Q 45. - Calculus

Question

Sketch the level curves $c = -3, -2, -1, 0, 1, 2, 3$ if they exist for the specified function. $f (x, y) = \frac{3x}{y^2}$

Step-by-Step Solution

Answer

These are all quadratic equations. Hence, they all represent parabolas.

Step 1: Given information

The given function is $f (x, y) = \frac{3x}{y^2}$

Step 2: Calculation

The goal is to draw the level curves for $c=-3,-2,-1,0,1,2,3$

Consider a function in two variables as $f(x, y)$ This function is in ${\mathrm{ℝ}}^2$

This function's graph will be in the ${\mathrm{ℝ}}^3$ format. Assume that the third variable is $z$ The equation of the graph is given as

$z=f(x, y)$

The graphs of this function's level curves are the graphs of the function for a constant value of $z$

Assume $z=c$ The level curve of the graph at $c$ is defined as the graph of the equation $f(x, y)=c$

Determine the equations for the various required level curves using the preceding definition.

$$\begin{gathered} \frac{3x}{y^2}=-3\Rightarrow y^2=-x \\ \frac{3x}{y^2}=-2\Rightarrow y^2=-\frac{3}{2}x \\ \frac{3x}{y^2}=-1\Rightarrow y^2=-3x \\ \frac{3x}{y^2}=0\Rightarrow x=0 \\ \frac{3x}{y^2}=1\Rightarrow y^2=3x \\ \frac{3x}{y^2}=2\Rightarrow y^2=\frac{3}{2}x \\ \frac{3x}{y^2}=3\Rightarrow y^2=x \end{gathered}$$

Step 3: Explanation

All of these equations are quadratic in nature. As a result, they're all parabolas. Plot these equations on the same $x y-$coordinate plane.