A spherical alien spaceship lands at rest on a Earth surface given by z = 2x^2 −3y^2 at a point A(2,1,5). Find the unit vector of the direct

I've found the following on-line 3d graphing facility that seems to be pretty good (though I had some difficulty controlling the orientation with the rotation tool!).  It can be found at: http://web.monroecc.edu/manila/webfiles/calcNSF/JavaCode/CalcPlot3D.htm 

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Small changes $$\delta x$$ and $$\delta y$$ in the x,y co-ordinates will bring about a small change $$\delta z$$ in the z co-ordinate given by

$$\delta z \approx \frac{\partial z}{\partial x}\delta x+ \frac{\partial z}{\partial y}\delta y.$$

So, $$\delta z \approx 4x \delta x - 6y \delta y.$$

We need $$\delta z$$ to be negative, in which case we need $$\delta x$$ to be negative and $$\delta y$$ to be positive. So, the move needs to be into the second quadrant of the xy plane.

If then we move a small distance $$h$$ at an angle $$\theta$$ to the negative x-axis, $$\delta x = -h\cos \theta,$$ and $$\delta y = h \sin \theta,$$ so $$\delta z \approx -h(8\cos\theta+6\sin\theta).$$

This will be a maximum when $$\tan \theta = 3/4$$ from which it follows that a vector in the required direction will be $$-4\hat{\imath}+3\hat{\jmath}$$ and a unit vector will be that divided by 5.

That looks impressive Bertie but where is my picture 

I did graph it and I got a very interesting landscape but I didn't know what to do with A.

I didn't find a 3D grapher that would plot both the graph and the point as well.

I'd really like to see them together.   Maybe I could get a 'feel' for the question that way.

That is hopeful thinking that is :)

Actually, this graph is not too bad:)

Here is a graph of the terrain around landing spot.  :)

Where did you find that grapher, Melody ????   ....pretty neat...

I just googled it    

Here is this one

http://www.math.uri.edu/~bkaskosz/flashmo/graph3d/

Here's a picture Melody:

The copper colour is the z = 2x² - 3y² surface.  The red blob is the point A = (2, 1, 5).  The blue mesh is the plane tangential to the surface at A (z = 8x - 6y - 5).  

The black line is the unit length vector lying in the tangential plane in the direction of steepest descent.

I took Bertie's solution for the vector in the horizontal plane, found where the end of this would meet the tangential plane, scaled the resulting vector from A to this point to be of unit length and plotted the black line to represent the unit vector in the tangential plane.

The black line goes from (2, 1, 5) to (1.92, 1.06, 4.01)  (approximately).

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Thank you Alan.

What graphing program did you use?

I used the graphing facilities in MATLAB.  Unfortunately, this is a commercial program, not free open-source.

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Thanks Alan,

I almost missed this post - how did that happen? 

Thanks for finding this grapher.  It looks really neat.    

I'll add that 3D grapher to our reference material.  I'll add mine too.  I only just thought to do that.