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

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: \begin{align} x'(t) &= 2x(t) - 1; x(0) = 0 \\ y'(t) &= 6y(t) - 2; y(0) = 0 \end{align}

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 \begin{align} x &= \frac{1}{2} -\frac{1}{2} e^{2t} \\ y &= \frac{1}{3} - \frac{1}{3} e^{6t} \end{align}

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!

### Leave a comment

## Info on how to comment

To make a comment, please send an email using the button below. Your email
address **won't be shared** (unless you include it in the body
of your comment). If you don't want your real name to be used next to your
comment, please specify the name you would like to use. If you want your name
to link to a particular url, include that as well.

**bold**, *italics*, and plain text are allowed in
comments. A reasonable subset of markdown is supported, including lists,
links, and fenced code blocks. In addition, math can be formatted using
`$(inline math)$`

or `$$(your display equation)$$`

.

**Please use plaintext email** when commenting. See
Plaintext Email and
Comments on this site for more. Note also that
**comments are expected to be open, considerate, and
respectful.**

Comments (3)2011-03-22 davidlowrydudaI have been informed that I had a small typo (two actually, and they cancelled each other out). $ (3y-1) = -e^{6t}$ and $ (2x - 1) = -e^{2t}$, but dividing cancels them out. Also note that $ e^t neq 0 $ anywhere, so we don't have to worry about that detail.

2011-03-28 dhruviktalaviyaCould you please elaborate on the treatment of constant (k and A)? I do not get how you changed the constant 'k' to 'A'...thanks

2011-03-28 davidlowrydudaSure. So we had $ e^{6t + k} $, and I think the question refers to how this becomes $ Ae^{6t}$.

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