Due to the amount of confusion and the large number of emails, I have written up the solution to Problem 1 from Test 2.

The Problem
Determine the path of steepest descent along the surface $z = 2 + x + 2y - x^2 - 3y^2$ from the point $(0,0,2).$

There are a few things to note – the first thing we must do is find which direction points ‘downwards’ the most. So we note that for a function $f(x,y) = z,$ we know that $\nabla f$ points ‘upwards’ the most at all points where it isn’t zero. So at any point $P,$ we go in the direction $-\nabla f.$

The second thing to note is that we seek a path, not a direction. So let us take a curve that parametrizes our path: ${\bf C} (t) = x(t) \hat{i} + y(t) \hat{j}.$

So $-\nabla f = (2x -1)\hat{i} + (6y -2)\hat{j}.$
As the velocity of the curve points in the direction of the curve, our path satisfies:

$x'(t) = 2x(t) -1; x(0) = 0$
$y'(t) = 6y(t) - 2; y(0) = 0$

These are two ODEs that we can solve by separation of variables (something that is, in theory, taught in 1502 – for more details, look at chapter 9 in Salas, Hille, and Etgen). Let’s solve the y one:

$y' = 6y - 2$
$\frac{dy}{dt} = 6y - 2$
$\frac{dy}{6y-2} = dt$
$ln(6y-2)(1/6) = t + k$ for a constant k
$6y = e^{6t + k} + 2= Ae^{6t}$ for a constant A
$y = Ae^{6t} + 1/3$ for a new constant A
$y(0) = 0 \Rightarrow A = -1/3$

Solving both yields:

$x = \frac{1}{2} -\frac{1}{2} e^{2t}$
$y = \frac{1}{3} - \frac{1}{3} e^{6t}$

Now let’s get rid of the t. Note that $(3y -1) = e^{6t}$ and $(2x -1) = e^{2t}$. Using these together, we can get rid of t by noting that $\dfrac{3y-1}{(2x - 1)^3} = 1.$ Rewriting, we get $3y = (2x-1)^3 + 1.$

So the path is given by $3y = (2x-1)^3 + 1$

Good luck on your next test!