Help me with a ski slope!

[OrangeOne]

A ski slope is described by the graph to the function h(x,y)= 10 - (x^2+ 2y^2). You are located on the point in the slope where (x,y)=1,1

To which direction should you point your skis ...
a) if you want to stand still and enjoy the view?
b) if you want to start going downhill as steep as possible?

I think it has to do with the direction derivative and gradient but for that I would need the direction-points I think..which I dont have...so this leaves me lost...
Please help!

 

[galactus]

The direction of max increase is \(\displaystyle {\nabla}z\)

The direction of min increase is \(\displaystyle -{\nabla}z\)

\(\displaystyle z=10-x^{2}-y^{2}\)

\(\displaystyle {\nabla}z=-2x\cdot i-2y\cdot j\)

To stand and watch, the skier can not ascend or descend, so there will be no change in directions perpendicular to the gradient.

That is, the direction of the gradient gives the slope of the line that is normal to the curve at (1,1).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

To descend down the steepest part, then the skier will go in the direction where the slope is falling at the fastest rate.

That is, the skier will ski in the direction in which the rate of change of z is minimum.

This is the direction opposite to \(\displaystyle {\nabla}z\). In the direction of \(\displaystyle -{\nabla}z\)

Therefore, since they are at (1,1), they will ski in the direction \(\displaystyle -{\nabla}z=2i+2j\).

 

[OrangeOne]

Thank you galactus, I understand the second part about going down the steepest part.

However, I dont understand which direction one needs to be pointing only to stand and watch.

Should I just plug in (1,1) into -2x, -2y to get the point?

The direction of max increase is \(\displaystyle {\nabla}z\)

The direction of min increase is \(\displaystyle -{\nabla}z\)

\(\displaystyle z=10-x^{2}-y^{2}\)

\(\displaystyle {\nabla}z=-2x\cdot i-2y\cdot j\)

To stand and watch, the skier can not ascend or descend, so there will be no change in directions perpendicular to the gradient.

That is, the direction of the gradient gives the slope of the line that is normal to the curve at (1,1).

I don't understand the last part. I know the gradient is always perpendicular to the curve, but I do not really understand what this means and what importance it has in practice. Thank you for helping me with this task. I really appreciate it.

 

[galactus]

Sorry, I did not explain myself very well.

Presume that the positive y axis in north and the positive x axis is east.

The skier will travel a level path in a direction perpendicular to \(\displaystyle {\nabla}z=-2xi-2yj\).

Find unit vectors in these directions. Then, the angle they make with the y axis can be found by using arc cos.


Example:

\(\displaystyle z=2000-2x^{2}-4y^{2}\) ft.

say \(\displaystyle {\nabla}z=80i-40j\) at (-20,5)

Unit vectors in directions are \(\displaystyle \pm\frac{(i+2j)}{\sqrt{5}}\)

They make an angle with the positive y axis of \(\displaystyle cos^{-1}(\frac{2}{\sqrt{5}})\approx 27^{o}\)

and the same angle with the negative y-axis.

Therefore, the bearings are \(\displaystyle N27E, \;\ S27W\)

 

[OrangeOne]

Hmm..thanks...but Im ashamed to say I still don't understand how to find the unit vectors...

 

[galactus]

This problem models a concentric ellipse. If you graph it you have an ellipse-shaped mountain.

If you are standing on a mountain looking up the steepest direction, then the steepest direction down is right behind you and the path around

the mountain, the level curve, is to your side at right angles to the "steepest" path. Like walking around the side of a hill.

The path around that is level and perpendicular to the steepest part.

So, say the path of steepest ascent is \(\displaystyle -2i-2j\), then the directional derivative will be 0 when it is perp. to this.

The unit vectors give the directional derivative from a point toward another point.

Implicitly differentiate z, we get \(\displaystyle \frac{dy}{dx}=\frac{-x}{y}\). Using (1,1), we get \(\displaystyle \frac{dy}{dx}=-1\)

The skier is standing at (1,1). A tangent vector is then (1,-1).

Note, if you sub these into z, you get the same elevation as (1,1). Thus, she is standing at the same elevation.

The unit vectors are \(\displaystyle u=(\frac{1}{\sqrt{2}}, \;\ \frac{-1}{\sqrt{2}})\).

The direction of steepest ascent is (-2,-2).

Because the dot product of steepest ascent and the unit vectors is 0, they are perpendicular.

\(\displaystyle {\nabla}z\cdot u=0\)