A person of height h = 1.7 m walks away from a lamp H = 3

A person of height $h = 1.7\ \text{m}$ walks away from a lamp $H = 3\ \text{m}$ above the ground at $6.34\ \text{km/h}$. How fast does the tip of the shadow of their head move?

Step 1: Set up the geometry

Put the origin on the ground directly below the lamp. At time $t$ the lamp is at $\bigl(0,H\bigr)$, the top of the head at $\bigl(x,h\bigr)$, and the shadow tip at $\bigl(L,0\bigr)$, with $L>x>0$.

Step 2: The correct pair of similar triangles

Two right triangles share the vertex at the shadow tip: $\text{large: legs } H \text{ and } L; \qquad \text{small: legs } h \text{ and } L-x$.

The vertical legs are the lamp and the person, the horizontal legs are their distances from the tip. Hence

\[ \frac{H}{L} = \frac{h}{L-x}. \tag{1} \]

It is worth pairing these carefully: writing $\dfrac{H}{L} = \dfrac{H-h}{L-x}$ instead would match the lamp's height against the difference of heights, which corresponds to no triangle in the figure.

Step 3: Solve for $L$

\[ H\bigl(L-x\bigr) = hL \;\Longrightarrow\; L\bigl(H-h\bigr) = Hx \;\Longrightarrow\; L = \frac{H}{H-h}\,x. \tag{2} \]

Step 4: Differentiate

The factor is constant, so the two speeds are simply proportional: $\frac{dL}{dt} = \frac{H}{H-h}\cdot\frac{dx}{dt}$.

Step 5: Substitute the numbers

\[ \frac{H}{H-h} = \frac{3}{3-1.7} = \frac{3}{1.3} = \frac{30}{13} \approx 2.3077, \]

\[ \frac{dL}{dt} = \frac{30}{13}\times6.34 \approx 14.63\ \frac{\text{km}}{\text{h}}. \]

Step 6: Verification

\[\text{Take } x = 10:\ L = \frac{3\cdot10}{1.3} \approx 23.077. \text{ Slope lamp}\to\text{head: } \frac{1.7-3}{10} = -0.13\]

\[\text{Slope head}\to\text{tip: } \frac{0-1.7}{23.077-10} = -0.13 \;\;\checkmark \text{ — the three points are collinear}\]

\[\text{Sanity: } \frac{dL}{dt}>\frac{dx}{dt} \text{ always, and } \to\infty \text{ as } h\to H \;\;\checkmark\]