Lukecrayonz
  • Lukecrayonz
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
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Lukecrayonz
  • Lukecrayonz
|dw:1332619702161:dw|
Lukecrayonz
  • Lukecrayonz
|dw:1332620290697:dw|
jim_thompson5910
  • jim_thompson5910
You're in degree mode when you should be in radian mode.

Looking for something else?

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

More answers

Lukecrayonz
  • Lukecrayonz
Ooh, sorry.
jim_thompson5910
  • jim_thompson5910
No worries, let me know what you get when you change modes.
Lukecrayonz
  • Lukecrayonz
|dw:1332620534494:dw|
Lukecrayonz
  • Lukecrayonz
My calculators graph looks strange, i'm not sure if it's supposed to look like this?
Lukecrayonz
  • Lukecrayonz
|dw:1332620690908:dw|
jim_thompson5910
  • 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
  • Lukecrayonz
Thanks!:)
jim_thompson5910
  • jim_thompson5910
you're welcome
Lukecrayonz
  • Lukecrayonz
QWait
Lukecrayonz
  • Lukecrayonz
Quickly graph these for me?
Lukecrayonz
  • Lukecrayonz
x=3-2t y=2+3t
Lukecrayonz
  • Lukecrayonz
I am nearly 100% sure my calculator isn't graphing these right
jim_thompson5910
  • jim_thompson5910
See attached. I'm using winplot to graph these.
1 Attachment
Lukecrayonz
  • Lukecrayonz
Alright, yeah, mine had random bumps in the graph.
jim_thompson5910
  • jim_thompson5910
hmm interesting...
Lukecrayonz
  • Lukecrayonz
|dw:1332621224978:dw|
jim_thompson5910
  • jim_thompson5910
very interesting...
Lukecrayonz
  • Lukecrayonz
Rough drawing obviously, but its like that.
Lukecrayonz
  • Lukecrayonz
|dw:1332621265829:dw|
Lukecrayonz
  • 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
  • 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
  • jim_thompson5910
So the rectangular equation is \[\Large y = -\frac{3}{2}x + \frac{13}{2}\]
Lukecrayonz
  • Lukecrayonz
So for writing my graph, do you suggest my graph or your graph?
Lukecrayonz
  • Lukecrayonz
@Hero your input on this?
jim_thompson5910
  • 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
  • Lukecrayonz
Any reason why my calculator is being weird then? I reset it, and the same results.
jim_thompson5910
  • 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
  • Lukecrayonz
What program are you using to graph?
jim_thompson5910
  • jim_thompson5910
Winplot
Lukecrayonz
  • Lukecrayonz
Different equation: x=4+2cos(theta), y=-1+2sin(theta)
Lukecrayonz
  • Lukecrayonz
Solve for theta using the x equation, x=4+cos(theta) Theta=positive/negative cos^-1(x-4)
Lukecrayonz
  • Lukecrayonz
Input theta for y=-1+2sin(theta)
Lukecrayonz
  • Lukecrayonz
and i get y=-1+2 sqrt(1-(-4+x)^2)
Lukecrayonz
  • Lukecrayonz
http://screensnapr.com/v/JPhaJv.png
jim_thompson5910
  • 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
  • Lukecrayonz
Then i input negative cos, and get the other half of the graph.
jim_thompson5910
  • 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
  • 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
  • jim_thompson5910
Alright then use my second method and be careful when drawing the triangle and making the appropriate labels.

Looking for something else?

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