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**seit****Member**- Registered: 2010-11-13
- Posts: 2

Suppose you are climbing a hill whole shape is given by the equation z=1000 - 0.005x^2 - 0.01y^2, where x, y and z are mesured in meters, and you are standing at a point with coordinates (60, 40, 966). The positive x-axis points east and the positive y-axis points north.

(a) If you walk due south, will you start to ascend or descend? At what rate?

(b) If you walk northwest, will you start to ascend or descend? At what rate?

(c) In which direction is the slope largest? What is the rate of ascent in that direction? At what angle above the horizontal does the path in that direction begin?

Please help me! Thanks.

*Last edited by seit (2010-11-13 05:39:44)*

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**seit****Member**- Registered: 2010-11-13
- Posts: 2

No one can help me! I'm disappointed.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,646

Hi seit;

seit wrote:

No one can help me! I'm disappointed.

Currently, I am the only one in here and there are other questions. I am swamped at the moment.

This is what you can be doing to help yourself instead of being disappointed at me or the forum. I am doing the best I can.

I presume you want the x axis running from left to right and the y axis up and down.

Since there are very general questions, you can be getting them for yourself. For instance the maximum value is 1000 and occurs at (0,0). Easy to see just from the function.

seit wrote:

a) If you walk due south, will you start to ascend or descend?

To be walking south in your system means for the y coordinate to be decreasing and the x coordinate to remain constant.

Obviously you are ascending until you get to (60,0). Then what happens?

seit wrote:

If you walk northwest, will you start to ascend or descend?

To walk Northwest means x is decreasing as y is increasing, both at the same rate. So you have this table.

Looks like you are descending.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,646

Hi;

The above work demonstrates the easiest parts of your questions could be answered with no math at all. Just a little computation and some effort.

My brother was kind enough to explain to me how this is done using vector analysis. He says:

jimmyR wrote:

This is a rather common book problem and is all over the internet. You both should really know how to do this one.

I will try to remember what he told me and do the problem by myself.

The gradient is the vector that points in the direction of steepest ascent or descent. It is computed by:

Taking partials of f(x,y) we get:

The vector is:

Now dot product g with the unit vector for south [0,-1]

It is positive you are going uphill as I already showed you with a gradient of .8

Now dot product g with the unit vector for northwest [-1/sqrt(2), 1/sqrt(2)].

So you are going downhill with a gradient of -.141421

For c) The slope is largest in the direction of the gradient g. The rate ascent is the norm of g,

It is positive so that is the steepest ascent. To get the angle with the horizontal you just take the arctan

of the norm of g which is 1.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**mano25****Member**- Registered: 2011-01-11
- Posts: 2

thanks for sharing useful information with us.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,646

Hi mano25;

Welcome to the forum and your welcome.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

**Online**

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