A man, 2m tall, walks at the rate of m/s towards a street light which is m above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he is m from the base of the light?
A man, 2m tall, walks at the rate of m/s towards a street light which is m above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he is m from the base of the light?

Let AB is the stature of streetlamp post and CD is the tallness of the man with the end goal that

\[AB\text{ }=\text{ }5\left( 1/3 \right)\text{ }=\text{ }16/3\text{ }m\text{ }and\text{ }CD\text{ }=\text{ }2\text{ }m\]

Let \[BC\text{ }=\text{ }x\] length (the distance of the man from the light post)

What’s more, \[CE\text{ }=\text{ }y\] is the length of the shadow of the man at any moment.

It’s seen from the figure that,

\[\text{ }ABE\text{ }\sim\text{ }\text{ }DCE\]  [by AAA closeness criterion]

Presently, taking proportion of their relating sides, we have

Separating the two sides w.r.t, t, we have

NCERT Exemplar Solutions Class 12 Mathematics Chapter 6 - 13

In this way, the length of shadow is diminishing at the pace of 1 m/s.

NCERT Exemplar Solutions Class 12 Mathematics Chapter 6 - 14

Presently, let \[u\text{ }=\text{ }x\text{ }+\text{ }y\]

(where, u = distance of the tip of shadow from the light post)

 

On separating the two sides w.r.t. t, we get

NCERT Exemplar Solutions Class 12 Mathematics Chapter 6 - 15

NCERT Exemplar Solutions Class 12 Mathematics Chapter 6 - 16

Hence, the tip of the shadow is moving at the pace of m/s towards the light post and the length of shadow diminishing at the pace of \[1\text{ }m/s.\]