A function that is useful in studying the air flow over mountains is

$$h(x,y) = \frac{h_0}{[(\frac{x}{a})^2+(\frac{y}{b})^2+1]^\frac{3}{2}} $$

where $h_0$, a, and b are all positive constants.

(a) Find $\nabla h$.

(b) Find the directional derivative of the point (x,y) in the direction of the vector $v=(v_1,v_2)$

(c) At what point(s) is the directional derivative equal to zero?

For part a), I found that $\nabla h$ is equal to $$\frac{-3h_0}{[(\frac{x}{a})^2+(\frac{y}{b})^2+1]^\frac{5}{2}}\langle\frac{x}{a^2},\frac{y}{b^2}\rangle $$

Using this, I found that for part b), I got that $D_uh(x,y) = \frac{-3h_0}{[(\frac{x}{a})^2+(\frac{y}{b})^2+1]^\frac{5}{2}}\langle\frac{x}{a^2},\frac{y}{b^2}\rangle \bullet\frac{1}{\sqrt{v_1^2+v_2^2}}\langle v_1, v_2 \rangle$

which simplifies to $D_uh(x,y) = \frac{-3h_0}{[(\frac{x}{a})^2+(\frac{y}{b})^2+1]^\frac{5}{2}{\sqrt{v_1^2+v_2^2}}}(\frac{xv_1}{a^2}+\frac{yv_2}{b^2})$

First of all, is this correct?

Secondly, given that I am correct up to this point, I am unsure on how to do part c). How can I solve for an x and y where $D_uh = 0$ when I do not have values for $v_1$ and $v_2$?

Any help here is appreciated.

  • $\begingroup$ I suposse you have to solve the equation $\frac{xv_{1}}{a^{2}}+\frac{yv_{2}}{b^{2}}=0$ $\endgroup$ – julio godoy Jun 23 '16 at 6:36
  • $\begingroup$ and the vector solution is $(-ya^{2},xb^{2})$, if the vector solution has unit lenght you have to normalize that vector $\endgroup$ – julio godoy Jun 23 '16 at 6:44
  • $\begingroup$ @juliogodoy Wouldn't that solution require that $v_1 = -v_2$? if you plug in those x and y values in the equation, you get $-yv_1 + xv_2 = 0$. What am I missing here? $\endgroup$ – dibdub Jun 23 '16 at 6:46
  • $\begingroup$ you can use also the fact that $u$ and $v$ are perpendicular then $u\cdot v = 0$ and if $u = (u_1,u_2)$ then $v=(-u_2,u_1)$ $\endgroup$ – Navaro Jun 23 '16 at 6:51
  • $\begingroup$ In that case wouldnt the answer be $(\frac{-yv_2}{b^2},\frac{xv_1}{a^2})$? *edit: Actually it would be $(\frac{-v_2}{b^2},\frac{v_1}{a^2})$. Is that correct? $\endgroup$ – dibdub Jun 23 '16 at 6:54

The first two answers are correct.

For the third question: the directional derivative is equal to $0$ if $v =(v_1,v_2)$ is perpendicular to $\nabla h$ , so: $$v\cdot \nabla h = 0 \Rightarrow \dfrac{v_1}{a^2}x+\dfrac{v_2}{b^2}y = 0 \Rightarrow y = -\left( \dfrac{b}{a} \right)^2\left(\dfrac{v_1}{v_2}\right)x\Rightarrow y=cx \quad\text{where } c=-\left( \dfrac{b}{a} \right)^2\left(\dfrac{v_1}{v_2}\right)$$ which is the equation of a straight line passing through the origin $O=(0,0)$

conclusion: every point $(x,y)$ satisfying the previous equation makes the directional derivative in the direction of $v=(v_1,v_2)$ equal to zero

  • $\begingroup$ Perfect, that makes sense. Thanks for all your help! $\endgroup$ – dibdub Jun 23 '16 at 7:47

The condition $x\frac{v_1}{a^2}+y\frac{v_2}{b^2}=0$ could be written $(P-O)\cdot\mathbf{u}=0$, where $P=(x,y)$, $O=(0,0)$ and $\mathbf{u}=\langle\frac{v_1}{a^2},\frac{v_2}{b^2}\rangle$.
This is the equation of the straight line passing through $O$ and orthogonal to $\mathbf{u}$.

The directional derivative is $0$ on each point of this line.

  • $\begingroup$ Could you elaborate on this? Why is it (P-O)? What does P represent? How would I present my answer for part c) then? $\endgroup$ – dibdub Jun 23 '16 at 6:45
  • $\begingroup$ @dubbler26: edited. $\endgroup$ – enzotib Jun 23 '16 at 7:03

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