A person 1.85 m tall walks towards a lamppost on level ground at a rate of 0.7 m/sec

Question

A person  tall walks towards a lamppost on level ground at a rate of  . The lamp on the post is  high. At which rate is the tip of the person's shadow moving toward the lamppost when the person is  from the post? Please enter your answer in decimal format with three decimal places.

Answer 1

A

| C

| | . E

B D

I can't insert a proper diagram, unfortunately. I tried my best above and you'll have to finish the sketch yourself :).

Make a taller segment AB for the lamppost. Make a shorter segment CD for the person. Construct a line for the ground, connecting base of the lamppost B, person's feet D, and extending it past the point D. Now make a line from the point A, representing the lamp on the lamppost, through the point C, representing the head of the person, all the way to the intersection with the ground line. The point of intersection E is the tip of the person's shadow.

You should get a large triangle ABE and a cut off tip of it, the smaller triangle CDE. AB =5, CD=1.85. BE = x is the distance between the tip of the person's shadow and the lamppost. We have to find the rate of change of x, or dx/dt. Let's label BD=y. It's the distance between the person and the post, and we know that dy/dt= - 0.7 ( it's negative, because the distance is decreasing).

These triangles are similar, so

AB/CD = AE/DE Notice that DE=BE - BD

5/1.85=x/(x-y)

5(x-y)=1.85x

3.15x=5y

Time to take derivatives, keeping in mind that both x and y change with time.

3.15 dx/dt = 5 dy/dt, dy/dt= - 0.7

dx/dt = -5*0.7/3.15 = - 1.1111

Hope it helps.