What are the steps in drawing the Contour maps for a differential equation having two stationary points..??

Question

I have a function z= x^(3) - 6x^(2) - y^(2) and I found the stationery points to be at (0,0) and (4,0). I just need to know the procedure of drawing the contour map for this function.

Answer 1

Start by marking the stationary points.

 

Method 1:

 

Next, get the gradient vector field

 

∇z = ⟨ ∂z/∂x , ∂z/∂y ⟩ = ∇z = ⟨ 3x² - 12x , -2y ⟩

 

The gradient is the zero vector at the two stationary points.

 

At any other point P, the gradient vector at P is perpendicular to the contour line that goes through P.

 

Method 2:

 

Get the differential equation for which the solution set is the set of all contours. That is, set z(x,y) = Constant, and find dy/dx by implicit differentiation.

 

x³ - 6x² - y² = C

 

3x² - 12x - 2y dy/dx = 0

 

dy/dx = (3x² - 12x) / (2y) = 3x(x - 4) / (2y)

 

Now evaluate dy/dx at many different points to draw a slope field.

 

At any point P, the contour through P must have the slope equal to the slope assigned at P.

 

Tip: knowing the points where dy/dx = 0 (horizontal slope), and dy/dx = undefined (vertical slope) is useful here. The slope marks at such points are called nullclines.