Question

polar graph restriction

Answer 1

for the graph \[r=\frac{2}{3\sin(\theta)-\cos(\theta)} \] find the restricttions on \[\theta\] and explain what this means in terms of the graph

Answer 2

We can't have the denominator equal to zero, so 3sin(theta) cannot equal cos(theta). Therefore if we divide by cos theta on both sides, and then by 3, tan theta cannot equal 1/3. Make sense?

Answer 3

The tangent describes the ratio between the sine of an angle and the cosine of an angle. This cannot equal 1/3, so the sine of theta cannot equal the cosine of theta. Think about what the cosine and sine of an angle look like in terms of polar co-ordinates and you will see what this means for the graph.

Answer 4

so the graph is undefined for theta=arctan(1/3) and will have no points corresponding to that angle?

Answer 5

@dpaInc ??

Answer 6

still trying to understand the problem of why \(\large \theta=arctan(1/3) \) "will have no effect"....
what do you mean by this?

Answer 7

im not sure .. the graph is is a linear line with cartesian equation y=1/3x+2/3
so what does the restriction actually mean?

Answer 8

i do see that the theta restriction is put on the polar equation, but i don't know what that restriction did (or didn't do) on its cartesian form...

Answer 9

so what effect does the restriction have on the shape of the graph?

Answer 10

i'm thinking a hole in the linear graph.... but i'm not sure....

Answer 11

the graph of cartesian has the same shape as its polar form yeah

Answer 12

i'm not sure of the cartesian coordinates of the hole... if it is even a hole...

Answer 13

yeah...

Answer 14

so is it a hole?

Answer 15

that's what i'm thinking... but idk exactly the cartesian coordinates of that hole.... (discontinuity)

Answer 16

a removable discontinuity....

Answer 17

i have another one.. parabola this time

Answer 18

do i just say discontinuity at that angle

Answer 19

yeah.. that's what i'd say.... hole or discontinuity... same thing...

Answer 20

but if you say discontinuity, i think it would be more precise if you say removable discontinuity.

Answer 21

but should the linear graph be continuous?

Answer 22

shouldnt*

Answer 23

ok.... i'm being more confident now that you're asking these questions that it is a hole.....

Answer 24

the polar equation has a restriction that \(\large \theta \ne arctan(\frac{1}{3}) \), but overall, the graph is just a line

Answer 25

does it have something to do with the fact that the tan(t)=1/3 for the graphs gradient?

Answer 26

i think it has more to do with doing the conversion (or simplification)...
take this for example.....

if you were to graph \(\large y=\frac{(x+1)(x-1)}{(x+1)} \), you would get the same thing as graphing y = x-1 BUT in the first equation there is a hole at x=-1 while in the second graph there is no hole.

Answer 27

oh is that like the limit

Answer 28

and another queston..
4y=x^2
i converted into polar and got r=4sin(t)/(cos(t))^2 = 0 . however i tried solving on calculator and got that as well as r=0? where did the r=0 come from

Answer 29

no... i'm just talking about simplifying equations... like you did when you "simplified" or converted from polar form to cartesian form....
in polar form, you had a restriction, but the actual graph it looks like is a regular line with one exception... it has a hole....

Answer 30

hang on.... lemme try...

Answer 31

ok... i got the same as you but where did the "= 0" come from?

Answer 32

the conversion from cartesian to polar is \(\large r=\frac{4sin\theta}{cos^2 \theta} \)

Answer 33

http://www.wolframalpha.com/input/?i=solve+%284rsin%28theta%29%3Drcos^2%28theta%29+for+r
they got r=0 as well

Answer 34

and theres a restriction when cos^2theta=0 so when theta=pi/2,3pi/2 is there also discontrinuity at these agnles

Answer 35

well, r=0 when \(\large \theta =2n\pi \), when n is an integer only... but when n is not that, then \(\large r\ne0 \)

Answer 36

discontinuity at the polar graph but in the equation 4y=x^2, there is no discontinuity

Answer 37

alright thanks for your help

Answer 38

yw...

Answer 39

another queston if you dont mind..
limacons have the form r=b+acos(theta)
"If b>=2a, the limaçon is convex. If 2a>b>a, the limaçon is dimpled. If b=a, the limaçon degenerates to a cardioid. "
why does the limacon go from dimpled to convex as b becomes >=2a? why does the 'dimple' disappear from as it exceeds 2a?
can you explain in terms of the graph of r against theta?