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 $latex z = 2 + x + 2y – x^2 – 3y^2 $ from the point $latex (0,0,2).$*

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 $latex f(x,y) = z, $ we know that $latex \nabla f $ points ‘upwards’ the most at all points where it isn’t zero. So at any point $latex P, $ we go in the direction $latex -\nabla f.$

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

So $latex -\nabla f = (2x -1)\hat{i} + (6y -2)\hat{j}.$

As the velocity of the curve points in the direction of the curve, our path satisfies:

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

$latex y'(t) = 6y(t) – 2; y(0) = 0$

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:

$latex y’ = 6y – 2$

$latex \frac{dy}{dt} = 6y – 2$

$latex \frac{dy}{6y-2} = dt$

$latex ln(6y-2)(1/6) = t + k$ for a constant k

$latex 6y = e^{6t + k} + 2= Ae^{6t}$ for a constant A

$latex y = Ae^{6t} + 1/3$ for a new constant A

$latex y(0) = 0 \Rightarrow A = -1/3 $

Solving both yields:

$latex x = \frac{1}{2} -\frac{1}{2} e^{2t} $

$latex y = \frac{1}{3} – \frac{1}{3} e^{6t} $

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

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

Good luck on your next test!

I have been informed that I had a small typo (two actually, and they cancelled each other out). $latex (3y-1) = -e^{6t}$ and $latex (2x – 1) = -e^{2t}$, but dividing cancels them out. Also note that $latex e^t neq 0 $ anywhere, so we don’t have to worry about that detail.

Could you please elaborate on the treatment of constant (k and A)? I do not get how you changed the constant ‘k’ to ‘A’…thanks

Sure. So we had $latex e^{6t + k} $, and I think the question refers to how this becomes $latex Ae^{6t}$.

We have the following: $latex e^{6t + k} = e^{6t}e^k$. Now $latex e^k$ is just some constant, and if we call it A we see that $latex e^{6t}e^k = Ae^{6t}$.