SOLUTION: "A man is walking away from a 5m tall streetlight at a rate of 2m/s. If the man is 2m tall, how fast is his shadow lengthening when he is 2m away from the streetlight? When he is 6

Question 982610: "A man is walking away from a 5m tall streetlight at a rate of 2m/s. If the man is 2m tall, how fast is his shadow lengthening when he is 2m away from the streetlight? When he is 6m away?"
Okay so I understand that I would want to make a graph and get tangents to get rate of change on his shadow for 2m and 6m (the section in the book I have is all stuff like this right now) I just do not understand how to draw a graph from this information... Any help please?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
If you draw a picture, you'll have 2 similar triangles. The larger triangle has a vertical leg of 5. The horizontal leg of this larger triangle, let's call it y.

The smaller triangle has a vertical component of 2. Let x be the distance from the base of the lamp to the base of the person. The remaining bit of length is y-x meters (add x and y-x to get y).

The two similar triangles form this proportion

cross multiply and isolate y to get

Now take the derivative of both sides with respect to t to get

The "dx/dt" portion is the speed of which the man is walking. So, dx/dt = 2. Notice how x itself (ignore dx/dt) is not in the dy/dt equation. What does this mean? It means that the speed of the shadow does not depend on how far the man is from the pole.

Now use dx/dt to find dy/dt

Therefore the shadow's rate of growth is 10/3 meters per second.
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