How to Calculate Unit Vector for Mountain Contourlines?

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Homework Help Overview

The discussion revolves around calculating a unit vector for walking downslope along contour lines of a mountain, given a specific gradient function and a starting point. The subject area involves concepts from calculus, particularly related to gradients and derivatives.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster seeks hints on determining the unit vector for walking at a 45-degree angle downslope. One participant identifies the problem as involving partial derivatives and suggests using the dot product to relate the gradient to the desired slope. There is also a discussion about the classification of the problem within precalculus versus calculus.

Discussion Status

Participants are exploring different interpretations of the problem and clarifying terminology related to calculus. Some guidance has been offered regarding the mathematical relationships involved, but no consensus has been reached on the approach to take.

Contextual Notes

There is a mention of differing educational terminology regarding calculus, which may influence the understanding of the problem's classification. The original poster expresses uncertainty about the terminology used in their country.

hexa
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I'll be greatful for any hint.

Imagine you walk over the contourlines of the map of a mountain (really! that's the question) with a gradient of h(x,y)=2xy, x^2). You are at point (1,3) and you want to walk downslope at an angle of 45 degrees. calculate the unit vector in order to find out in which direction to walk.

Hexa
 
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What's this doing in precalculus? This is a partial derivative problem. If the gradient is (2xy, x2), then the derivative in the direction of unit vector (a, b) is just the dot product, 2xya+ x2b. If you are walking downslope at an angle of 45 degrees, then the slope must be tan(45)= 1. You want 2xya+ x2b= 1 and, of course, a2+ b2= 1. Solve for a and b in terms of x and y.
 
Thanks a lot for your help. I'll work on with this.

Calculus: not a term used in this country, so I'm not quiet sure where the border is between precalculus and calculus.
 
Then you might say "analysis" or "applied analysis". Essentially, derivatives and integrals are calculus. The basics of limits might be in calculus or pre-calculus.
 

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