1. a) Plot the graph of the curve r = cos(theda) in the x, y plane. Remember that r can be negative.
b) Find an equation for the x, y coordinates
c) Describe the curve geometrically.
This is the solution I obtained:
a) See attachment labeled "Part A". Not sure if that's how the graph is supposed to look. It's not perfect, but just a rough sketch.
b). For the equation, I did something like this. Since x=rcos(theda) and x= ysin(theda), you can set the equation to be...
x^2 + y^2 = r^2, if you let x=r^2, you can set it to
x^2 + y^2 = x..then
x^2 + y^2 -x =0, which is the equation
c) I took out the equation x^2 + y^2 -x =0 and did the square method:
(x-1/2)^2 -(1/2)^2 +y^2 = 0
and (x-1/2)^2 + y^2 = (1/2)
basically when you graph 0 and 1/2, you get almost the same exact graph in part A, showing that using the polar coordinate and square method gives the same technique
Can someone verify this and show that it's correct? I'm not sure if I did the entire thing right.
your graph is off a bit ...
... it's "theta" , not "theda"