Find the points on the surface y^2 = 9 + xz that are closest to the origin.

Find The Points On The Surface Y2 9 Xz That Are Closest To The Origin.

This question aims to learn the basic methodology for optimizing a mathematical function (maximizing or minimizing).

Critical points are the points where the value of a function is either maximum or minimum. To calculate the critical point(s), we equate the first derivative’s value to 0 and solve for the independent variable. We can use the second derivative test to find maxima/minima. For the given question, we can minimize the distance function of the desired point from the origin as explained in the below answer.

Expert Answer

Given:

\[ y^{ 2 } \ = \ 9 \ + \ x \ z \]

Let $ ( x, \ y, \ z ) $ be the point that is nearest to the origin. The distance of this point from the origin is calculated by:

\[ d = \sqrt{ x^{ 2 } + y^{ 2 } + z^{ 2 } } \]

\[ \Rightarrow d^{ 2 } = x^{ 2 } + y^{ 2 } + z^{ 2 } \]

\[ \Rightarrow d^{ 2 } = x^{ 2 } + 9 + x z + z^{ 2 } \]

To find this point, we simply need to minimize this $ f(x, \ y, \ z) \ = \ d^{ 2 } $ function. Calculating the first derivatives:

\[ f_x = 2x + z \]

\[ f_z = x + 2z \]

Finding critical points by putting $ f_x $ and $ f_z $ equal to zero:

\[ 2x + z = 0\]

\[ x + 2z = 0\]

Solving the above system yields:

\[ x = 0\]

\[ z = 0\]

Consequently:

\[ y^{ 2 } = 9 + xz = 9 + (0)(0) = 0 \]

\[ \Rightarrow = y = \pm 3 \]

Hence, the two possible critical points are $ (0, 3, 0) $ and $ (0, -3, 0) $. Finding the second derivatives:

\[ f_{xx} = 2 \]

\[ f_{zz} = 2 \]

\[ f_{xz} = 1 \]

\[ f_{zx} = 1 \]

Since all second derivatives are positive, the calculated critical points are at  a minimum.

Numerical Result

Points Closest to the origin = $ (0, 0, 5)$ and $ (0, 0, -5) $

Example

Find the points on the surface  $ z^2 = 25 + xy $ nearest to the origin.

Here, the distance function becomes:

\[ d = \sqrt{ x^{ 2 } + y^{ 2 } + z^{ 2 } } \]

\[ \Rightarrow d^{ 2 } = x^{ 2 } + y^{ 2 } + z^{ 2 } \]

\[ \Rightarrow d^{ 2 } = x^{ 2 } + y^{ 2 } + 25 + xy \]

Calculating first derivatives and equating to zero:

\[ f_x = 2x + y \Rightarrow 2x + y = 0\]

\[ f_y = x + 2y \Rightarrow x + 2y = 0\]

Solving the above system yields:

\[ x = 0 \text{and} y = 0\]

Consequently:

\[ z^{ 2 } = 25 + xy = 25 \]

\[ \Rightarrow = z = \pm 5 \]

Hence, the two possible critical points are $ (0, 3, 0) $ and $ (0, -3, 0) $. Finding the second derivatives:

\[ f_{xx} = 2 \]

\[ f_{yy} = 2 \]

\[ f_{xy} = 1 \]

\[ f_{yx} = 1 \]

Since all second derivatives are positive, the calculated critical points are at a minimum.

Points Closest to the origin = $ (0, 0, 5) $ and $ (0, 0, -5) $

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