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### Course: Multivariable calculus > Unit 3

Lesson 1: Tangent planes and local linearization# Tangent planes

Just as the single variable derivative can be used to find tangent lines to a curve, partial derivatives can be used to find the tangent plane to a surface.

## Background

## What we're building to

- A
**tangent plane**to a two-variable function is, well, a plane that's tangent to its graph.$f(x,y)$

- The equation for the tangent plane of the graph of a two-variable function
at a particular point$f(x,y)$ looks like this:$({x}_{0},{y}_{0})$

## The task at hand

Think of a scalar-valued function with a two-coordinate input, like this one:

Intuitively, it's common to visualize a function like this with its three-dimensional graph.

Remember, you can describe this graph more technically by describing it as a certain set of points in three-dimensional space. Specifically, it is all the points that look like this:

Here, $x$ and $y$ can range over all possible real numbers.

A

**tangent plane**to this graph is a plane which is tangent to the graph. Hmmm, that's not a good definition. This is hard to describe with words, so I'll just show a video with various different tangent planes.**Key question**: How do you find an equation representing the plane tangent to the graph of the function at some specific point

## Representing planes as graphs

Well, first of all, which functions $g(x,y)$ have graphs that look like planes?

The slope of a plane in any direction is constant over all input values, so both partial derivatives ${g}_{x}$ and ${g}_{y}$ would have to be constants. The functions with constant partial derivatives look like this:

Here, $a$ , $b$ , and $c$ are each some constant. These are called

**linear functions**. Well, technically speaking they are**affine functions**since linear functions must pass through the origin, but it's common to call them linear functions anyway.**Question**: How can you guarantee that the graph of a linear function passes through a particular point

One clean way to do this is to write our linear function as

**Concept check**: With

Writing $g(x,y)$ like this makes it clear that $g({x}_{0},{y}_{0})={z}_{0}$ . This guarantees that the graph of $g$ must pass through $({x}_{0},{y}_{0},{z}_{0})$ :

The other constants ${a}$ and ${b}$ are free to be whatever we want. Different choices for ${a}$ and ${b}$ result in different planes passing through the point $({x}_{0},{y}_{0},{z}_{0})$ . The video below shows how those planes change as we tweak ${a}$ and ${b}$ :

## Equation for a tangent plane

Back to the task at hand. We want a function $T(x,y)$ that represents a plane tangent to the graph of some function $f(x,y)$ at a point $({x}_{0},{y}_{0},f({x}_{0},{y}_{0}))$ , so we substitute $f({x}_{0},{y}_{0})$ for ${z}_{0}$ in the general equation for a plane.

As you tweak the values of ${a}$ and ${b}$ , this equation will give various planes passing through the graph of $f$ at the desired point, but only one of them will be a

*tangent*plane.Of all the planes passing through $({x}_{0},{y}_{0},f({x}_{0},{y}_{0}))$ , the one tangent to the graph of $f$ will $a$ and $b$ .

**have the same partial derivatives as**$f$ . Pleasingly, the partial derivatives of our linear function are given by the constants**Try it!**Take the partial derivatives of the equation for above.$T(x,y)$

Therefore setting ${a={f}_{x}({x}_{0},{y}_{0})}$ and ${b={f}_{y}({x}_{0},{y}_{0})}$ will guarantee that the partial derivatives of our linear function $T$ match the partial derivatives of $f$ . Well, at least they will match for the input $({x}_{0},{y}_{0})$ , but that's the only point we care about. Putting this together, we get a usable formula for the tangent plane.

## Example: Finding a tangent plane

**Problem**:

Given the function

find the equation for a plane tangent to the graph of $f$ above the point $({\displaystyle \frac{\pi}{6}},{\displaystyle \frac{\pi}{4}})$ .

The tangent plane will have the form

**Step 1**: Find both partial derivatives of

**Step 2**: Evaluate the function

Putting these three numbers into the general equation for a tangent plane, you can get the final answer

## Summary

- A
**tangent plane**to a two-variable function is, well, a plane that's tangent to its graph.$f(x,y)$

- The equation for the tangent plane of the graph of a two-variable function
at a particular point$f(x,y)$ looks like this:$({x}_{0},{y}_{0})$

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