Multivariable calculus

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Homework Statement


The height of a hill is given as the following:
[tex]
h \left( x,y \right) =40\, (\left( 4+{x}^{2}+3\,{y}^{2} \right) ^{-1})
[/tex]

There's a stream passes the point (1,1,5) which is on the surface of h. The stream follows the steepest descent. Find the equation of the stream.

Homework Equations


I take this is relevant to the tangent hyperplane of a surface.

Tangent plane at point a is: f(a)+gradient(f(a)) dot (x-a)


The Attempt at a Solution


I think the path of the stream is the intersection of the surface h and a plane orthogonal to the tangent plane at point (1,1,5), but I'm not sure.
 

Answers and Replies

  • #2
Dick
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The tangent plane has many different orthogonal planes. You are thinking in too many dimensions. Just regard h as a function of x,y. Then the gradient of h will point in the direction of most rapid change of h.
 
  • #3
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So I take the gradient of h at point (1,1), which gives me grad(a)=(-10,-30). But the path of the stream isn't going to be a straight line, is it? How do I get the equation of the stream using only the gradient and the point at which the gradient is taken?
 
  • #4
Dick
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No, it isn't going to be a straight line. Can you see how to use the gradient vector to compute the slope of steepest descent direction at an arbitrary point (x,y)? Then parametrize the solution curve as (x(t),y(t)). The slope of this solution curve is y'(t)/x'(t)=dy/dx. Equate the two slopes and solve the differential equation.
 
Last edited:
  • #5
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Forgive me I'm not used to differential equations. Here's what I've got:
The gradient of a point (x,y) on the surface is (-80x/_blah_, -240y/_blah_). So the slope at (x,y) is -240y/-80x=-3y/x. this should be equal to dy/dx. Move sides =>dy/y=3dx/x, integrate both sides, ln(y)=ln(x)+C. Am I on the right track?

Thanks.
 
  • #6
Dick
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Forgive me I'm not used to differential equations. Here's what I've got:
The gradient of a point (x,y) on the surface is (-80x/_blah_, -240y/_blah_). So the slope at (x,y) is -240y/-80x=-3y/x. this should be equal to dy/dx. Move sides =>dy/y=3dx/x, integrate both sides, ln(y)=ln(x)+C. Am I on the right track?

Thanks.

You are very much on the right track. But what happened to the '3'?
 
  • #7
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You are very much on the right track. But what happened to the '3'?

Oops...that was a typo. It's supposed to be ln(y)=3ln(x)+C.

However, I haven't used the fact that the stream passes through point (1,1,5) yet...Can I just plug in the point into the equation: ln(1)=3ln(1)+C, then C=0. Therefore, is the equation of the stream ln(y)=3ln(x)?

Thanks.
 
  • #8
Dick
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Oops...that was a typo. It's supposed to be ln(y)=3ln(x)+C.

However, I haven't used the fact that the stream passes through point (1,1,5) yet...Can I just plug in the point into the equation: ln(1)=3ln(1)+C, then C=0. Therefore, is the equation of the stream ln(y)=3ln(x)?

Thanks.

Yes and yes. If you exponentiate both sides you can get a simpler expression.
 
  • #9
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Great! Thanks a lot man.
 

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