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)

Asked on by pphok

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llltkl | College Teacher | (Level 3) Valedictorian

Posted on

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).

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