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Theres gotta be an easier way

  • Thread starter joemama69
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  • #1
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Homework Statement



Determine the points on the surface xy^2z^3 = 2 that are closest to the origin

Homework Equations





The Attempt at a Solution



is there an easier way to do this than to plug it into the distance formula and taking the derivative set to 0.
 

Answers and Replies

  • #2
HallsofIvy
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Use the square of the distance formula.

Also, and this may be what you are looking for, instead of using xy2z3= 2 to replace one variable with the other two, use "Lagrange multipliers". If we write f(x,y,z)= x2+ y2+ z2, the square of the distance to the origin, and g(x,y,z)= xy2z3= 2, then max or min values of f, for points that satisfy g(x,y,z)= 2 must have [itex]\nabla f[/itex] parallel to [itex]\nabla g[/itex]- one must be a multiple of the other. Setting [itex]\nabla f= \lambda g[/itex] and comparing the components, together with g(x,y,z)= 2, gives 4 equations to solve for x, y, z, and [itex]\lambda[/itex].

Tip: since you are not interested in the value of [itex]\lambda[/itex], and [itex]\lambda[/itex] is simply multiplied by the functions of x, y, and z, often a best first thing to do is to divide one equation by another to eliminate [itex]\lambda[/itex].
 
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  • #3
matt grime
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I case you need to look that up, Halls meant Lagrange multipliers, not Laplace.
 
  • #4
HallsofIvy
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Thanks, matt, I have editted that.
 
  • #5
399
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ok i got a little stuck, heres my work, BTW k = lagrange multiplier

f(x,y,z) = x2 + y2 + z2
g(x,y,z) = xy2z3 = 2

grad f = 2xi + 2yj + 2zk
k grad g = k(y2z3i + 2xyz3j + 3z2xy2k

k = 2x/y2z3
k = 1/xz3
k = 2/3zxy2

and im lost
 
  • #6
matt grime
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So you know that 2x/y^2z^3 = 1/xz^3 ie. 2x^2=y^2, and you can subs that in and see what happens.
 
  • #7
399
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ok so i can sove for each of the variables

y = [tex]\sqrt{2x^2}[/tex]

x = [tex]\sqrt{(y^2)/2}[/tex]

z = [tex]\sqrt{(3y^2)/2}[/tex]

how does that help
 

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