Question

Need some help please.
x = Pi
Convert the Cartesian equation to polar equation and sketch the graph.

I know that x = r cos (theta) and y = r sin(theta)

Answer 1

\[r \cos(\theta)+0 \times r \sin(\theta) = \pi\]
what \[\theta for \cos = \pi?\]
\[\arccos(\cos(\theta) = \arccos(\pi) \]\[\cos^-1(\cos(\theta) = \cos^\]
x = -1
y = 0

Is this correct?

Answer 2

x=pi is a vertical line thru x=pi, right?

Answer 3

yeah

Answer 4

how do i turn to polar equation?

Answer 5

hmm, i vaguely recall a rule for it. let me see if I can relocate it

Answer 6

okay. thanks!

Answer 7

http://tutorial.math.lamar.edu/Classes/CalcII/PolarCoordinates.aspx

towards the bottom:

r cos(t) = pi ; i beleive

Answer 8

makes sense since x = r cos(t)

Answer 9

yes. that does make sense. so the y = 0?

Answer 10

the cartesian form of x=4 has no y part to worry about does it?

Answer 11

well, x = pi ... has no y parts to consider

Answer 12

so the graph is just the picture of a vertical line at \[x=\pi\]?

Answer 13

yes

Answer 14

http://www.wolframalpha.com/input/?i=x%3Dpi

Answer 15

So what is the polar equation? \[x=r \cos(\pi)\]
?

Answer 16

there are no xs in polar equations ...

Answer 17

Can I ask you another one?
\[4x ^{2}+9y^{2}=16\]

Same thing.

Answer 18

x = pi is a cartesian equation

rcos(t) = pi is its equivalent polar equation

Answer 19

hmm, an ellipse in polar form eh

Answer 20

thinking take sqrt of everything?

Answer 21

need to turn to polar again...

Answer 22

there is something more to do with eccentricity and stuff for those

http://mysite.du.edu/~jcalvert/math/ellipse.htm

has a nice little rundown of it

Answer 23

okay. thank you sir.

Answer 24

wish i was more proficient with these, but good luck :)

Answer 25

prob gonna repost it.
thanks!