Q 47. - Calculus

Question

Sketch the level curves $c = -3, -2, -1, 0, 1, 2, 3$ if they exist for the specified function. $f (x, y) = 4 - (x^2 + y^2)$

Step-by-Step Solution

Answer

These are all equations of circles.

Step 1: Given information

The given function is $f(x,y)=4-\left(x^2+y^2\right)$

Step 2: Calculation

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

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

The graph of this function will be in ${\mathrm{ℝ}}^3$ Assume the third variable to be $z$ 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.

$$\begin{gathered} 4-\left(x^2+y^2\right)=-3\Rightarrow x^2+y^2=7 \\ 4-\left(x^2+y^2\right)=-2\Rightarrow x^2+y^2=6 \\ 4-\left(x^2+y^2\right)=-1\Rightarrow x^2+y^2=5 \\ 4-\left(x^2+y^2\right)=0\Rightarrow x^2+y^2=4 \\ 4-\left(x^2+y^2\right)=1\Rightarrow x^2+y^2=3 \\ 4-\left(x^2+y^2\right)=2\Rightarrow x^2+y^2=2 \\ 4-\left(x^2+y^2\right)=3\Rightarrow x^2+y^2=1 \end{gathered}$$

Step 3: Explanation

These are all equations of circles. Plot these equations on the same $x y-$plane.