Find the point P on the curve $z^2=x^2+y^2$ nearest to point$ A(2,4,0)$. Then find the distance of point P to the origin Ans. Let P (x,y,z) be any point on the curve so the distance between P and A is $f(xyz)=D^2=(x-2)^2+(y-4)^2+(z-0)^2$ or $f(xyz)=D^2=(x-2)^2+(y-4)^2+x^2+y^2$ the minimum of this function occur at (1,2) so the point nearest to curve would be$ (1,2,\sqrt5)$. And the distance from the origin to this point would be$ \sqrt10$. Am I doing it all right? I don't have the answer so I'm not sure. Thank you
Nearest distance to the curve
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calculus
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0I haven't checked the work, but the method is sound. – 2017-02-15
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0The distance of $P$ from origin is $\sqrt(10)$. Everything in the steps are good. – 2017-02-15
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0Ofcourse it's $\sqrt(10)$. Thanks for checking – 2017-02-15
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0You have a surface, not a curve. – 2017-02-15
2 Answers
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Everything is fine. But just for completeness do the second partial derivative test to ensure that point $(1,2,\sqrt 5)$ is indeed the minimizer of $f(x,y,z)$. (https://en.wikipedia.org/wiki/Second_partial_derivative_test#Functions_of_two_variables)
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For this problem, it‘s fairly easy to check your own work geometrically. The surface is a right cone, so the two points and the origin form a right isosceles triangle. The $z$-coordinate of the point on the surface should then be half of the distance of $A$ from the origin, its $x$- and $y$-coordinates should be half those of $A$, and the distance of of $A$ from the origin should be $\sqrt2$ times the distance that you‘re asked to find.