Finding direction where the rate of change is fastest

munkhuu1
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Homework Statement


surface of mountain is modeled by h(x,y)=5000-0.001x^2-.004y^2. Climber is at (500,300,4390) and he drops bottle which is at (300,450,4100)
What is his elevation rate of change if he heads for the bottle?
In what direction should he proceed to reach z=4100 feet the fastest, so that he can be on a constant elevation to reach his bottle?

Homework Equations





The Attempt at a Solution


i solved for first question and got -.334 where i just found gradient and multiplied by unit vector
and for the 2nd question I am little confused.
 
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The gradient is a vector and contains two pieces of information- its "length" and its direction. In which direction does the gradient here point?
 
is the direction just gradient of (500,300) which is <-1,-2.4>
or do i find the difference of the 2 points and find gradient of (200,-150) = <-.6,-1.2>
 
I'm sorry, I assumed from your saying that you had found the gradient that you knew what "gradient" meant!

The gradient of the function f(x,y,z) is the vector
\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}
Isn't that what you did? The "direction" you want to go is the direction of that vector. It may be, what you wrote was ambiguous, that you want a compass direction, not including the "downward" part. If that is the case, find the angle that
\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j} makes with "north", the positive y-axis.
 
HallsofIvy said:
I'm sorry, I assumed from your saying that you had found the gradient that you knew what "gradient" meant!

The gradient of the function f(x,y,z) is the vector
\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}
Isn't that what you did? The "direction" you want to go is the direction of that vector. It may be, what you wrote was ambiguous, that you want a compass direction, not including the "downward" part. If that is the case, find the angle that
\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j} makes with "north", the positive y-axis.

yea i did found the gradient which was <-1,-2.4> i just wasnt sure how to get direction from this.
 
What do you mean by "direction"? A direction in three dimensions is given by either a unit vector or by the "direction cosines" (the cosines of the angles a line in that direction makes with each coordinate axis which is the same as the components of the unit vector).

But, as I said, you may want the two dimensional compass direction the person should take.
 
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