Question

Draw a Contour Map of the function showing several level Curves.

\(f\left( {x,y} \right) = \ln \left( {{x^2} + 4{y^2}} \right)\)

Short answer

Given:\(f\left( {x,y} \right) = \ln \left( {{x^2} + 4{y^2}} \right)\)

To Draw: Contour Map of the given function showing several level curves.

Step-by-step solution

Assuming A Constant

To complete the drawing, graph equations with varying values of c as a value of the function

\(\ln \left( {{x^2} + 4{y^2}} \right) = c\)

\(\)

Determining Equations Of Level Curve

To graph this equation, notice that the equation can be put into the form of an ellipse where the coefficient in front of the x and y terms are the x and y dimensions of the ellipse, respectively;

\(\begin{aligned}{l}\ln \left( {{x^2} + 4{y^2}} \right) &= c\\{x^2} + 4{y^2} &= {e^c}\\{x^2} + {(2y)^2} &= {e^c}\\{\left( {\frac{x}{{{e^{\frac{c}{2}}}}}} \right)^2} + {\left( {\frac{{2y}}{{{e^{\frac{c}{2}}}}}} \right)^2} &= 1\end{aligned}\)

The equation where\(c = 4\)is;

\(\begin{aligned}{l}{\left( {\frac{x}{{{e^{\frac{4}{2}}}}}} \right)^2} + {\left( {\frac{{2y}}{{{e^{\frac{4}{2}}}}}} \right)^2} &= 1\\{\left( {\frac{x}{{{e^2}}}} \right)^2} + {\left( {\frac{{2y}}{{{e^2}}}} \right)^2} &= 1\end{aligned}\)

Thus the level curve look like Elliptical Curves.

Drawing Contour Map

Contour Map for \(f\left( {x,y} \right)\) looks like: