- anonymous

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

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- Owlfred

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!

- anonymous

Ok well do you have any ideas about how to think about this?
What would the curve
x = cos(theta)
y = sin(theta) parameterize?

- anonymous

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 ?)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

Ah. Ok

- anonymous

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.

- anonymous

Do these numbers look familiar?

- anonymous

yes

- anonymous

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

- anonymous

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.

- anonymous

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.

- anonymous

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

- anonymous

So now imagine that we had instead
x = 2 + cos(theta)
y = 3 + sin(theta)
What would this change?

- anonymous

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

- anonymous

Would we still be describing a circle?

- anonymous

Would it still have radius 1?

- anonymous

Would it still be centered at the origin?

- anonymous

no it would be greater

- anonymous

ohhh and the origin would be shifted as well??

- anonymous

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?

- anonymous

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

- anonymous

I meant plug into my new version of the function.

- anonymous

Just to see how changing different pieces can change the function

- anonymous

oh sorry, hold on

- anonymous

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

- anonymous

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

- anonymous

right

- anonymous

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

- anonymous

so only the radius is changing? maybe :)

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

oh!!! sorry

- anonymous

Do you have a graphing calculator?

- anonymous

yes.

- anonymous

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

- anonymous

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

- anonymous

right.

- anonymous

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

- anonymous

genius!

- anonymous

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

- anonymous

makes the radius bigger

- anonymous

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

- anonymous

Right.

- anonymous

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.

- anonymous

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

- anonymous

ahh, its starting to make sense!

- anonymous

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.

- anonymous

and still centered at 1,-2

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

using x=1 y = -2

- anonymous

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

- anonymous

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

- anonymous

you are a genius though! I appreciate the help

- anonymous

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

- anonymous

so just got to refresh =)

- anonymous

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

- anonymous

thanks again!

- anonymous

One sec

- anonymous

k

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

you went above beyond, many thanks!

- anonymous

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

- anonymous

will do!

Looking for something else?

Not the answer you are looking for? Search for more explanations.