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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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,
Differentiate both sides with respect to t,
`(dx)/(dt)` is the rate of change of position, i.e. velocity of the man
Putting the value,
`(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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