A boat 10 m high floating at a uniform speed of 13 meters per minute(m/min) away from a lamp post 15 m high. Then the rate at which the length of shadow of the boat increases is:

A lamp post stands 15 m tall. A boat of height 10 m moves away from the lamp post at a speed of 13 m/min.

Let $x$ = distance of boat from lamp post

Let $s$ = length of the boat's shadow

Given: $\frac{dx}{dt} = 13$ m/min

The lamp casts light over the boat, creating two similar triangles:

• Large triangle: from lamp post top to shadow tip (height 15 m, base $x + s$)
• Small triangle: from boat top to shadow tip (height 10 m, base $s$)

Using similar triangles:

$\frac{15}{x + s} = \frac{10}{s}$

$15s = 10(x + s)$

$15s = 10x + 10s$

$5s = 10x$

$s = 2x$

Differentiating both sides with respect to time:

$\frac{ds}{dt} = 2 \cdot \frac{dx}{dt}$

$\frac{ds}{dt} = 2 \times 13$

$\frac{ds}{dt} = 26$ m/min

The shadow increases at a rate of 26 m/min.