Question

Sketch Both A Contour Map And Graph Of The Function and Compare Them.

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

Short answer

Hence, the level of curve \(f\left( {x,y} \right) = k\) are just the traces of graph of f is plane \(z = k\) .

Step-by-step solution

Defining The Level Curves

\(f\left( {x,y} \right) = k\), Where k is aConstant

The level curves of\(f\left( {x,y} \right)\)are

\(\sqrt {36 - 9{x^2} - 4{y^2}} = k,k = - 3, - 2, - 1,0,1,2....\)

Sketching The Curves With \(k = 1,2,3,4......\) 

Plotting The Contour Map

Using, Maple to check the contour map showing several curves.

Maple input

\(\begin{aligned}{l} > with(plot)\\ > contourplot\left( {\sqrt {36 - {x^2} - 4{y^2}} ,x = 3.3,y = - 3.3} \right)\end{aligned}\)

A level curve\(f\left( {x,y} \right) = k\)is the set of all points in the domain of f, at which f takes on a given value k.

Observing, the two graphs, the relation between the level curves andHorizontal Traces.

Hence, the level of curve \(f\left( {x,y} \right) = k\) are just the traces of graph of f is plane \(z = k\) .