What Is the Steepness of the Hill at a Specific Point?

[oddiseas]

Homework Statement

The height of a certain hill is given by h(x,y)=10(3xy-4x^2-2y^2-11x+17y+9)

(a) How steep is the slope in the x and in the y direction (in meters per kilometer) at the point 1 km North and 1 km
East of Dandenong?
(b) In what direction is the slope the steepest at that point and how steep is the slope in that
direction?

Homework Equations

b)grad(1,1)=-160i-160j
|grad(1,1)|=226km/km

now for part a what does it mean when they ask how steep is the slope in the x and y direction?
grad(1,1)=-160i-160j

are they asking for the co-ordinates? or the magnitude of each separate component?
anyway this has me a bit confused.

The Attempt at a Solution

 

[AJ Bentley]

Presumably, you are also told where Dandenong is in relation to the hill?

 

[oddiseas]

The height of a certain hill is given by h(x,y)=10(3xy-4x^2-2y^2-11x+17y+9)
where y is the distance north of dandenong and x is the distance east of dandenong.

 

[AJ Bentley]

OK, the question is asking about the gradient in the x direction and the gradient in the y direction at the point x=1, y=1.
This is clearly an exercise in partial differentials.

I have no idea what your 'relevant equations' are - they don't look relevant at all (EDIT - in fact they look like the answer values).
To obtain the gradient in x, you simply differentiate with respect to x, treating y as a constant. Similarly for y.
Then you plug in the values for the point (1,1) to those differentials. The result is the gradient (slope) in the direction chosen.

Then having obtained those values, you are asked to find the direction of maximum slope at that point and give a value to that slope.

 

[paranormal]

The slope of a hill can be described by its gradient, which is a vector that represents both the direction and magnitude of the steepest increase in elevation. In this case, the gradient of the hill at the point 1 km North and 1 km East of Dandenong can be calculated using the given height function.

(a) To find the slope in the x and y directions, we can take the partial derivatives of the height function with respect to x and y, respectively. This will give us the slope in meters per kilometer in each direction.

h_x(x,y) = 30y - 40x - 11
h_y(x,y) = 30x - 4y + 17

At the point (1,1), these partial derivatives become:

h_x(1,1) = 30 - 40 - 11 = -21
h_y(1,1) = 30 - 4 + 17 = 43

Therefore, the slope in the x direction is -21 meters per kilometer, and the slope in the y direction is 43 meters per kilometer.

(b) The direction of the steepest slope can be found by taking the gradient of the height function and finding its magnitude.

grad(x,y) = (h_x(x,y), h_y(x,y))
= (-21, 43)

The magnitude of this vector is given by:

|grad(x,y)| = √((-21)^2 + (43)^2) = √(441 + 1849) = √2290 = 47.85

Therefore, the steepest slope at the point (1,1) is in the direction of the vector (-21, 43), which has a magnitude of 47.85 meters per kilometer.