2s$ is the same in any unit of length, so if $l$ grows at 5 km/h then $s$ grows at half that, in km/h.
   A positive rate of change means the quantity is increasing. Basics: rate of change as a derivative

6. What the answer does not depend on

   The rate $\frac{5}{2}$ km/h is the same at every instant: it does not matter how far from the post the man is. The tip of the shadow, at distance $\text{AS} = 3s$ from the post, moves at $\frac{d}{dt}(3s) = 3\cdot\frac{5}{2} = \frac{15}{2}$ km/h.

   The relation $l = 2s$ is linear, so its derivative relation has constant coefficients and a constant walking speed gives a constant shadow speed. This is why the question can ask for "the rate" without naming an instant.
   $\text{AS} = l + s$⁠, so also $\frac{d}{dt}\text{AS} = 5 + \frac{5}{2} = \frac{15}{2}$ km/h, the same value.

Every fact used, and where it comes from