Contents
$ z = xy +\dfrac { 1 } { x } +\dfrac{1}{y}$
This article aims to find the point on the surface at which the tangent plane is horizontal.
Point on surface
This article uses the concept of the surface at which the tangent plane is horizontal. To answer these questions, we must realize that the horizontal plane is tangent to the curve in space at maximum, minimum, or saddle points. Tangent planes to a surface are planes that touch the surface at a point and are “parallel” to the surface at a point.
Area of surface
Parallel lines
Expert Answer
Determine partial derivatives with respect to $ x $ and $ y $ and set them equal to zero. Solve for $ x $ partial with respect to $ y $ and put the result back into partial with respect to $ y $ and put the result back into partial with respect to $ x $ to solve for $ y $, $ y $ can’t be zero because we can’t have a zero denominator in it, so $ y $ must be $ 1 $. Put $ 1 $ in the equation for $ y $ to find $ x $.
\[ z = x y + \dfrac { 1 } { x } + \dfrac { 1 } { y } \]
\[f_{ x } ( x , y ) = y – \dfrac { 1 } { x ^ { 2 } } = 0 \]
\[f_{ y } ( x , y ) = x – \dfrac { 1 } { y ^ { 2 } } = 0 \]
\[ x = \dfrac { 1 } { y ^ { 2 } } \]
\[ y – \dfrac { 1 } { \ dfrac { 1 } { y ^ { 2 } } } = 0 \]
\[-y^{2}+y = 0\]
\[y(-y+1)=0\]
\[y=1\]
\[x = \dfrac{1}{1^{2}}= 1\]
Insert the point $(1,1)$ into $z$ and find the $3rd$ coordinate.
\[ z(1,1) = 1.1 + \dfrac{1}{1}+\dfrac{1}{1} = 3\]
\[(x,y,z) = (1,1,3) \]
Numerical Result
The point on the surface at which the tangent plane is horizontal $ (x,y,z)=(1,1,3)$.
Example
Find the point(s) on the surface at which the tangent plane is horizontal.
$ z = xy -\dfrac{1}{x} -\dfrac{1}{y}$
Solution
Determine partial derivatives with respect to $ x $ and $ y $ and set them equal to zero. Solve for $ x $partial with respect to $ y $ and put the result back into partial with respect to $ y $ and put the result back into partial with respect to $ x $ to solve for $ y $, $ y $ can’t be zero because we can’t have a zero denominator in it, so $ y $ must be $ 1 $. Put $ 1 $ in the equation for $ x $ to find $ x $.
\[z = xy-\dfrac{1}{x}-\dfrac{1}{y} \]
\[f_{x}(x,y) = y+\dfrac{1}{x^{2}} = 0\]
\[f_{y}(x,y) = x+\dfrac{1}{y^{2}} = 0\]
\[x = -\dfrac{1}{y^{2}}\]
\[y+\dfrac{1}{\dfrac{1}{y^{2}}}= 0 \]
\[y^{2}+y = 0\]
\[y(y+1)=0\]
\[y=-1\]
\[x = -\dfrac{1}{-1^{2}}= -1\]
Insert the point $(1,1)$ into $z$ and find the $3rd$ coordinate.
\[ z(1,1) = (-1).(-1) – \dfrac{1}{-1}-\dfrac{1}{-1} = 3\]
\[(x,y,z) = (-1,-1,3) \]