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Hill Climb

Suppose you climb a hill whose shape is given by z=1400-0.09x^2-0.02y^2 and you are at the point (100,100,300)?

In which direction (unit vector) should you proceed initially in order to reach the top of the hill fastest?

The equation of the curve is
(0.09)x^2 + (0.02)y^2 + z = 1400
Equation of a plane tangential to the point (100, 100, 300) is
9x + 2y + (1/2)(z + 300) = 1400
=> 18x + 4y + z = 2500.

Let a plane perpendicular to it and passing through (100, 100, 300) and (0, 0, 1400) be
ax + by + z = c
(0, 0, 1400) lies on it
=> c = 1400
=> ax + by + z = 1400.
(100,100,300) lies on it
=> a + b + 3 = 14
=> b = 11 - a
18x + 4y + z = 2500 is perpendicular to ax + by + z = 1400
=> 18a + 4b + 1 = 0
=> 18a + 4(11 - a) + 1 = 0
=> 14a = - 45
=> a = - 45/14 and b =199/14
=> plane of geodesic is
=> 45x - 199y - 14z = - 19600.

Direction of the line of intersection of the planes
18x + 4y + z = 2500 and 45x – 199y – 14z = -19600 is the initial direction of the path
=> Initial direction is along
(18, 4, 1) x (45, -199, -14)
= (143, 297, - 3762)
=> initial direction towards +ve z-axis
= (-143, -297 + 3762)
Unit vector along the initial direction
= 1/√(143^2 + 297^2 + 3762^2) * (-143, -297, 3762)
= (- 0.0379, - 0.0786, 0.9962).

Angle between this unit vector and the xy-plane, z = 0 having its normal n = (0, 0, 1)
= 90° - arccos[(0, 0, 1) . (-0.0379, - 0.0787, 0.9962)
= 90° - arccos(0.9962)
= 90° - 5°
= 85°.

Edit:
I got 2 thumbs down indicating some error. I could not find any on rechecking. I shall appreciate anyone telling me the error and posting the correct solution.

Edit:
I had posted the above answer inspired by Dr. D, who deleted his answer suspecting it to be incorrect. As I got 2 thumbs down, I requested Mr. Duke to check my answer for a possible error. Both Dr. D and Mr. Duke are my valuable contacts and prominent contributors to the challenging problems in YA. Mr. Duke replied as under not being sure about what is wanted by the asker.
"I am not quite sure what exactly the Asker wants by 'to reach the top of the hill fastest?'. The surface is an elliptic paraboloid:

http://en.wikipedia.org/wiki/Elliptic_paraboloid

opened downwards, the top is (0,0,1400). You have considered the shortest arc on the surface (geodesic), connecting (100,100,300) with the top, but I suspect the question is about the direction of the steepest ascent:

http://en.wikipedia.org/wiki/Gradient_descent

given simply by the gradient in the initial point:

http://en.wikipedia.org/wiki/Gradient#Interpretations

We have grad(z) = (-0.18x, -0.04y), which value in (100,100,300) is (-18,-4), collinear to the unit vector (-9/sqrt(85), -2/sqrt(85)) - perpendicular to the ellipse in the plane z=1100. The gradient is not constant, it slowly changes its direction finally pointing the origin (see the illustration of gradient ascent/descent in the Article). So I suppose the above unite vector is the answer to the question of the desired direction, and the required angle is complement to the angle between both normals:
(18,4,1) and (0,0,1) - approx. 87 deg. according Windows Calculator - similarly to Your calculation.

Once again, I am not sure, the time is about to expire, I checked my mail somewhat late too, maybe it would be better to email the Asker personally to give a precise formulation and to extend the answer period.

I would be very glad if the above can help You in any way to decide."

Edit:
Mr. Duke raised a doubt that the asker might have asked the unit vector and angle in the direction of the steepest ascent. I had sent an email to the asker and am awaiting reply. On re-reading the question, the word "fastest" means "shortest" path which is geodesic path.

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