A man $2$ m tall walks at $5$ km/h away from a lamp post $6$ m high. The length of his shadow increases at the rate of

1. By similar triangles $\dfrac{MS}{AS}=\dfrac{MN}{AB}=\dfrac{2}{6}$, giving $AS=3s$ where $s=$ shadow. 2. Distance walked $l=AS-s=3s-s=2s$. 3. Differentiate: $\dfrac{dl}{dt}=2\dfrac{ds}{dt}$. 4. Given $\dfrac{dl}{dt}=5$, so $\dfrac{ds}{dt}=\dfrac{5}{2}$ km/h. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.32_