Recent content by alfredska

That looks correct.

After "This gives:", you need to check your math. I see (1-\lambda)^2(-1-\lambda)-4(1-\lambda)-4(1-\lambda) To find eigenvectors, see if this resource helps: http://www.sosmath.com/matrix/eigen2/eigen2.html Essentially, you will set up A \left(\begin{matrix} x \\ y \\ z...

The vector corresponding to NW is \hat{r}=\{-1/\sqrt{2},1/\sqrt{2}\}. Then the instantaneous rate of change in that direction, starting from point (8,6) is: \vec{\nabla}A(8,6)\cdot \hat{r} = ?

Great, so you can either state these in vector format: \hat{r}=-(3/5)\hat{x}+(4/5)\hat{y} (and similarly for -\hat{r}), or in angular format as measured from the positive x-axis.

No, I suggest that your answer is wrong. So far, I agree with r_1=-(3/4)r_2, but you somehow go awry when plugging this into r_1^2+r_2^2=1.

Check your math.

Check your math.

Your vector's length is |\hat{r}|=\sqrt{r_x^2+r_y^2}=1. Square both sides and 1^2=1.

So that's one equation in two unknowns. Now use that \hat{r} is a unit vector: r_x^2+r_y^2=1. Then you have two equations and two unknowns.

Where is the equal sign? And why are there two dx's?

See if this contour plot of A(x,y) helps. The blue vector represents the unit vector pointed in the direction of \vec{\nabla}A(x,y). The black vectors represent unit vectors in the direction of no ascent - they correspond to each \hat{r} you need to find.

I'm just throwing a guess in here, but if your goal is to find the velocity of something which is under the influence of a non-constant gravitational field, integrating its acceleration is not the easiest way to go. You would typically apply conservation of energy between initial and final...

The length of \vec{r} is of no concern, only its direction matters. In fact, here's what you really need to solve: \vec{\nabla}A(8,6)\cdot\hat{r}=0. Is this less ambiguous?

Now that you have \vec{\nabla}A(8,6), dot it with \vec{r}=\{r_x,r_y\}. In solving for r_x and r_y, you can use r_x^2+r_y^2=1.

I see no attachment.