Directional derivatives

• Apr 5th 2011, 10:28 PM
Glitch
Directional derivatives
The question:
A skier is on a mountain described by the equation $\displaystyle h(x, y) = 2000 - x^4/10^8 - y^2/10^2$ at the point (100, 1). He skis down the mountain, always moving in the direction of the steepest descent.
a) In what direction does he start moving?
b) Describe the curve along which he skis. [You will need to solve a separable first order ODE.] [HINT: Let the skier's position at time t be (x(t), y(t), z(t)) where z(t) = h(x(t), y(t)). What does the condition say about the skier's velocity in the xy-plane, ($\displaystyle \dot{x}, \dot{y}$)?]

My attempt:
I can almost solve a), I ended up getting $\displaystyle \frac{10^{-2}}{2}(-2i - j)$ which is in the opposite direction. I'm not sure why it's the wrong direction, my working is:

P = (100, 1)
$\displaystyle \frac{dh}{dx} = \frac{-4x^3}{10^8}$
$\displaystyle \frac{dh}{dy} = \frac{-2y}{10^2}$

$\displaystyle \nabla f(x, y) = \frac{-4x^3}{10^8}i + \frac{-2y}{10^2}j$

When I subbed in the point, I got my answer. Where am I getting the signs wrong?

For b), I have no idea how to attempt it. I think I'm missing some intuition here. Any ideas?

Thanks.
• Apr 6th 2011, 04:00 PM
Glitch
No ideas?
• Apr 6th 2011, 04:17 PM
NOX Andrew
My answer for part (a) is slightly different.

$\displaystyle h_x = -\dfrac{4x^3}{10^8} \implies h_{x}(100,1) = -\dfrac{4}{10^2}$

$\displaystyle h_y = -\dfrac{2y}{10^2} \implies h_{y}(100,1) = -\dfrac{2}{10^2}$

$\displaystyle \nabla h(100,1) = -\dfrac{4}{10^2} \mathbf{i} - \dfrac{2}{10^2} \mathbf{j} = -\dfrac{2}{10^2}\left(2\mathbf{i} + \mathbf{j}\right) = -\dfrac{1}{50}\left(2\mathbf{i} + \mathbf{j}\right)$

• Apr 6th 2011, 05:03 PM
Plato
Quote:

Originally Posted by Glitch
No ideas?

Do you realize that any form of 'bumping" is out-lawed at this site?
You can be banded for doing so.
• Apr 6th 2011, 05:34 PM
Glitch
Quote:

Originally Posted by Plato
Do you realize that any form of 'bumping" is out-lawed at this site?
You can be banded for doing so.

Really? That's a bit harsh. All I wanted to know is whether people had even considered helping. If the question is too difficult, I can ask elsewhere.
• Apr 6th 2011, 08:28 PM
minifhncc
Quote:

Originally Posted by Glitch
When I subbed in the point, I got my answer. Where am I getting the signs wrong?

Your calculations show that the skier starts moving UP the mountain in the direction of the vector (-2,-1). ie. (2,1) DOWN the mountain.

Quote:

For b), I have no idea how to attempt it. I think I'm missing some intuition here. Any ideas?
For (b): dx/dt = k 4x^3/10^8, dy/dt= k 2y/10^2.

So dy/dx=10^6 y / 2x^3

Integrate etc and you should end up with y=e^((-2.5 *10^5)/x^2 + 25)...