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# A flat plane on a 3D cartesian grid is described by the equation z = 2.73x - 1.51y - 1.99

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A flat plane on a 3D cartesian grid is described by the equation z = 2.73x - 1.51y - 1.99

• If you were standing at (x,y) = (1.3,4.8), in which horizontal direction is the steepest slope DOWNWARDS?
• In that direction, what is the absolute angle between the surface and the horizontal plane?

Consider the positive x-axis to be zero degrees, the positive y-axis to be 90 degrees, the negative x-axis to be 180 degrees etc.

• Too dec places please. thanks again guys

Direction of steepest descent
(degrees)
Absolute angle of steepest descent
(degrees)

Oli96  Sep 12, 2014

#2
+19076
+5

If you were standing at (x,y) = (1.3,4.8), in which horizontal direction is the steepest slope DOWNWARDS  ?

In a flat plane:  The gradient is independent from the place ( co-ordinates ).

The gradient points in every point of the space in the direction of the strongest increase.

f(x,y) z = 2.73x - 1.51y - 1.99

$$\dfrac{\partial f(x,y)}{\partial x} = 2.73 \quad and \quad \dfrac{\partial f(x,y)}{\partial y} = -1.51\\ Gradient = \left( \begin{array}{r} 2.73 \\ -1.51 \end{array} \right)\\$$

horizontal direction is the steepest slope UPWARDS  =

$$\tan^{-1}(\frac{-1.51}{2.73}) = -28\ensurement{^{\circ}}.9475759928 + 360 \ensurement{^{\circ}} =331\ensurement{^{\circ}}.052424007$$

horizontal direction is the steepest slope DOWNWARDS  = $$331\ensurement{^{\circ}} .052424007 - 180\ensurement{^{\circ}} = 151\ensurement{^{\circ}}.052424007$$

Consider the positive x-axis to be zero degrees, the positive y-axis to be 90 degrees, the negative x-axis to be 180 degrees etc.

In that direction, what is the absolute angle between the surface and the horizontal plane?

slope: $$\tan^{-1}( \sqrt{1.51^2+2.73^2 } ) = \tan^{-1}(3.11977563296) = 72\ensurement{^{\circ}}.2274815207$$

heureka  Sep 12, 2014
Sort:

#1
+26547
+5

The rate of change of z with respect to x is just 2.73, which is positive, so the slope in the positive x-direction is up, and in the negative x-direction is down.

The rate of change of z with respect to y is -1.51 which is negative, so the slope in the positive y-direction is down and in the negative y-direction is up

Go with heureka's answer - I've changed my mind on this twice already!!

Alan  Sep 12, 2014
#2
+19076
+5

If you were standing at (x,y) = (1.3,4.8), in which horizontal direction is the steepest slope DOWNWARDS  ?

In a flat plane:  The gradient is independent from the place ( co-ordinates ).

The gradient points in every point of the space in the direction of the strongest increase.

f(x,y) z = 2.73x - 1.51y - 1.99

$$\dfrac{\partial f(x,y)}{\partial x} = 2.73 \quad and \quad \dfrac{\partial f(x,y)}{\partial y} = -1.51\\ Gradient = \left( \begin{array}{r} 2.73 \\ -1.51 \end{array} \right)\\$$

horizontal direction is the steepest slope UPWARDS  =

$$\tan^{-1}(\frac{-1.51}{2.73}) = -28\ensurement{^{\circ}}.9475759928 + 360 \ensurement{^{\circ}} =331\ensurement{^{\circ}}.052424007$$

horizontal direction is the steepest slope DOWNWARDS  = $$331\ensurement{^{\circ}} .052424007 - 180\ensurement{^{\circ}} = 151\ensurement{^{\circ}}.052424007$$

Consider the positive x-axis to be zero degrees, the positive y-axis to be 90 degrees, the negative x-axis to be 180 degrees etc.

In that direction, what is the absolute angle between the surface and the horizontal plane?

slope: $$\tan^{-1}( \sqrt{1.51^2+2.73^2 } ) = \tan^{-1}(3.11977563296) = 72\ensurement{^{\circ}}.2274815207$$

heureka  Sep 12, 2014

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