To analyze the problem, we need to represent the level curves (courbes de niveau) for the following functions, one by one. Each function $f(x, y)$ will involve determining level curves of the form $f(x, y) = k$, where $k$ is a constant.

1. $f(x, y) = x^2y$

The level curves are given by setting $f(x, y) = k$, so: $x^2y = k$ This can be rewritten as: $y = \frac{k}{x^2}$

• Analysis: For different values of $k$, this represents a family of hyperbolas, where $k > 0$ lies in the first and third quadrants, and $k < 0$ lies in the second and fourth quadrants. At $k = 0$, the curve is simply $y = 0$ (the x-axis).

• Steps to Plot:

  1. Choose specific $k$ values, e.g., $k = -2, -1, 0, 1, 2$.
  2. For each $k$, plot the hyperbolas $y = \frac{k}{x^2}$ over both positive and negative $x$.

2. $f(x, y) = x^3$

The level curves are given by $f(x, y) = k$, so: $x^3 = k$ This simplifies to: $x = \sqrt[3]{k}$

• Analysis: Here, the level curves are vertical lines at specific $x$ values. For $k > 0$, $x$ is positive, and for $k < 0$, $x$ is negative. For $k = 0$, the curve is simply $x = 0$ (the y-axis).

• Steps to Plot:

  1. Choose $k$ values, e.g., $k = -2, -1, 0, 1, 2$.
  2. Plot the corresponding vertical lines at $x = \sqrt[3]{k}$.

3. $f(x, y) = x^2 + 9y^2$

The level curves are given by $f(x, y) = k$, so: $x^2 + 9y^2 = k$

• Analysis: This is an ellipse centered at the origin. The semi-major axis and semi-minor axis depend on $k$:

  • If $k > 0$, the ellipse has a horizontal radius $\sqrt{k}$ along $x$ and a vertical radius $\sqrt{k/9}$ along $y$.
  • If $k = 0$, the ellipse collapses to a point at the origin.

• Steps to Plot:

  1. Choose $k$ values, e.g., $k = 1, 4, 9,$ etc.
  2. For each $k$, plot the corresponding ellipse using the radii.

4. $f(x, y) = \sqrt{36 - 9x^2 - 4y^2}$

The level curves are given by $f(x, y) = k$, so: $\sqrt{36 - 9x^2 - 4y^2} = k$ Squaring both sides: $36 - 9x^2 - 4y^2 = k^2$ Rearranging: $9x^2 + 4y^2 = 36 - k^2$

• Analysis: This is an ellipse, provided $36 - k^2 > 0$. For $k^2 = 36$, the ellipse collapses to a single point. If $k^2 > 36$, there are no real solutions.

• Steps to Plot:

  1. Choose $k$ values such that $k^2 < 36$, e.g., $k = 0, 2, 4, 5$.
  2. For each $k$, the semi-major axis is $\sqrt{\frac{36 - k^2}{9}}$ along $x$, and the semi-minor axis is $\sqrt{\frac{36 - k^2}{4}}$ along $y$.
  3. Plot the ellipses accordingly.

Would you like me to generate visual plots for these functions? Let me know if you'd like more details.

5 Questions to Expand Understanding:

1. How do the level curves change as the value of $k$ increases for each function?
2. What is the geometric interpretation of level curves in general?
3. Why does $f(x, y) = x^3$ only produce vertical lines?
4. Can you identify any symmetries in the level curves for $f(x, y) = x^2 + 9y^2$?
5. What happens when $k^2 = 36$ for $f(x, y) = \sqrt{36 - 9x^2 - 4y^2}$?

Tip:

For level curves, always analyze the symmetry and type of curve (ellipse, hyperbola, etc.) to simplify graphing.