4.1 Functions of several variables  (Page 3/12)

Level curves

If hikers walk along rugged trails, they might use a topographical map that shows how steeply the trails change. A topographical map contains curved lines called contour lines . Each contour line corresponds to the points on the map that have equal elevation ( [link] ). A level curve of a function of two variables $f\left(x,y\right)$ is completely analogous to a contour line on a topographical map.

Returning to the function $g\left(x,y\right)=\sqrt{9-x^2-y^2},$ we can determine the level curves of this function. The range of $g$ is the closed interval $\left[0,3\right].$ First, we choose any number in this closed interval—say, $c=2.$ The level curve corresponding to $c=2$ is described by the equation

To simplify, square both sides of this equation:

Now, multiply both sides of the equation by $\mathrm{-1}$ and add $9$ to each side:

This equation describes a circle centered at the origin with radius $\sqrt{5}.$ Using values of $c$ between $0\text{and}3$ yields other circles also centered at the origin. If $c=3,$ then the circle has radius $0,$ so it consists solely of the origin. [link] is a graph of the level curves of this function corresponding to $c=0,1,2,\text{and}3.$ Note that in the previous derivation it may be possible that we introduced extra solutions by squaring both sides. This is not the case here because the range of the square root function is nonnegative.

A graph of the various level curves of a function is called a contour map    .

Making a contour map

Given the function $f\left(x,y\right)=\sqrt{8+8x-4y-4x^2-y^2},$ find the level curve corresponding to $c=0.$ Then create a contour map for this function. What are the domain and range of $f?$

To find the level curve for $c=0,$ we set $f\left(x,y\right)=0$ and solve. This gives

$0=\sqrt{8+8x-4y-4x^2-y^2}.$

We then square both sides and multiply both sides of the equation by $\mathrm{-1}\text{:}$

$4x^2+y^2-8x+4y-8=0.$

Now, we rearrange the terms, putting the $x$ terms together and the $y$ terms together, and add $8$ to each side:

$4x^2-8x+y^2+4y=8.$

Next, we group the pairs of terms containing the same variable in parentheses, and factor $4$ from the first pair:

$4\left(x^2-2x\right)+\left(y^2+4y\right)=8.$

Then we complete the square in each pair of parentheses and add the correct value to the right-hand side:

$4\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=8+4\left(1\right)+4.$

Next, we factor the left-hand side and simplify the right-hand side:

$4{\left(x-1\right)}^2+{\left(y+2\right)}^2=16.$

Last, we divide both sides by $16\text{:}$

$\frac{{\left(x-1\right)}^2}{4}+\frac{{\left(y+2\right)}^2}{16}=1.$

This equation describes an ellipse centered at $\left(1,\mathrm{-2}\right).$ The graph of this ellipse appears in the following graph.

We can repeat the same derivation for values of $c$ less than $4.$ Then, [link] becomes

$\frac{4{\left(x-1\right)}^2}{16-c^2}+\frac{{\left(y+2\right)}^2}{16-c^2}=1$

for an arbitrary value of $c.$ [link] shows a contour map for $f\left(x,y\right)$ using the values $c=0,1,2,\text{and}3.$ When $c=4,$ the level curve is the point $\left(\mathrm{-1},2\right).$

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