Contour map of the function showing several level curves

[cosmos42]

Homework Statement

ƒ(x,y) = ln(x²+4y²)

Homework Equations

I'm not really sure but I solved for y

The Attempt at a Solution

 

[haruspex]

What do you get for y at x=0 from x²+4y²=1? Where is/are that/those on your diagram?

I'm not sure whether you are supposed to plot curves for equal increments in f, as one would on a cartographic map.

 

[HallsofIvy]

"Level curves" are curves on which the expression has a specific value. I would not solve for y. Instead, I would look at the equation [itex]ln(x^2+ 4y^2)= C[/itex] and recognize that, taking the exponential of both sides, [itex]x^2+ 4y^2= e^C[/itex] which is equivalent to [itex]\frac{x^2}{e^C}+ \frac{y^2}{e^C/4}= 1[/itex]. You should immediately see that, for every C, this is an ellipse, centered at the origin, with x intercepts at [itex]\left(-e^{C/2}, 0\right)[/itex] and [itex]\left(e^{C/2}, 0\right)[/itex], y intercepts at [itex]\left(0, -\frac{e^{C/2}}{2}\right)[/itex] and [itex]\left(0, \frac{e^{C/2}}{2}\right)[/itex].

 

[haruspex]

"Level curves" are curves on which the expression has a specific value. I would not solve for y. Instead, I would look at the equation [itex]ln(x^2+ 4y^2)= C[/itex] and recognize that, taking the exponential of both sides, [itex]x^2+ 4y^2= e^C[/itex] which is equivalent to [itex]\frac{x^2}{e^C}+ \frac{y^2}{e^C/4}= 1[/itex]. You should immediately see that, for every C, this is an ellipse, centered at the origin, with x intercepts at [itex]\left(-e^{C/2}, 0\right)[/itex] and [itex]\left(e^{C/2}, 0\right)[/itex], y intercepts at [itex]\left(0, -\frac{e^{C/2}}{2}\right)[/itex] and [itex]\left(0, \frac{e^{C/2}}{2}\right)[/itex].

Yes, that was a bit confusing in the OP, but judging from the drawing I would say cosmos42 understood about different values for C.
I was trying to guide cosmos into realising that the curves were centred wrongly.