Consider the parametric equations x=4cos^2theta and y=2sin theta.
Use a graphing utility to complete the table: (will draw)
Plot the points (x, y) generated in part (a) and sketch a graph of the parametric equations

- Lukecrayonz

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- Lukecrayonz

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- Lukecrayonz

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- jim_thompson5910

You're in degree mode when you should be in radian mode.

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## More answers

- Lukecrayonz

Ooh, sorry.

- jim_thompson5910

No worries, let me know what you get when you change modes.

- Lukecrayonz

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- Lukecrayonz

My calculators graph looks strange, i'm not sure if it's supposed to look like this?

- Lukecrayonz

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- jim_thompson5910

You have the correct table. The graph should look like the following (see attached). So you have the correct graph.

##### 1 Attachment

- Lukecrayonz

Thanks!:)

- jim_thompson5910

you're welcome

- Lukecrayonz

QWait

- Lukecrayonz

Quickly graph these for me?

- Lukecrayonz

x=3-2t
y=2+3t

- Lukecrayonz

I am nearly 100% sure my calculator isn't graphing these right

- jim_thompson5910

See attached. I'm using winplot to graph these.

##### 1 Attachment

- Lukecrayonz

Alright, yeah, mine had random bumps in the graph.

- jim_thompson5910

hmm interesting...

- Lukecrayonz

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- jim_thompson5910

very interesting...

- Lukecrayonz

Rough drawing obviously, but its like that.

- Lukecrayonz

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- Lukecrayonz

Hmm, well, the directions for this question:
Sketch the curve represented, indicate the direction of the curve. Use a graphing utility to confirm your result. Then eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve

- jim_thompson5910

well you can prove it's a straight line showing the following
x=3-2t
x-3 = -2t
t = -1/2x + 3/2
y=2+3t
y = 2 + 3(-1/2x + 3/2)
y = 2 - 3/2x + 9/2
y = -3/2x + 13/2
which is in the form y = mx+b, a linear equation

- jim_thompson5910

So the rectangular equation is \[\Large y = -\frac{3}{2}x + \frac{13}{2}\]

- Lukecrayonz

So for writing my graph, do you suggest my graph or your graph?

- Lukecrayonz

@Hero your input on this?

- jim_thompson5910

My graph is more accurate as I've shown above that it's actually a linear equation (so it's a straight line) and Winplot confirms this.

- Lukecrayonz

Any reason why my calculator is being weird then? I reset it, and the same results.

- jim_thompson5910

Not all parametric equations result in nasty curves. Remember that calculators graph by plotting points and then connecting those points the best it can. So my guess is that it's not plotting enough points to get an accurate picture.

- Lukecrayonz

What program are you using to graph?

- jim_thompson5910

Winplot

- Lukecrayonz

Different equation: x=4+2cos(theta), y=-1+2sin(theta)

- Lukecrayonz

Solve for theta using the x equation, x=4+cos(theta)
Theta=positive/negative cos^-1(x-4)

- Lukecrayonz

Input theta for y=-1+2sin(theta)

- Lukecrayonz

and i get y=-1+2 sqrt(1-(-4+x)^2)

- Lukecrayonz

http://screensnapr.com/v/JPhaJv.png

- jim_thompson5910

Different way to do it
x=4+2cos(theta)
x-4 = 2cos(theta)
(x-4)/2 = cos(theta)
cos(theta) = (x-4)/2
cos^2(theta) = ((x-4)^2)/4
---------------------------------
y=-1+2sin(theta)
y+1=2sin(theta)
(y+1)/2=sin(theta)
sin(theta) = (y+1)/2
sin^2(theta) = ((y+1)^2)/4
----------------------------------
cos^2(theta) + sin^2(theta) = 1
((x-4)^2)/4 + ((y+1)^2)/4 = 1
(x-4)^2 + (y+1)^2 = 4
Therefore, this is a circle with center (4, -1) with radius 2.
Graph is attached

##### 1 Attachment

- Lukecrayonz

Then i input negative cos, and get the other half of the graph.

- jim_thompson5910

You should have x = 4+2cos(theta) which means that
theta = cos^-1( (x-4)/2)
So the equation should be
y = -1 + 2*sin(theta)
y = -1 + 2*sin(cos^-1( (x-4)/2))
y = -1 + 2*(1/2)*sqrt((4-(x-4)^2))
y = sqrt(4-(x-4)^2) - 1
This graphs the upper half
The lower half is then
y = -sqrt(4-(x-4)^2) - 1

- Lukecrayonz

Hmm, alright, I get this. Theres two sections of this section, and the way I just did it was the second way, and the way you did it was the correct way for the way they asked, but now I have to solve all the problems in the way I did it.

- jim_thompson5910

Alright then use my second method and be careful when drawing the triangle and making the appropriate labels.

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