Multivariable calculus problem

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The discussion centers on finding points on the surface defined by the equation xy^2z^3=2 that are closest to the origin. Participants clarify that the origin (0, 0, 0) is not on the surface, so it cannot be the closest point. The Lagrange multiplier method is suggested as a viable approach to solve the problem, along with using partial derivatives to minimize the distance function x^2 + y^2 + z^2 under the given constraint. Additionally, an alternative method involves expressing one variable in terms of the others and substituting it into the distance function for simplification. The conversation emphasizes the importance of correctly applying mathematical methods to find the solution.
komarxian
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Homework Statement


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

Homework Equations

The Attempt at a Solution


x,y,z=/= 0, as when x,y,z = 0 it is untrue. Right?? Otherwise, I am very unsure as to how to approach this problem. Should I be taking partial derivatives to fin the maximum and minimums?
 
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komarxian said:

Homework Statement


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

Homework Equations

The Attempt at a Solution


x,y,z=/= 0, as when x,y,z = 0 it is untrue. Right??
The surface doesn't go through the origin, so it should be fairly obvious that (0, 0, 0) is not the point on the surface that is closest to the origin.
komarxian said:
Otherwise, I am very unsure as to how to approach this problem. Should I be taking partial derivatives to fin the maximum and minimums?
If you're working from a textbook, there should be some examples of how to find the point or points in question. Basically, you want to find any points (x, y, z) that minimize the value of ##xy^2z^3 - 2## One approach uses partial derivatives.
 
Mark44 said:
T Basically, you want to find any points (x, y, z) that minimize the value of ##xy^2z^3 - 2## One approach uses partial derivatives.

I don't think this is true. You want to find points that minimize the value of ##x^2+y^2+z^2## subject to the constraint that ##xy^2z^3 - 2 = 0##.
 
komarxian said:

Homework Statement


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

Homework Equations

The Attempt at a Solution


x,y,z=/= 0, as when x,y,z = 0 it is untrue. Right?? Otherwise, I am very unsure as to how to approach this problem. Should I be taking partial derivatives to fin the maximum and minimums?
Besides the Lagrange multiplier method suggested in #2, you could use the constraint surface to solve for one of the variables in terms of the other two; then substitute that expression into the distance function (squared) ##x^2 + y^2 + z^2##. For example, it is particularly easy to use the surface equation to solve for ##x = x(y,z)## in terms of ##y## and ##z##; then you end up with an unconstrained minimization of some function ##F(y,z) = x(y,z)^2 + y^2 + z^2 ## in ##y## and ##z## alone. You can solve that using standard partial-derivative methods. (But, if you happen to know about "geometric programming" you will realize that your function ##F(y,z)## is a three-term posynomial in two variables, so is a "zero degree-of-difficulty" Geometric programming problem which is solvable without calculus, using just simple algebra.)
 
Last edited:
phyzguy said:
I don't think this is true. You want to find points that minimize the value of ##x^2+y^2+z^2## subject to the constraint that ##xy^2z^3 - 2 = 0##.
Yes, you're right. I didn't think things all the way through before writing.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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