Question

A 5.1-ft-tall person walks away from a 12-ft lamppost at a constant rate of 3.9 ft/sec. What is the rate that the tip of the person's shadow moves away from the lamppost when the person is 9 ft away from the lampost?

Answer 1

Answer:

6.78ft/sec

Step-by-step explanation:

From the question, dx/dt= 3.9 ft/sec

We are looking for Dy/dt

From geometry,vof this case the relationship between x and y is needed here, there is two similar triangle that exhibited by the man and the lamb

12/y= 5.1/(y-x)

Then ,cross multiply, we have

12(y-x)=5.1y

12y-12x=5.1y

12y-5.1y=12x

6.9y=12x

y=( 12/6.9)x

Differentiating implicitly the bother sides with respect to t, we have

Dy/dt= ( 12/6.9)dx/dt

But dx/dt= 3.9 ft/sec

Then Dy/dt= ( 12/6.9)× 3.9

Dy/dt=6.78ft/sec

Hence, the rate that the tip of the person's shadow moves away from the lamppost when the person is 9 ft away from the lampost is 6.78ft/sec

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