Question

The equations x=1+2cos(theta)
y=-2+3sin(theta)
parameterize what curve in the xy-plane? Determine its Cartesian equation.

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

Ok well do you have any ideas about how to think about this?

What would the curve
x = cos(theta)
y = sin(theta) parameterize?

Answer 3

I'm really not sure, I'm getting tripped up by "parameterize." I know cos curve and sin curve are the same but just start on different intervals (cos starts at y=1 sin start at x=0 ?)

Answer 4

Ah. Ok

Answer 5

Well lets imagine that we have something simple like

y = sin(theta)
x = cos(theta)

Then when we plug in theta = 0 we get x = 1, y = 0 or the point (1,0)
If we plug in \(\pi \over 6\) we get x = \(\sqrt{3} \over 2\), y = \(1 \over 2\)
If we plug in \(\pi \over 4\) we get x = \(\sqrt{2} \over 2\), y = \(\sqrt{2} \over 2\)
If we plug in \(\pi \over 3\) we get x = \(1 \over 2\), y = \(\sqrt{3} \over 2\)
If we plug in \(\pi \over 2\) we get x = 0, y = 1
etc.

Answer 6

Do these numbers look familiar?

Answer 7

yes

Answer 8

so we would take the values we plugged in to eqaute the value for the curve? (sorry I'm very lost thus far in Calc III) The numbers look familiar from using them in calc II with trig sub

Answer 9

x = cos(theta), y = sin(theta)
is the standard parameterization of a circle. As we plug in various values for theta we will get corrisponding x and y points that describe a circle around the origin with radius 1.

Answer 10

If you imagine a radius fixed at the origin making an angle theta from the x axis the x and y value of the outer tip will be the x and y values we get from our parameterization.

Answer 11

oh, I think I'm understanding it a little better

Answer 12

So now imagine that we had instead

x = 2 + cos(theta)
y = 3 + sin(theta)

What would this change?

Answer 13

the different values would just be shifted by the coefficient right?

Answer 14

Would we still be describing a circle?

Answer 15

Would it still have radius 1?

Answer 16

Would it still be centered at the origin?

Answer 17

no it would be greater

Answer 18

ohhh and the origin would be shifted as well??

Answer 19

Well try plugging in a few common points \(\pi, 0, \pi/2, \pi/4, etc\). What do you get for some of the x/y values?

Answer 20

1 + 2cos(0) = 3
-2 + 3sin(0) = -2

1+2cos(pi/2) =1
-2 + 3 sin(pi/2) = -1

Answer 21

I meant plug into my new version of the function.

Answer 22

Just to see how changing different pieces can change the function

Answer 23

oh sorry, hold on

Answer 24

they seem to move the same only deviate by the number being added?

Answer 25

cos (pi/4) and sin (pi/4) are equivalent?

Answer 26

right

Answer 27

So what does that tell you about adding a constant value to the original function? What did it do?

Answer 28

so only the radius is changing? maybe :)

Answer 29

Did the radius change?
x = 2 + cos(theta)
y = 3 + sin(theta)

theta x,y
At 0 3,3
At \(\pi \) 1,3
At \( \pi/2\) 2,4
At \(3\pi/2\) 2,2

Answer 30

no.. is it the length of the curve thats changing while the radius remains constant? thats just a wild guess :(

Answer 31

No.. The radius is 1. All that has happened is we've changed where we're drawing the circle. Now instead of being centered around 0,0 we are centered around 2,3. It's still just an ordinary circle.

Answer 32

oh!!! sorry

Answer 33

Do you have a graphing calculator?

Answer 34

yes.

Answer 35

http://www.wolframalpha.com/input/?i=x+%3D+2%2Bcos%28theta%29%2C+y+%3D+3%2Bsin%28theta%29

Answer 36

so its a constant circle as long as cos(theta) and sin(theta) are unchanged and anything added it too that shifts the origin?

Answer 37

right.

Answer 38

Now lets watch what happens when we put a coefficient on the cos and sin

Answer 39

genius!

Answer 40

x = 2 + 3cos(theta)
y = 3 + 3sin(theta)

Answer 41

makes the radius bigger

Answer 42

http://www.wolframalpha.com/input/?i=x+%3D+2%2B3cos%28theta%29%2C+y+%3D+3%2B3sin%28theta%29

Answer 43

Right.

Answer 44

And finally, if we make the sin and cos have different coefficients. we will end up having one grow faster than the other which will skew our circle making it an ellipse.

Answer 45

x=1+2cos(theta)
y=-2+3sin(theta)

Answer 46

ahh, its starting to make sense!

Answer 47

The length of the major axis being 6 (in the y direction) and the minor axis being 4 (in the x direction) which are twice the respective radial coefficients.

Answer 48

and still centered at 1,-2

Answer 49

okay, so when it asks to "parameterize" it would be x=1 and y = -2 ?

Answer 50

No. It gave the parameterization. It's asking for the Cartesian equation.

Answer 51

so would Cartesian equation be the eq of a circle (x - a)^2 + (y - b)^2 = r^2

Answer 52

using x=1 y = -2

Answer 53

No, it'd be the equation of an ellipse..

\[{(x-h)^2 \over a^2} + {(y-k)^2 \over b^2} = 1\]

Where h,k is the center, and a and b are half the lengths of the major and minor axis of the ellipse.

In your case h = 1, k = -2, a = 2, b = 3

Answer 54

Yes, because you just said it was an elipse too, blahh. I wish this stuff would click easier

Answer 55

you are a genius though! I appreciate the help

Answer 56

well this is stuff you should have covered in pre-calc.

Answer 57

so just got to refresh =)

Answer 58

i never took precalc ! I went straight into calc I. I should def find a resource and review it.

Answer 59

thanks again!

Answer 60

One sec

Answer 61

k

Answer 62

http://www.khanacademy.org/video/polar-coordinates-1?playlist=Precalculus
Start here, and watch (click next video) up to and including
http://www.khanacademy.org/video/parametric-equations-4?playlist=Precalculus

Answer 63

There are about 7 videos that are pretty short but should be a good intro.

Answer 64

excellent! I've watched some of their videos in the past, I will watch these right now

Answer 65

you went above beyond, many thanks!

Answer 66

oh, and click the 'good answer' button so it knows you're done with this question =)

Answer 67

will do!