> restart:
with(plots):

In class we said that the graph of #16d from section 1.3 should be a bowl which opens up but which is twisted. I used Maple to graph this function below.

> plot3d(5/4*x^2+3/2*x*y+5/4*y^2,x=-4..4,y=-5..5,view=0..10,axes=boxed);

How is the graph twisted? Below is the graph of a similar function with out the xy term.

> z:=x^2+5*y^2;

> plot3d(x^2+5*y^2,x=-4..4,y=-5..5,view=0..15,axes=boxed);

So what did I mean in class when I said the middle term will twist the graph, and how can we predict this twist? Next let's look at cross sections of these function. Problem #16d from section 1.3 with z=5/4 will be in red and the function I created with z=2 will be in violet.

> text:=implicitplot(5/4*(x+3/5*y)^2+4/5*y^2=5/4,x=-1.5..1.5,y=-1.5..1.5):
other:=implicitplot(x^2+5*y^2=2,x=-1.5..1.5,y=-1.5..1.5,color=violet):
display(text,other);

Finally how can we predict the twist? To answer this I plotted a few more cross section along with the lines of symmetry. What are these lines of symmetry and where did they come from in terms of the equation for the original function given in #16d from section 1.3?

> A:=implicitplot((x+3/5*y)^2+16/25*y^2=1,x=-1.5..1.5,y=-1.5..1.5):
B:=plot(-x,x=-1.5..1.5,y=-1.5..1.5,color=blue):
C:=plot(5/3*x,x=-1.5..1.5,y=-1.5..1.5,color=green):

> display(A,B,C);

Above I guessed possible symmetry lines and then plotted them against a cross section of the graph. Eventually I found what I thought was the equation for the symmetry lines. Let's test these guesses with a few more cross sections:

> A[1]:=implicitplot((x+3/5*y)^2+16/25*y^2=1,x=-3..3,y=-3..3):
A[2]:=implicitplot((x+3/5*y)^2+16/25*y^2=2,x=-3..3,y=-3..3):
A[3]:=implicitplot((x+3/5*y)^2+16/25*y^2=3,x=-3..3,y=-3..3):
A[4]:=implicitplot((x+3/5*y)^2+16/25*y^2=4,x=-3..3,y=-3..3):
A[5]:=implicitplot((x+3/5*y)^2+16/25*y^2=5,x=-3..3,y=-3..3):
B[1]:=plot(-x,x=-3..3,y=-3..3,color=blue):
C[1]:=plot(5/3*x,x=-3..3,y=-3..3,color=green):

> display(A[1],A[2],A[3],A[4],A[5],B[1],C[1]);

>