1. ## Polar Boundaries.

A window is in the shape of a 2′ by 3′ rectangle surmounted by a
semicircle. Describe the boundary of the window as r = f (θ) in a polar
coordinate system whose location and orientation you specify. The f (θ)
likely will need to be deﬁned “piecewise".

2. This is a cool problem. I'm thinking of the halfway point of the top side of the rectangle to be the centre of your coordinate system. Theta ranges from 0 to 2pi as usual, in counter-clockwise direction.

You'll have four pieces. The first, between 0 and pi, is easy -- almost trivial. The next three are more difficult; you will have to define straight lines. Just use the equations for converting between cartesian and polar coordinates.

To determine the borders of the piecewise bits, draw a picture and look at the angles -- draw lines between the corners of the rectangle and the centre point, and look at the angles that they make.

3. I think I know what you mean. I have the top part, which would just be defined in terms of sines and cosines like the top of unit circle.

I'm a little unsure about the line part. So I have a line from (0,0) which is the top of the rectangle. I am supposing I draw a line from there to the left side of the rectangle that has length 3. So I have a triangle with one leg as 1 (since the other side of the rectangle is 2) and then I have a hypotenuse and some part of the side of the rectangle that is length 3. I am not sure how to formulate this precisely.

4. The centre point of the coordinate axes should be such that f(theta) = 1.5

for the rest, for example the line x = -1.5 is the line -1.5 = r cos(theta)

the line y = 2 is -2 = r sin(theta)

the line x = 1.5 is 1.5 = r cos(theta)

these are four different parts. first part goes 0 < theta <= pi, the rest you should be able to work out by drawing a picture.

if (0,0) is the bottom left corner of the window in cartesian coordinates, then the point (1.5, 2) is the point (0,0) in the new coordinate system.

5. Thanks, but I still don't understand why f(theta) should be 1.5 Also tuition would tell me the point (0,0) should be (1, 2) so that it is halfway between since we have a 2 by 3 and a semicircle..

6. the easiest way to do it by far is to construct the new axes so f(theta) = 1.5

the semicircle has diameter 3, so radius 1.5...

7. Oh that's the confusion...I should have posted the picture to start with...the diameter is 2.

http://www.math.rutgers.edu/~greenfi...fstuff/w6C.pdf