Multivariable calculus, level curves

184 Views Asked by At

Let $$f(x,y) = 3x^2 + 4xy + 3y^2$$ for all $(x,y)$ in $R^2$. To study f it can be useful to implement the variable substitution $$u = x + y \qquad \textrm{and} \qquad v= x-y .$$

a) Sketch some of the level curves of $f$. b) Sketch the graph of $f$. c) Determine a parametrisation of the curve that is given by the intersection of $f$ and the plane with equation $z=x+3y$.

So I'm stuck here on the first assignment. It feels like every way I twist and turn it I still end up with some unwanted $y$'s on the wrong side, it was a while since doing calculus so I'm not really used to screwing around with these kind of things.

I tried using $(x+y)^2 = x^2 + 2xy + y^2$ and then I end up with $x^2 + y^2 + u^2 = c$ (using the u-substitution) but I don't know where to go from there. Any help at all greatly appreciated! Thanks in advance :-)

2

There are 2 best solutions below

4
On

Hint We can solve simultaneously for $u, v$ in terms of $x, y$: $$x = \tfrac{1}{2}(u + v), \qquad y = \tfrac{1}{2}(u - v).$$ Substituting these expressions in $f(x, y)$ gives a formula $f(x(u, v), y(u, v))$ for $f$ in the coordinates $u, v$.

0
On

Hint:

Note that a level curve is represented by the equation: $$ 3x^2+4xy+3y^2=k $$ That is an ellipse with center in the origin because $B^2-4AC= (4)^2-4\cdot(3)\cdot(3)<0$.

Now use the Principal axis theorem to find the axis of the ellipse and take some point $P=(x_P,y_P)$ on these axis to find some level curve.