Q.A man, $2$ m tall, walks at the rate of $1\tfrac{2}{3}$ m/s towards a street light which is $5\tfrac{1}{3}$ m above the ground. At what rate is the tip of his shadow moving? At what rate is the length of his shadow changing when he is $3\tfrac{1}{3}$ m from the base of the light?

Idea: Use similar triangles to relate the shadow length $s$ and the man's distance $x$ from the light, then differentiate with respect to time.

Light height $H=5\tfrac{1}{3}=\frac{16}{3}$ m, man's height $h=2$ m. Let $x$ be his distance from the base and $s$ the shadow length. The light-tip and the man-tip triangles are similar:

Cross-multiplying, $\frac{16}{3}s=2(x+s)\Rightarrow\frac{10}{3}s=2x\Rightarrow s=\frac{3}{5}x.$

He walks toward the light at $1\tfrac{2}{3}=\frac{5}{3}$ m/s, so $\frac{dx}{dt}=-\frac{5}{3}$ m/s.

Shadow length: $\frac{ds}{dt}=\frac{3}{5}\frac{dx}{dt}=\frac{3}{5}\left(-\frac{5}{3}\right)=-1$ m/s. …

The intuition

A man walking toward a street light casts a shadow that shrinks. The lamp-top, the man's head, and the tips of the two shadows form two similar right triangles: a big one from the lamp to the shadow tip, and a small one from the man's head to the same tip. Similar triangles give a fixed relation between the shadow length and the man's distance, and differentiating in time turns "how fast he walks" into "how fast the shadow changes."

Set up with similar triangles

Let $x$ = the man's distance from the base of the light and $s$ = the length of his shadow. The lamp is $H=5\tfrac{1}{3}=\frac{16}{3}$ m high and the man is $h=2$ m tall. The large and small triangles share the shadow tip, so

Work the steps

1. Relate $s$ and $x$.

2. The given rate. He moves toward the light, so $x$ decreases:

3. Rate of change of the shadow length.

$\frac{ds}{dt}=\frac{3}{5}\frac{dx}{dt}=\frac{3}{5}\left(-\frac{5}{3}\right)=-1\ \text{m/s}.$ …

Method: Related Rates via Similar Triangles (Shadow / Lamp-Post Problems)

A person walking near a light source casts a shadow whose length and tip position are governed by similar triangles: the large triangle (light to shadow tip) and the small triangle (person's head to shadow tip) share the same tip and are similar, since both sit on the same horizontal ground and have a vertical side.

Steps

Step 1: Define the variables and sketch the two similar triangles.

Let $x$ = distance of the person from the base of the light, $s$ = length of the shadow, $H$ = height of the light, and $h$ = height of the person (adjusted for the height of the person's hand/eye if the problem specifies it).

Step 2: Set up the similar-triangles ratio.

Step 3: Cross-multiply and solve for $s$ in terms of $x$.

This produces a linear relation $s=mx$ for some constant $m$ built from $H$ and $h$ — the key simplification that makes both $\frac{ds}{dt}$ and $\frac{dx}{dt}$ constant multiples of each other.

Step 4: Differentiate with respect to time.

$\frac{ds}{dt}=m\,\frac{dx}{dt}.$ …

Common Mistakes

Mistake 1: Confusing "rate of the shadow's length" with "rate of the shadow tip's position"

Why it's wrong: the question asks for two distinct quantities — $\frac{ds}{dt}$ (how fast the shadow shortens) and $\frac{d}{dt}(x+s)$ (how fast the tip itself moves) — and giving only one, or reporting the same value for both, misses half the question. Correct approach: set up both $s=\frac{3}{5}x$ and $x+s=\frac{8}{5}x$ separately, then differentiate each.

Mistake 2: Setting up the similar-triangles ratio against the wrong side

Why it's wrong: writing $\frac{H}{x}=\frac{h}{s}$ (comparing the lamp height to $x$ alone) ignores that the larger triangle's base is the total distance from the lamp to the shadow tip, $x+s$, not just $x$. Correct approach: $\frac{H}{x+s}=\frac{h}{s}$, since both triangles share the same tip vertex. …