Could Someone please help me with this Topic.The Lecture literally made no sense to me.The professor was like compare the values from R theta graph and draw the X-y coordinate graph from it.My syllabus Cover's Limacons, Spirals and rose petals ?
#Graphing Polar Coordinate Equations
gloomy star
sly grailBOT
gloomy star
Here are a few questions from Thomas Calc
I would love to learn how to draw other kinds of graph too
dim mirage
Do you understand how you would graph something like r = theta on an r theta graph?
gloomy star
yes
it would be a striaght line inclined at 45 degrees to x axis and passing through origin
dim mirage
do you know what r and theta would correspond to in the x-y plane?
gloomy star
r would be the distance between the origin and the point(x,y) and theta would be the tan inverse of the slope of the line joining (x,y) and origin
i think i defined theta in the wrong way
but i get the idea
dim mirage
informally, theta is just the angle of the line attached from the origin to the (x, y) point to the positive x axis
gloomy star
yes
dim mirage
so for r = theta in the x-y plane, you can imagine starting at theta = 0, and then slowly increasing theta. What you'll essentially get is a vector that's attached to the origin and constantly rotating, and the length of the vector is constantly increasing, with a length exactly equal to the angle measured in radians
so basically, all of the curves you sketch are just vectors rotating around the origin, and the relationship between r and theta tells you how the length of the vector changes as it rotates
gloomy star
i understand why we are having a spiral for r=theta graph in the x-y plane
but i am still not able to graph it
dim mirage
Are you just being asked to roughly draw it?
gloomy star
yes
well
yes
we just have to get the shape right
like if its a spiral or a circle or a limacon etc
or a rose petal ?
gloomy star
yes
i just happened to remember that a+bcostheta would give us a limacon
but
i don't know how to graph it
even roughly
dim mirage
I'd start at theta = 0
and see what the radius of the vector would be
so for 1 + cos(theta)
when theta = 0 it would start on the x axis with a length of 1 + cos(0) = 2
gloomy star
I am still stuck at r=theta graph
dim mirage
I think as long as it looks roughly like a spiral and you mark the values of the intersections with the x and y axes you should be fine
gloomy star
sorry,new to this topic and graphing on my own without using desmos
so im bad at it
timber zodiacBOT
dim mirage
yeah looks good
dim mirage
I'm not sure if they'd want you to mark points of intersection
might be a good habit to get in to
gloomy star
Could i just say (0,pi/2) or do have to expand pi to a certain number of digits ?
dim mirage
0, pi/2 is totally fine
dim mirage
ye
dim mirage
sure, the r = 1 + cos(theta) one?
timber zodiacBOT
dim mirage
should be (0, 1)
gloomy star
AHH
im getting confused btw r-theta and x-y plane
im mixing them both
lol
got it
dim mirage
nice
gloomy star
not able to figure out how to draw the graph in 2nd quadrant
dim mirage
so basically, as theta moves in to the second quadrant, cos(theta) will decrease and subtract away from the magnitude
gloomy star
so this is good?
dim mirage
for the first 2 quadrants ye
dim mirage
nah what you drew is good enough
imo
I'm guessing that there are different classes of spirals you're being asked to draw
gloomy star
but in the 1st and 4th quadrant ,the graph goes above 1 and -1 respectively
in my graph,it wasnt ?
dim mirage
ah wait yeah you're right
gloomy star
ouch
dim mirage
since r decreases from 2 to 1, and only hits 1 when it hits the y axis
gloomy star
Guess i will get it right after practicing
dim mirage
something you could do is try sampling one or two points
in each quadrant
so you know roughly at which places your curve should hit
dim mirage
sure
gloomy star
at theta = 0 , we would be getting r = -1 which would be (-1,0) in the x-y plane right ?
dim mirage
yup
dim mirage
ngl I'm sorta confused
r can't ever be negative
since it's a magnitude
the theta tells you which quadrant the vector is in
gloomy star
oh
(-r,theta) = (r,theta+pi)
or theta - pi
like
first we find (r,theta)
and then
for
(-r,theta)
we rotate the line joining origin and (r,theta) by pi degrees clock wise or anticlockwise
brb
@dim mirage ?
dim mirage
ok so
I'm pretty sure that if you take r = -1 to mean the negative solution to the square root of x^2+y^2
you get two values of x when y = 0, x = 0 and x = -3
gloomy star
I'm pretty sure that if you take r = -1 to mean the negative solution to the squ...
yes ig
gloomy star
you get two values of x when y = 0, x = 0 and x = -3
x=-3 ?
dim mirage
yeah, and that's also what the graph shows
dim mirage
I basically expressed r = 1-2cos(theta) in terms of x and y
because I had no idea how to interpret r = -1
gloomy star
okay ?
dim mirage
so I figured out that the curve intersects the y axis when x = 0 and x = -3
gloomy star
how does a curve intersect the y axis when x= - 3 ?
because the equation of y axis is x=0
dim mirage
sorry
I meant to say x axis
so when y = 0, x = 0, x = -3
I just have no idea what r = -1 is supposed to mean when theta = 0
that's so strange
especially when r is supposed to be 0 at that angle
gloomy star
so when y = 0, x = 0, x = -3
i don't understand where are going with this
gloomy star
I just have no idea what r = -1 is supposed to mean when theta = 0
should i send a picture of the textbook explain (r,theta) and (-r,theta) thingy
?
dim mirage
ahh okay
dim mirage
brb getting food
gloomy star
alright
gloomy star
I have to sleep, Can we please do this tomorrow ?
dim mirage
sure
gloomy star
Thanks!
gloomy star
Status:I'm getting almost similar graphs for most of the questions
dim mirage
AIght I'm back
so for that question yesterday
r=1-2cos(theta)
desmos straight up graphed it wrong
and it confused me so much lol
https://www.desmos.com/calculator/ms3eghkkgz use this for reference to check your answers
when I put in the equation in terms of x and y it graphed it wrong
but this one does it properly
gloomy star
r=1-2cos(theta)
I was able to get it right today
It's just that the shapes are similiar
But
gloomy star
but in the 1st and 4th quadrant ,the graph goes above 1 and -1 respectively
Some times I'm making this mistake again and again
Not that specific one
Similar to those
I'm getting most of the graphs right though
gloomy star
@dim mirage
i was not able to understand this method
Could you help?
i was able to plot it by using the method you mentioned where we plot a few points and then draw the shape
dim mirage
<@146346462182113280>
As far as I understand this, r^2 isn't defined below 0. So you can sorta imagine that when you take the square root only values in the 1st and 3rd quadrant will be defined
since we're only looking at real values of r^2, r is basically 0 in the second and 4th quadrants
you can also think of it in terms of the theta + pi transformation
so if you graph the first quadrant, you'll find that for all (r, theta) in that quadrant, you'll have (-r, theta+pi), and theta+pi covers all angles in the third quadrant
gloomy star
so if you graph the first quadrant, you'll find that for all (r, theta) in that ...
I do understand what you said,but I am not able to relate it to the way they solved it
dim mirage
I do understand what you said,but I am not able to relate it to the way they sol...
try and imagine with just r^2 = sin(theta)
try drawing the r theta graph of that
gloomy star
It would be similar to the graph in B?
Dimpled limacons are kind of hard to figure out
dim mirage
sorta except sin of theta is only greater than or equal to zero in the first two quadrants
gloomy star
👍
timber zodiacBOT
Cnidarian
gloomy star
Nice
Need to graph this
But I am not able to imagine the vector properly thus I'm not able to graph
Get it?
Please help
Not able to do c,d,e
dim mirage
$r= e^{\theta/10}$
For this one the radius will simply grow exponentially as theta grows, so just mark the points where the spiral intersects with the axes
will look something like this
dim mirage
well, for the first intersection with the y axis
you'd just mark it as (0, e^(pi/20)), which is like (0, 1.17ish)
gloomy star
you'd just mark it as (0, e^(pi/20)), which is like (0, 1.17ish)
Okay,I will try
Thanks
dim mirage
the hardest part is the first quadrant
when theta is very close to zero r will be very close to infinity
so the spiral will sorta fly in from the left
gloomy star
Yes what I meant is
How to figure out where the graph meets x axis and y axis
Oh wait
I'm dumb
You just take theta equals pi/2 pi 3pi/2 so on
Right?
dim mirage
ye
gloomy star
Nice
dim mirage
0+kpi for x axis, 0+kpi/2 for y
gloomy star
Similar for the e part too right