Multivariable calculus question

  • #1

Homework Statement



Sketch the following curves on separate number planes.

y= 2x^2

x^2 + 2y^2= 4

Homework Equations





The Attempt at a Solution



For the first one, would the intersection of the curve with the y axis be at y=2 ( I have attached a diagram of my solution). The minimum point would be at (0,2)???

For the second one, what do we do with the 2 which is the coefficient of y^2. Would the 4 be the radius of the circle.

Thanks in advance.
 

Attachments

  • Curve.JPG
    Curve.JPG
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Answers and Replies

  • #2
34
0
In the first one. If the curve was intersecting the y-axis, then x = 0 at the point of intersection, and so y = 0 aswell, so it intersects are (0,0). Also, the minimum point of the function is at (0,0).

For the second one, are you sure that the equation you have describes a circle? I would call it an ellipse myself.
 
  • #3
oh alrite, but then what happens with the 2 in x^2 for the first equation. Yea, the second one is an ellipse so would 4 be the radius?? Cheers
 
  • #4
34
0
[tex] y = 2x^2 [/tex]
if x = 0; [tex] y = 2*(0^2) = 0 [/tex]

As for the ellipse. have you tried working out the values for x and y when the other one is set to 0? That will tell u where it crosses each axis.

edit: This might help - http://en.wikipedia.org/wiki/Ellipse
 
Last edited:

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