Week 6 Discussion Notes

Solution.

$x = a$:

We can just substitute $x = a$ into the equation of the surface to get

$a$ is fixed, so this tells us that this trace is a parabola opening upwards, and $a$ simply translates the parabola:

$y = b$:

Similarly,

so we still get a parabola, but increasing $b$ shifts the parabola more than the above:

$z = c$:

This one is a little different from previous ones:

Before, the traces gave us one variable as a function of the other, e.g., $z = g\p{x}$ and $z = g\p{y}$. For this one, you can't solve for $x$ or $y$ without breaking things up into cases (i.e., without writing $\pm$ everywhere). But the equation is still something familiar to us: it's the equation of an ellipse:

Solution.

Generally, you want to try a couple paths to get a feel for the problem. If you find two different limits, then the limit doesn't exist, but if every path you try gives you the same answer, you might want to start proving that the limit exists.

I usually try the following paths first: a horizontal line, a vertical line, or a $45^\circ$ line.

Horizontal line through $\p{0, 0}$:

You can parametrize this by $\p{x, 0}$, so the limit on this path becomes

Vertical line through $\p{0, 0}$:

This is $\p{0, y}$, so you get

$45^\circ$ line through $\p{0, 0}$:

The path is $\p{x, x}$, so you end up with

We found two paths where the limits don't agree: $\p{x, 0}$ gives $0$ but $\p{x, x}$ gives $\frac{1}{5}$, so

Solution.

Usually, if you see sines or cosines, you can try the squeeze theorem since

for any $\theta$. In our problem, we get

since $\sqrt{x^2 + y^2}$ is continuous at $\p{0, 0}$. So, by the squeeze theorem,