slope of hill

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August 17th 2008, 09:28 PM
sk1001

slope of hill

The height of a certain hill (in meters) is given by
h(x,y)
= 10 ( 2x y - 3 x^2- 4 y^2- 18 x + 28 y + 12),
where y is the distance (in kilometers) North, x the distance East

(a) How steep is the slope (in meters per kilometer) at a point 1 km north and 1 km east and in what direction is the slope the steepest at that point?

I know this has something to do with partial derivatives, but I am not 100% sure of the step proces..
August 17th 2008, 10:01 PM
mr fantastic

Quote:

Originally Posted by sk1001

The height of a certain hill (in meters) is given by

h(x,y)

= 10 ( 2x y - 3 x^2- 4 y^2- 18 x + 28 y + 12),

where y is the distance (in kilometers) North, x the distance East

(a) How steep is the slope (in meters per kilometer) at a point 1 km north and 1 km east and in what direction is the slope the steepest at that point?

I know this has something to do with partial derivatives, but I am not 100% sure of the step proces..

(a) $\nabla f \cdot \hat{l}$ evaluated at (1, 1) where $\hat{l}$ is a unit vector in the given direction.

(b) $\nabla f$ evaluated at (1, 1).

I suggest you revise your notes on the directional derivative.
August 17th 2008, 10:48 PM
sk1001

I've calculated grad (http://www.mathhelpforum.com/math-he...fb9e51b8-1.gif), should I dot with the unit vector before evaluating, or should I evaluate grad at 1,1 then dot with unit vector?
August 17th 2008, 11:55 PM
mr fantastic

Quote:

Originally Posted by sk1001

I've calculated grad (http://www.mathhelpforum.com/math-he...fb9e51b8-1.gif), should I dot with the unit vector before evaluating, or should I evaluate grad at 1,1 then dot with unit vector?

I'd first evaluate $\nabla f$ at (1, 1) ....
August 18th 2008, 12:20 AM
sk1001

Quote:

Originally Posted by mr fantastic

I'd first evaluate $\nabla f$ at (1, 1) ....

I tried and returned an answer of 0, somehow I'm not convinced this is correct!

I found:
partial derivative of h with respect to x to be 20y - 60x - 180
and
partial derivative of h with respect to y to be 20x - 80y + 280

Evaluated at (1,1) these are -220 and 220 respectively.

Then the at (1,1) is (1/sqrt(2), 1/sqrt(2)).

Dotting gives (-220)(1/sqrt(2)) + (220)(1/sqrt(2)) yielding 0!

Where have I gone wrong?