# A man 6 ft tall walks at a rate of 2 ft/sec away from a lamppost that is 24 ft high. At what rate is the length of his shadow changing when he is 55 ft away from the lamppost? (Do not round your...

A man 6 ft tall walks at a rate of 2 ft/sec away from a lamppost that is 24 ft high. At what rate is the length of his shadow changing when he is 55 ft away from the lamppost? (Do not round your answer)

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### 1 Answer

Let x be the distance of the man from the lamp post and y be the distance of the tip of his shadow from the lamp post (please refer to the attached image).

Triangles `Delta SBL` and `DeltaHFL` are similar.

By the properties of similar triangles,

SB/HF=BL/FL

`rArr 24/6=y/(y-x)`

`rArr 4/1=y/(y-x)`

`rArr 4(y-x)=y`

`rArr 3y=4x`

Differentiate both sides with respect to t,

`3*(dy)/(dt)=4*(dx)/(dt)`

`(dx)/(dt)` is the rate of change of position, i.e. velocity of the man

=2 ft/sec.

Putting the value,

`3*(dy)/(dt)=4*2`

`(dy)/(dt)=8/3=2 2/3` ft/sec

The length of his shadow is changing at a rate of `2 2/3` ft/sec when he is 55 ft away from the lamppost.

(Distance of the man from the lamppost is not required for the calculation).