Coordinate System Setup:

Let the lamp $S$ be at a fixed position relative to the sidewalk. Based on the problem description:

• Let the lamp post be at $x = -l$. The lamp $S$ is at coordinates $(-l, 0, H)$.
• The boy walks along the edge of the sidewalk, which we align with the $y$-axis ($x=0, z=0$).
• The boy’s position at time $t$ is $\vec{r}_b = y \hat{j} = v_0 t \hat{j}$ (assuming $y=0$ at $t=0$).
• The boy’s head $B$ is at $(0, y, h)$.

(a) Velocity vector of the shadow of his head

The position of the shadow’s head, $P$, is found using similar triangles or the vector line equation. The light travels in a straight line from $S(-l, 0, H)$ through the boy’s head $B(0, y, h)$ to the ground $z=0$.

Let $P$ have coordinates $(x_s, y_s, 0)$. By similar triangles in the vertical planes:

$$ \frac{H}{H-h} = \frac{\text{Distance from } S \text{ to } P}{\text{Distance from } S \text{ to } B} $$

Using the section formula for coordinates:

$$ y_s = \frac{H}{H-h} y $$ $$ x_s = -l + \frac{H}{H-h} (0 – (-l)) = -l + \frac{H l}{H-h} = l \left( \frac{H}{H-h} – 1 \right) = \frac{l h}{H-h} $$

Notice that $x_s$ is constant. The shadow moves along a line parallel to the boy’s path. To find the velocity vector $\vec{v}_s$, we differentiate the position with respect to time:

$$ \vec{v}_s = \frac{d}{dt} (x_s \hat{i} + y_s \hat{j}) $$ $$ \vec{v}_s = 0 \hat{i} + \frac{H}{H-h} \frac{dy}{dt} \hat{j} $$

Since $\frac{dy}{dt} = v_0$, the velocity vector is:

(b) Time rate of change of length of shadow

The length of the shadow $L$ is the distance from the boy’s feet $(0, y, 0)$ to the shadow of his head $(x_s, y_s, 0)$.

$$ \vec{L}_{vec} = (x_s – 0) \hat{i} + (y_s – y) \hat{j} $$ $$ \Delta x = \frac{lh}{H-h}, \quad \Delta y = \frac{Hy}{H-h} – y = y \left( \frac{h}{H-h} \right) $$

The magnitude is:

$$ L = \sqrt{ (\Delta x)^2 + (\Delta y)^2 } = \frac{h}{H-h} \sqrt{ l^2 + y^2 } $$

We need the rate of change $\frac{dL}{dt}$. Using the chain rule:

$$ \frac{dL}{dt} = \frac{h}{H-h} \cdot \frac{d}{dt} \left( \sqrt{l^2 + y^2} \right) $$ $$ \frac{dL}{dt} = \frac{h}{H-h} \cdot \frac{1}{2\sqrt{l^2 + y^2}} (2y) \frac{dy}{dt} $$

Substituting $\frac{dy}{dt} = v_0$: