Example similar to the take home final

By request, I present an example like the take home final.  This is the graph of r =2cos(2theta).  I calculated the values at all 24 points that have lines radiating from the origin and marked them, then I played connect the dots, smoothing the curve instead of drawing straight lines.  Here are the values at all 24 points.

r = 2cos(2*0) = 2

r = 2cos(2*pi/12) = sqrt(3)

r = 2cos(2*pi/6) = 1

r = 2cos(2*pi/4) = 0

r = 2cos(2*pi/3) =-1

r = 2cos(2*5pi/12) =-sqrt(3)

r = 2cos(2*pi/2) =-2

r = 2cos(2*7pi/12) =-sqrt(3)

r = 2cos(2*2pi/3) =-1

r = 2cos(2*3pi/4) = 0

r = 2cos(2*5pi/6) = 1

r = 2cos(2*11pi/12) = sqrt(3)

r = 2cos(pi) = 2

r = 2cos(2*13pi/12) = sqrt(3)

r = 2cos(2*7pi/6) = 1

r = 2cos(2*5pi/4) = 0

r = 2cos(2*4pi/3) =-1

r = 2cos(2*17pi/12) =-sqrt(3)

r = 2cos(2*3pi/2) =-2

r = 2cos(2*19pi/12) =-sqrt(3)

r = 2cos(2*5pi/3) =-1

r = 2cos(2*7pi/4) = 0

r = 2cos(2*11pi/6) = 1

r = 2cos(2*23pi/12) = sqrt(3)

The result is called a four leaf rose.With the problem from the take home, because we have the formula r = 2 + cos(3*theta), the radius will never get to be less than zero, so it will not cross the origin mulitple times like this one does.

The take home section of the final

Here is the take-home section of the final as a picture file.  You should be able to download it, but if not, send me an e-mail at mhubbard(at)peralta(dot)com, substituting the correct symbols for the words in parentheses.

I can also send you a list of the topics for the final if you did not get it in class today.  You are allowed to have notes on two 3" x 5" cards.

See you on Wednesday, Dec. 14 at 8:00 am, one hour earlier than usual.