# 4.4 Tangent planes and linear approximations

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• Determine the equation of a plane tangent to a given surface at a point.
• Use the tangent plane to approximate a function of two variables at a point.
• Explain when a function of two variables is differentiable.
• Use the total differential to approximate the change in a function of two variables.

In this section, we consider the problem of finding the tangent plane to a surface, which is analogous to finding the equation of a tangent line to a curve when the curve is defined by the graph of a function of one variable, $y=f\left(x\right).$ The slope of the tangent line at the point $x=a$ is given by $m=f\prime \left(a\right);$ what is the slope of a tangent plane? We learned about the equation of a plane in Equations of Lines and Planes in Space ; in this section, we see how it can be applied to the problem at hand.

## Tangent planes

Intuitively, it seems clear that, in a plane, only one line can be tangent to a curve at a point. However, in three-dimensional space, many lines can be tangent to a given point. If these lines lie in the same plane, they determine the tangent plane at that point. A more intuitive way to think of a tangent plane is to assume the surface is smooth at that point (no corners). Then, a tangent line to the surface at that point in any direction does not have any abrupt changes in slope because the direction changes smoothly. Therefore, in a small-enough neighborhood around the point, a tangent plane touches the surface at that point only.

## Definition

Let ${P}_{0}=\left({x}_{0},{y}_{0},{z}_{0}\right)$ be a point on a surface $S,$ and let $C$ be any curve passing through ${P}_{0}$ and lying entirely in $S.$ If the tangent lines to all such curves $C$ at ${P}_{0}$ lie in the same plane, then this plane is called the tangent plane    to $S$ at ${P}_{0}$ ( [link] ).

For a tangent plane to a surface to exist at a point on that surface, it is sufficient for the function that defines the surface to be differentiable at that point. We define the term tangent plane here and then explore the idea intuitively.

## Definition

Let $S$ be a surface defined by a differentiable function $z=f\left(x,y\right),$ and let ${P}_{0}=\left({x}_{0},{y}_{0}\right)$ be a point in the domain of $f.$ Then, the equation of the tangent plane to $S$ at ${P}_{0}$ is given by

$z=f\left({x}_{0},{y}_{0}\right)+{f}_{x}\left({x}_{0},{y}_{0}\right)\left(x-{x}_{0}\right)+{f}_{y}\left({x}_{0},{y}_{0}\right)\left(y-{y}_{0}\right).$

To see why this formula is correct, let’s first find two tangent lines to the surface $S.$ The equation of the tangent line to the curve that is represented by the intersection of $S$ with the vertical trace given by $x={x}_{0}$ is $z=f\left({x}_{0},{y}_{0}\right)+{f}_{y}\left({x}_{0},{y}_{0}\right)\left(y-{y}_{0}\right).$ Similarly, the equation of the tangent line to the curve that is represented by the intersection of $S$ with the vertical trace given by $y={y}_{0}$ is $z=f\left({x}_{0},{y}_{0}\right)+{f}_{x}\left({x}_{0},{y}_{0}\right)\left(x-{x}_{0}\right).$ A parallel vector to the first tangent line is $\text{a}=\text{j}+{f}_{y}\left({x}_{0},{y}_{0}\right)\text{k;}$ a parallel vector to the second tangent line is $\text{b}=\text{i}+{f}_{x}\left({x}_{0},{y}_{0}\right)\text{k}.$ We can take the cross product of these two vectors:

$\begin{array}{cc}\hfill a\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}b& =\left(j+{f}_{y}\left({x}_{0},{y}_{0}\right)k\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(i+{f}_{x}\left({x}_{0},{y}_{0}\right)k\right)\hfill \\ & =|\begin{array}{ccc}i\hfill & j\hfill & k\hfill \\ 0\hfill & 1\hfill & {f}_{y}\left({x}_{0},{y}_{0}\right)\hfill \\ 1\hfill & 0\hfill & {f}_{x}\left({x}_{0},{y}_{0}\right)\hfill \end{array}|\hfill \\ & ={f}_{x}\left({x}_{0},{y}_{0}\right)i+{f}_{y}\left({x}_{0},{y}_{0}\right)j-k.\hfill \end{array}$

This vector is perpendicular to both lines and is therefore perpendicular to the tangent plane. We can use this vector as a normal vector to the tangent plane, along with the point ${P}_{0}=\left({x}_{0},{y}_{0},f\left({x}_{0},{y}_{0}\right)\right)$ in the equation for a plane:

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20/(×-6^2)
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