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:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 26 m/min
Let x m be the distance of the boat from the lamp post and s m the length of the boat’s shadow at time t, so that \(\frac{dx}{dt} = 13\) m/min.
The lamp, the top of the boat and the tip of the shadow lie on one straight line. So the triangle formed by the lamp post (height 15, base x + s) and the triangle formed by the boat (height 10, base s) are similar triangles:
\(\displaystyle \frac{15}{x + s} = \frac{10}{s}\)
\(15s = 10x + 10s\)
\(5s = 10x\)
\(s = 2x\)
Differentiating with respect to t gives the rate of change of the shadow:
\(\displaystyle \frac{ds}{dt} = 2\,\frac{dx}{dt}\)
\(= 2 \times 13 = 26\) m/min
NCERT (2026-27 edition):
Application of Derivatives
