I got this question Which I have no idea how to solve or even how to think when sketching, this is the question:
Let $f(x,y)=x^2+4xy+y^2$ for all $(x,y) \in \mathbb R^2$. To study $f$ it can be useful to do a change of variables like this: $u=x+y$ and $v=x-y.$
a)sketch some of the contour-lines/isopleth's of the function $f(x,y)$.
b)sketch the graph of the function $f(x,y).$
c)Find a parametrization of the curve that is given by the intersection of the function $f$ and the plane that is given by $z=x+3y.$
I kind of have an idea of what contour lines are, but not sure how to sketch them since it seems very hard in this specific case. And sketching the function itself seems also awfully complicated. I have however plotted it in some $3d$ graph program so I know kind of how to graph looks. I have no idea how to solve C either, so I'd appreciate all help that is given to me.
Thanks in advance.
$$z=f(x,y)=x^2+4xy+y^2=(x+2y)^2-(\sqrt{3}y)^2$$
can be written under the form:
$$z=(x+ay)(x-by) \ \ \text{with} \ \ a=2-\sqrt{3} \ \ \ \text{and} \ \ \ b=2+\sqrt{3}$$
This surface, that can be written $z=XY$ (up to an affine change of basis), is therefore a hyperbolic paraboloid (https://en.wikipedia.org/wiki/Paraboloid). See figure.
Concerning the intersection with plane $z=x+3y$, a good way is to transform cartesian coordinates $(x,y)$ into polar coordinates $(r,\theta)$, with:
$$x=r \cos(\theta), y=r \sin(\theta) \ \ \ $$
In this way, the double equation for intersections points :
$$z=x+3y=x^2+4xy+y^2$$
is transformed into:
$$z=r(\cos(\theta)+3\sin(\theta))=r^2(1+2\sin(2\theta)) \ \ (1)$$
from which we get
$$r=r(\theta)=\dfrac{\cos(\theta)+3\sin(\theta)}{1+2 sin(2\theta)}$$
Thus a parametric representation of the curve can be given under the form:
$$x=r(\theta)\cos(\theta), y=r(\theta)\sin(\theta), z=r(\theta)(cos(\theta)+3\sin(\theta)).$$