Given a man 180cm tall walks at a rate of 2 m/sec away; from a source of light that is 9 m above the ground

To find the rate at which the length of his shadow increases when he is 3m away from the pole

Let AB be the lamp post and let MN be the man of height 180cm or 1.8m.

Let AL = l meter and MS be the shadow of the man

Suppose the of the shadow MS = s (shown in the below figure)

Given as the man walks at the speed of 2m/sec

So, \( \dfrac{dl}{dt}\) = 2m/sec ...(i)

Therefore, the rate at which the length of the man's shadow increases will be \( \dfrac{ds}{dt}\)

Considering ΔASB,

Then considering ΔMSN,

Therefore, from equation (ii) and (iii)

1 = 3.5s

By applying derivative with respect to time on both sides

\( \dfrac{ds}{dt}=0.57m/sec\)

Thus, the rate at which the length of his shadow increases by 0.57 m/sec, and it is independent to the current distance of the man from the base of the light.

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