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anonymous
 5 years ago
The equations x=1+2cos(theta)
y=2+3sin(theta)
parameterize what curve in the xyplane? Determine its Cartesian equation.
anonymous
 5 years ago
The equations x=1+2cos(theta) y=2+3sin(theta) parameterize what curve in the xyplane? Determine its Cartesian equation.

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! 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
 5 years ago
Best ResponseYou've already chosen the best response.0Ok well do you have any ideas about how to think about this? What would the curve x = cos(theta) y = sin(theta) parameterize?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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 ?)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 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
 5 years ago
Best ResponseYou've already chosen the best response.0Do these numbers look familiar?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0x = 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
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh, I think I'm understanding it a little better

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So now imagine that we had instead x = 2 + cos(theta) y = 3 + sin(theta) What would this change?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the different values would just be shifted by the coefficient right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Would we still be describing a circle?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Would it still have radius 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Would it still be centered at the origin?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it would be greater

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhh and the origin would be shifted as well??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 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
 5 years ago
Best ResponseYou've already chosen the best response.01 + 2cos(0) = 3 2 + 3sin(0) = 2 1+2cos(pi/2) =1 2 + 3 sin(pi/2) = 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I meant plug into my new version of the function.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just to see how changing different pieces can change the function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they seem to move the same only deviate by the number being added?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos (pi/4) and sin (pi/4) are equivalent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So what does that tell you about adding a constant value to the original function? What did it do?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so only the radius is changing? maybe :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did 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
 5 years ago
Best ResponseYou've already chosen the best response.0no.. is it the length of the curve thats changing while the radius remains constant? thats just a wild guess :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No.. 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
 5 years ago
Best ResponseYou've already chosen the best response.0Do you have a graphing calculator?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=x+%3D+2%2Bcos%28theta%29%2C+y+%3D+3%2Bsin%28theta%29

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so its a constant circle as long as cos(theta) and sin(theta) are unchanged and anything added it too that shifts the origin?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now lets watch what happens when we put a coefficient on the cos and sin

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x = 2 + 3cos(theta) y = 3 + 3sin(theta)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0makes the radius bigger

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=x+%3D+2%2B3cos%28theta%29%2C+y+%3D+3%2B3sin%28theta%29

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And 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
 5 years ago
Best ResponseYou've already chosen the best response.0x=1+2cos(theta) y=2+3sin(theta)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh, its starting to make sense!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0and still centered at 1,2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, so when it asks to "parameterize" it would be x=1 and y = 2 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. It gave the parameterization. It's asking for the Cartesian equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so would Cartesian equation be the eq of a circle (x  a)^2 + (y  b)^2 = r^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, it'd be the equation of an ellipse.. \[{(xh)^2 \over a^2} + {(yk)^2 \over b^2} = 1\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, because you just said it was an elipse too, blahh. I wish this stuff would click easier

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you are a genius though! I appreciate the help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well this is stuff you should have covered in precalc.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so just got to refresh =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i never took precalc ! I went straight into calc I. I should def find a resource and review it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.khanacademy.org/video/polarcoordinates1?playlist=Precalculus Start here, and watch (click next video) up to and including http://www.khanacademy.org/video/parametricequations4?playlist=Precalculus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There are about 7 videos that are pretty short but should be a good intro.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0excellent! I've watched some of their videos in the past, I will watch these right now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you went above beyond, many thanks!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, and click the 'good answer' button so it knows you're done with this question =)
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