A man of height 1.5m walks away from the lamp post of height 4.5 m at the rate of 20 cm/sec. How fast is the shadow lengthening when the man is 42 cm from the post?
Help in application of derivatives.
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Given the lamppost is 4.5m tall, the man is 1.5m tall and the distance from the lamppost is 42cm. The man is walking away at 20cm/sec.
Draw a right triangle with leg 4.5. From the other leg draw a perpendicular of length 1.5 meeting the hypotenuse of the right triangle. The 4.5 unit length represents the pole and the 1.5 unit length the man.
The triangles formed are similar by AA~. Let s represent the distance from the man's shoes to the tip of the shadow and x the distance of the man from the pole. Then `4.5/1.5=(x+s)/s` . Then 3s=x+s or x=2s.
Differentiating both sides with respect to time we get `(dx)/(dt)=2(ds)/(dt)`
Here `(dx)/(dt)=20` and we want to find `(ds)/(dt)` .
`(ds)/(dt)=20/2=10` so the shadow is lengthening at 10cm/sec.
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