banner

Find the point(s) on the surface at which the tangent plane is horizontal.

$ 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. 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.

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.

\[ z(1,1) = (-1).(-1) – \dfrac{1}{-1}-\dfrac{1}{-1} = 3\]

\[(x,y,z) = (-1,-1,3) \]

5/5 - (4 votes)