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!

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