A person 1.88m tall is walking away from a lamp-post at 1.2 m/s. What rate is their shadow length changing

The light of the lamp-post is 5 meters above the ground.

Also, at what speed in m/s is the tip of the person's shadow moving across the ground?

I wont be able to draw the picture, but I can solve it for you. If you have trouble with the picture (and this problem in general) see this video:http://www.youtube.com/watch?v=kBfSwZAUo…


Let x= the distance between the lamp and the person, y= distance between the person and the tip of his shadow, L= distance from the lamp to the tip of the persons shadow.

(a) Find the rate of change of the shadow length

By similar triangles, we have

(x+y)/5 = y/(1.88)

5y=1.88x+1.88y

5y-1.88y=1.88x

3.22y= 1.88x

y= (1.88)x/(3.22)

y=0.6026x. (rounded)

Take the derivative with respect to time to get

dy/dt = 0.6026 dx/dt

Since dx/dt is the rate at which the man walks away, we have the rate of change of the shadow's length (dy/dt) is

dy/dt = 0.06026(1.2) m/s

dy/dt=0.072312 m/s

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(b) what speed in m/s is the tip of the person's shadow moving across the ground?

That is, we seek dL/dt.

By similar triangles, we have

L/5 = y/1.88

1.88L =5y

L= (5)y/(1.88)

L=2.66y (rounded)

Take the derivative of both sides with respect to time to get

dL/dt =2.66 dy/dt

Since dy/dt = 0.072312 m/s , we have

dL/dt = 2.66(0.072312 m/s ) =

dL/dt = 0.20 m/s (rounded)