Gradients/Laplacians.. Hill Height problem:

[mooeypoo]

Hey again guys,

 

My second physics course also deals a lot with some math I haven't really dealt with before. This time, though, it's a bit more practical question rather than the 'principle'. I actually understand the principle, but the question doesn't work for me - I think I got confused somewhere, and the book's not much help.

 

My professor is writing ineligible script on the board and in the hw assignments, apparently he forgot that there are non-Americans in the course (I'm nto the only one) who have trouble reading script. Damnit.

 

In any case, here's the problem (hopefully I translated the script right, took me a while):

 

The height of a certain hill (in feet) is given by

 

[math]h(x,y)=10(3xy-4x^2-3y^2+28x-17y+12)[/math]

 

a. Where is the top of the hill located?

b. How high is the hill?

c. How steep is the slope (in feet per mile) at a point 1 mile north and 1 mile east of the town. In what direction is the slope steepest, at that point?

So, I know this is a question about gradients; the magnitude of the gradient will give the height, and the direction of the gradient will give the direction of the maximum increase (max slope of the hill).

So, I tried calculating the gradient:

 

Top of the hill is:

[math]\bigtriangledown h=0[/math]

 

So:

[math]\frac{\partial h}{\partial x} = 10(3y-8x+28)[/math]

 

[math]\frac{\partial h}{\partial y} = 10(3x-6y-17)[/math]

 

[math]\bigtriangledown h = 10((3y-8x+28)\hat{x} + (3x-6y-17)\hat{y}) [/math]

 

But... I have 2 variables.. when I equated this to 0, I couldn't find the actual coordinates..

 

Help? What am I doing wrong here?

[timo]

Nothing till now (I've not checked the numbers, though) except the last step:

[math]\bigtriangledown h = 10((3y-8x+28)\hat{x} + (3x-6y-17)\hat{y}) = 10 \left( \begin{array}{c} 3y-8x+28 \\ 3x-6y-17 \end{array} \right) \stackrel{!}{=} \vec 0 = \left( \begin{array}{c} 0 \\ 0 \end{array} \right)[/math].

[Bignose]

[math]\bigtriangledown h=0[/math]

 

Basically, what Atheist said -- the root of the issue is this equation which is written wrong.

 

it should be:

 

[math]\nabla h = \mathbf{0}[/math]

 

A vector ([math]\nabla h[/math]) must, must, must be equal to another vector ([math]\mathbf{0}[/math]) -- in this case the zero vector. You have a vector set equal to a scalar which is essentially meaningless.

 

By setting it equal to the zero vector, you get two equations in two unknowns which should be solvable.

 

p.s. you may want to use \nabla in the LaTeX code instead of \bigtriangledown -- the nabla lines up with the text nicer. It's a little thing, but it is something that I noticed as soon as I read you post.