# Homework Help: Related Rates

1. Apr 1, 2005

### erik05

There are two questions that I'm stuck on and I was wondering if someone could help me. I'm more worried about the process of finding the right answer than the answer itself. Thanks in advance.

1) A man who is 2 m tall is walking straight away from a lamppost at a rate of 8 km/h. The lamppost is 7 m high. How fast is the end of his shadow moving when he is 7 m from the foot of the lamppost? Ans: 11.2 km/h

I know that you have to use the property of similiar triangles but I'm not too sure how to set it up. Any suggestions?

2) Two ships start out from harbour at the same time; one travels northeast and the other southeast. If the first has a speed of 16 kn and the second 12 kn, how fast are they separating after 1 h? Ans: 20 kn

Would it help to know what kn stood for? Because I have no idea. Anyways, I started out with a coordinate grid with the harbour at (0,0) and the first ship at (x,16) and the second ship at (x,-12). Don't really know if this is the right to approach this but I had to start with something. Thanks, and have a good night.

2. Apr 1, 2005

### whozum

Draw a triangle inscribed in a triangle, sharing the two non vertical sides
The vertical end is the lamppost, the vertical centerline is the man. You want ds/dt (speed of shadow), given dx/dt = 8

Well the main step here is drawing a relationship between how far the man is and where his shadow is. Using trig properties, take a couple stabs at possible relationships between his distance from the post and the distance of the end of his shadow to the post.

3. Apr 2, 2005

### erik05

Okay, I got the first question..thanks whozum. Could anyone help me out on the second question?

4. Apr 2, 2005

### p53ud0 dr34m5

this is simple geometry. it is a right triangle, and the ships path are the two sides. the hypotenuse is the distance between them.

so, $(\frac{dx}{dt})^2+(\frac{dy}{dt})^2=(\frac{ds}{dt})^2$(dx/dt=12kn and dy/dt=16kn).

so, $12^2+16^2=(\frac{ds}{dt})^2~~144+256=(\frac{ds}{dt})^2~~400=(\frac{ds}{dt})^2~~\frac{ds}{dt}=20$
so, the ships are moving away from each other at 20kn.

Last edited: Apr 2, 2005