anonymous
  • anonymous
The equations x=1+2cos(theta) y=-2+3sin(theta) parameterize what curve in the xy-plane? Determine its Cartesian equation.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Owlfred
  • 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
  • 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
  • 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
  • anonymous
Ah. Ok
anonymous
  • 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
  • anonymous
Do these numbers look familiar?
anonymous
  • anonymous
yes
anonymous
  • 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
  • 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
  • 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
  • anonymous
oh, I think I'm understanding it a little better
anonymous
  • anonymous
So now imagine that we had instead x = 2 + cos(theta) y = 3 + sin(theta) What would this change?
anonymous
  • anonymous
the different values would just be shifted by the coefficient right?
anonymous
  • anonymous
Would we still be describing a circle?
anonymous
  • anonymous
Would it still have radius 1?
anonymous
  • anonymous
Would it still be centered at the origin?
anonymous
  • anonymous
no it would be greater
anonymous
  • anonymous
ohhh and the origin would be shifted as well??
anonymous
  • 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
  • anonymous
1 + 2cos(0) = 3 -2 + 3sin(0) = -2 1+2cos(pi/2) =1 -2 + 3 sin(pi/2) = -1
anonymous
  • anonymous
I meant plug into my new version of the function.
anonymous
  • anonymous
Just to see how changing different pieces can change the function
anonymous
  • anonymous
oh sorry, hold on
anonymous
  • anonymous
they seem to move the same only deviate by the number being added?
anonymous
  • anonymous
cos (pi/4) and sin (pi/4) are equivalent?
anonymous
  • anonymous
right
anonymous
  • anonymous
So what does that tell you about adding a constant value to the original function? What did it do?
anonymous
  • anonymous
so only the radius is changing? maybe :)
anonymous
  • 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
  • anonymous
no.. is it the length of the curve thats changing while the radius remains constant? thats just a wild guess :(
anonymous
  • 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
  • anonymous
oh!!! sorry
anonymous
  • anonymous
Do you have a graphing calculator?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=x+%3D+2%2Bcos%28theta%29%2C+y+%3D+3%2Bsin%28theta%29
anonymous
  • 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
  • anonymous
right.
anonymous
  • anonymous
Now lets watch what happens when we put a coefficient on the cos and sin
anonymous
  • anonymous
genius!
anonymous
  • anonymous
x = 2 + 3cos(theta) y = 3 + 3sin(theta)
anonymous
  • anonymous
makes the radius bigger
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=x+%3D+2%2B3cos%28theta%29%2C+y+%3D+3%2B3sin%28theta%29
anonymous
  • anonymous
Right.
anonymous
  • 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
  • anonymous
x=1+2cos(theta) y=-2+3sin(theta)
anonymous
  • anonymous
ahh, its starting to make sense!
anonymous
  • 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
  • anonymous
and still centered at 1,-2
anonymous
  • anonymous
okay, so when it asks to "parameterize" it would be x=1 and y = -2 ?
anonymous
  • anonymous
No. It gave the parameterization. It's asking for the Cartesian equation.
anonymous
  • anonymous
so would Cartesian equation be the eq of a circle (x - a)^2 + (y - b)^2 = r^2
anonymous
  • anonymous
using x=1 y = -2
anonymous
  • 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
  • anonymous
Yes, because you just said it was an elipse too, blahh. I wish this stuff would click easier
anonymous
  • anonymous
you are a genius though! I appreciate the help
anonymous
  • anonymous
well this is stuff you should have covered in pre-calc.
anonymous
  • anonymous
so just got to refresh =)
anonymous
  • anonymous
i never took precalc ! I went straight into calc I. I should def find a resource and review it.
anonymous
  • anonymous
thanks again!
anonymous
  • anonymous
One sec
anonymous
  • anonymous
k
anonymous
  • 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
  • anonymous
There are about 7 videos that are pretty short but should be a good intro.
anonymous
  • anonymous
excellent! I've watched some of their videos in the past, I will watch these right now
anonymous
  • anonymous
you went above beyond, many thanks!
anonymous
  • anonymous
oh, and click the 'good answer' button so it knows you're done with this question =)
anonymous
  • anonymous
will do!

Looking for something else?

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