Explorations in math and programming
David Lowry-Duda

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!

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Comments (3)
  1. 2011-03-22 davidlowryduda

    I 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.

  2. 2011-03-28 dhruviktalaviya

    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

  3. 2011-03-28 davidlowryduda

    Sure. 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}$.