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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

Best Answer 

 #2
avatar+20598 
+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
 #1
avatar+27223 
+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
avatar+20598 
+5
Best Answer

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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