Question

Math Analysis:
Sketch the polar function
16. r = 1-cos 0

Answer 1

It is a cardioid. Why?

Answer 2

idk what a cardioid is

Answer 3

http://www.wolframalpha.com/input/?i=PolarPlot [+1+-+Cos[\[Theta]]%2C+{\[Theta]%2C+0%2C+2+\[Pi]}]

Answer 4

Copy and past the whole link above in your browser

Answer 5

ohh okay :) but i just need to know how to plot it for my test

Answer 6

and show my work

Answer 7

Take several points for different value of \[ \theta\]

\[
\theta=0,\, r =0\\
\theta=\pm \frac \pi 3,\, r =\frac 1 2\\


\]
etc
and plot these points in polar coordinates.

Answer 8

this is what my teacher did
\[\theta | 0 | 45| 90| 135|180|225|270|315|360 \
\[r|0|0.3|1|1.7|2|1.7|1|0.3|0\]

Answer 9

he put that info on the chart i can't really draw it in sorry

Answer 10

See the cardiode with the points for
\[
\theta \in \left\{0,\frac{\pi }{6},\frac{\pi }{3},\frac{\pi
}{2},\frac{2 \pi }{3},\frac{5 \pi }{6},\pi
,\frac{7 \pi }{6},\frac{4 \pi }{3},\frac{3 \pi
}{2},\frac{5 \pi }{3},\frac{11 \pi }{6},2 \pi
\right\}
\]

Answer 11

Did you get it now?

Answer 12

how did you get those points thow

Answer 13

my teacher did it by degrees

Answer 14

would i just continually add 45 degrees up until 360 to plot my points?

Answer 15

If you want more points do it every 30 degrees, like I did for my graph above.
Where you able to see the graph.?

Answer 16

wait so how did you automatically know that the equation was a cardioid

Answer 17

my teacher did every 45 degrees

Answer 18

That is fine, you get fewer points.
\[
r = 1 \pm \sin(\theta)\\
r = 1 \pm \cos(\theta)\\
\]
are cardioid

Answer 19

cardioids look like those bean like graphs rite?

Answer 20

cardioids look like heart.
Have you heard about cardiac arrest?

Answer 21

yes i have lol

Answer 22

ohh i get it now :)

Answer 23

so then, since I know that the equation is a cardioid it would look like a heart but then how did my teacher know how to plot the points for it?

Answer 24

If you plot several points you get the shape of the heart.
Try ploting
\[
r = 1 + \sin(\theta)
\]

Answer 25

so i just randomly guess my points..? o-o

Answer 26

by sketching the graph

Answer 27

oh sorry :(
i will gives you in a minute ;33

Answer 28

You take some values of theta that you know the cosine of and you compute the corresponding r and you plot them.

Answer 29

so if i want my theta values to be every 45 degrees i would do what again..?

Answer 30

sorry
\[
(\theta, r) \in \left(
\begin{array}{cc}
0 & 0 \\
\frac{\pi }{4} & 1-\frac{1}{\sqrt{2}} \\
\frac{\pi }{2} & 1 \\
\frac{3 \pi }{4} & 1+\frac{1}{\sqrt{2}} \\
\pi & 2 \\
\frac{5 \pi }{4} & 1+\frac{1}{\sqrt{2}} \\
\frac{3 \pi }{2} & 1 \\
\frac{7 \pi }{4} & 1-\frac{1}{\sqrt{2}} \\
2 \pi & 0 \\
\end{array}
\right)
\]

Answer 31

what if i get an equation and I can't guess what it is right away?

Answer 32

and i can't guess what shape of graph it is

Answer 33

where did u get 1/square root of 2? o-o

Answer 34

Take more points to try to guess the sahpe.

Answer 35

\[
\cos(45) = \frac 1 {\sqrt 2} = \frac {\sqrt 2} 2
\]

Answer 36

why don't you just leave it at square root of 2/2? o:

Answer 37

would i get the cosine of the degree or the sine?

Answer 38

It is the same number.

Answer 39

cosine and sine are equal when the angle is 45

Answer 40

how about when its 135 degrees

Answer 41

cos( 135) =cos(45)

Answer 42

Try it with your calculator

Answer 43

Have you had trigonometry before?

Answer 44

yeah. im in it atm actually lol

Answer 45

ok thx i get it now! :P