# Sketching level curves

• Dec 9th 2011, 04:37 AM
SyNtHeSiS
Sketching level curves
Sketch the level curves for the function $\displaystyle z = ln(x^2 + y^2)$.

Attempt:

$\displaystyle k = ln(x^2 + y^2)$

$\displaystyle z = ln(k^2 + y^2)$

$\displaystyle z = ln(x^2 + k^2)$

Now I am not sure how I would draw each of these graphs due to the $\displaystyle ln(x^2 + y^2)$ part. (Worried)

Also when asked to sketch the level curves, does that mean you just let z = k and draw one graph? (as opposed to also letting x = k and y = k)
• Dec 9th 2011, 07:10 AM
alexmahone
Re: Sketching level curves
Quote:

Originally Posted by SyNtHeSiS
Sketch the level curves for the function $\displaystyle z = ln(x^2 + y^2)$.

Attempt:

$\displaystyle k = ln(x^2 + y^2)$

$\displaystyle z = ln(k^2 + y^2)$

$\displaystyle z = ln(x^2 + k^2)$

Now I am not sure how I would draw each of these graphs due to the $\displaystyle ln(x^2 + y^2)$ part. (Worried)

Also when asked to sketch the level curves, does that mean you just let z = k and draw one graph? (as opposed to also letting x = k and y = k)

$\displaystyle z=k$

$\displaystyle \ln (x^2+y^2)=k$

$\displaystyle x^2+y^2=e^k$

So, the level curves are circles in the xy-plane that are centered at the origin.
• Dec 10th 2011, 12:17 AM
SyNtHeSiS
Re: Sketching level curves
Thanks :). Would drawing the graph give you a circular paraboloid (or do you have to draw the traces for $\displaystyle x = k$and $\displaystyle y = k,$ before drawing $\displaystyle x = ln(x^2 + y^2))$?
• Dec 10th 2011, 12:23 AM
alexmahone
Re: Sketching level curves
Quote:

Originally Posted by SyNtHeSiS
Thanks :). Would drawing the graph give you a circular paraboloid (or do you have to draw the traces for $\displaystyle x = k$and $\displaystyle y = k,$ before drawing $\displaystyle x = ln(x^2 + y^2))$?

No, because the equation of a circular paraboloid is $\displaystyle x^2+y^2=a^2z$.
• Dec 16th 2011, 01:51 AM
SyNtHeSiS
Re: Sketching level curves
Oh ok. What steps would you use to draw the correct graph?
• Dec 16th 2011, 01:58 AM
alexmahone
Re: Sketching level curves
Quote:

Originally Posted by SyNtHeSiS
Oh ok. What steps would you use to draw the correct graph?

$\displaystyle z=\ln (x^2+y^2)=\ln r^2=2\ln r$

$\displaystyle r=e^{z/2}$

I have attached a graph generated using MATLAB.
• Dec 16th 2011, 06:11 AM
HallsofIvy
Re: Sketching level curves
Any function of the form $\displaystyle z= f(x^2+ y^2)= f(r)$ has circular symmetry. In particular, $\displaystyle z= ln(x^2+ y^2)= ln(r^2)= 2ln(r)$ is a logarithm graph. Draw that and rotate around the z-axis.