Inverse Trigonometric Functions / Rates of change with 2 points of references

1. Jun 11, 2004

kioria

Hi,

A questions regarding trigonometric (or inverse) functions. I just can't get it right.

Q1: A picture 2 m high is hung on the wall with its bottom 6 m above the observer's eye level. How far should the viewer stand for the picture to subtend the largest possible vertical angle with the viewer's eye?

Thanks

Also, I would like to know more about (in basic level, I am a first year) the rates of change. I cannot give a specific topic of what I am thinking, but I can give an example:

E1: A light house (P) is located at the middle of a long straight beach with its light-beam rotating at 3 rpm. And 2000 m directly below the location of P, a ship (S) travels towards the P at 20 km/s. How fast is the light moving at point S at this moment?

(Please don't try to do this, as I made this thing up).

If you know the topic, please reply and just leave the topic name (and by any chance you know a great site explaining it, then please tell me the site as well!!)

Thanks :)

2. Jun 11, 2004

HallsofIvy

Draw a picture. Draw the picture itself, the lines from the top and bottom of the picture to the viewer's eye, and the line from the from the viewer's eye perpendicular to the wall. You now have two right triangles. The base of both triangles has length x, the unknown distance from the viewer to the wall.
The height of the larger triangle is 8 m (that's one heck of a high wall!) so tan(θ)= 8/x (θ is the larger angle at the viewer's eye).
The height of the smaller triangle is 6 m so tan(φ)= 6/x (φ is the smaller angle at the viewer's eye.)

The angle subtended is θ-&phi= tan-1(8/x)- tan-1(6/x).

Can you find x to maximize that?

It's not clear what you want to know. Typically this is referred to as "related rates" and is based on the chain rule. Set up a "static" formula relating the two variables and differentiate with respect to time to get a formula relating the rates of change of the two variables.