If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### 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.

## What we're building to

• A tangent plane to a two-variable function $f\left(x,y\right)$ is, well, a plane that's tangent to its graph.
• The equation for the tangent plane of the graph of a two-variable function $f\left(x,y\right)$ at a particular point $\left({x}_{0},{y}_{0}\right)$ looks like this:
$T\left(x,y\right)=f\left({x}_{0},{y}_{0}\right)+{f}_{x}\left({x}_{0},{y}_{0}\right)\left(x-{x}_{0}\right)+{f}_{y}\left({x}_{0},{y}_{0}\right)\left(y-{y}_{0}\right)$

Think of a scalar-valued function with a two-coordinate input, like this one:
$f\left(x,y\right)=-{x}^{2}-{y}^{2}+3$
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:
$\left(x,y,f\left(x,y\right)\right)=\left(x,y,-{x}^{2}-{y}^{2}+3\right)$
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 $\left({x}_{0},{y}_{0},f\left({x}_{0},{y}_{0}\right)\right)$ in three-dimensional space?

## Representing planes as graphs

Well, first of all, which functions $g\left(x,y\right)$ 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:
$g\left(x,y\right)=ax+by+c$
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 $\left({x}_{0},{y}_{0},{z}_{0}\right)$ in space?
One clean way to do this is to write our linear function as
$g\left(x,y\right)=a\left(x-{x}_{0}\right)+b\left(y-{y}_{0}\right)+{z}_{0}$
Concept check: With $g$ defined this way, compute $g\left({x}_{0},{y}_{0}\right)$.

Writing $g\left(x,y\right)$ like this makes it clear that $g\left({x}_{0},{y}_{0}\right)={z}_{0}$. This guarantees that the graph of $g$ must pass through $\left({x}_{0},{y}_{0},{z}_{0}\right)$:
$\left({x}_{0},{y}_{0},g\left({x}_{0},{y}_{0}\right)\right)=\left({x}_{0},{y}_{0},{z}_{0}\right)$
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 $\left({x}_{0},{y}_{0},{z}_{0}\right)$. 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 $T\left(x,y\right)$ that represents a plane tangent to the graph of some function $f\left(x,y\right)$ at a point $\left({x}_{0},{y}_{0},f\left({x}_{0},{y}_{0}\right)\right)$, so we substitute $f\left({x}_{0},{y}_{0}\right)$ for ${z}_{0}$ in the general equation for a plane.
$\phantom{\rule{1em}{0ex}}T\left(x,y\right)=f\left({x}_{0},{y}_{0}\right)+a\left(x-{x}_{0}\right)+b\left(y-{y}_{0}\right)$
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 $\left({x}_{0},{y}_{0},f\left({x}_{0},{y}_{0}\right)\right)$, the one tangent to the graph of $f$ will have the same partial derivatives as $f$. Pleasingly, the partial derivatives of our linear function are given by the constants $a$ and $b$.
• Try it! Take the partial derivatives of the equation for $T\left(x,y\right)$ above.
Therefore setting $a={f}_{x}\left({x}_{0},{y}_{0}\right)$ and $b={f}_{y}\left({x}_{0},{y}_{0}\right)$ 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 $\left({x}_{0},{y}_{0}\right)$, but that's the only point we care about. Putting this together, we get a usable formula for the tangent plane.
$\begin{array}{r}\phantom{\rule{1em}{0ex}}T\left(x,y\right)=f\left({x}_{0},{y}_{0}\right)+{f}_{x}\left({x}_{0},{y}_{0}\right)\left(x-{x}_{0}\right)+{f}_{y}\left({x}_{0},{y}_{0}\right)\left(y-{y}_{0}\right)\end{array}$

## Example: Finding a tangent plane

Problem:
Given the function
$\phantom{\rule{1em}{0ex}}f\left(x,y\right)=\mathrm{sin}\left(x\right)\mathrm{cos}\left(y\right)$,
find the equation for a plane tangent to the graph of $f$ above the point $\left(\frac{\pi }{6},\frac{\pi }{4}\right)$.

The tangent plane will have the form
$\begin{array}{r}\phantom{\rule{1em}{0ex}}T\left(x,y\right)={f}_{x}\left({x}_{0},{y}_{0}\right)\left(x-{x}_{0}\right)+{f}_{y}\left({x}_{0},{y}_{0}\right)\left(y-{y}_{0}\right)+f\left({x}_{0},{y}_{0}\right)\end{array}$
Step 1: Find both partial derivatives of $f$.
${f}_{x}\left(x,y\right)=$
${f}_{y}\left(x,y\right)=$

Step 2: Evaluate the function $f$ as well as both these partial derivatives at the point $\left(\frac{\pi }{6},\frac{\pi }{4}\right)$:
$f\left(\pi /6,\pi /4\right)=$
${f}_{x}\left(\pi /6,\pi /4\right)=$
${f}_{y}\left(\pi /6,\pi /4\right)=$

Putting these three numbers into the general equation for a tangent plane, you can get the final answer
$T\left(x,y\right)=$

## Summary

• A tangent plane to a two-variable function $f\left(x,y\right)$ is, well, a plane that's tangent to its graph.
• The equation for the tangent plane of the graph of a two-variable function $f\left(x,y\right)$ at a particular point $\left({x}_{0},{y}_{0}\right)$ looks like this:
$T\left(x,y\right)=f\left({x}_{0},{y}_{0}\right)+{f}_{x}\left({x}_{0},{y}_{0}\right)\left(x-{x}_{0}\right)+{f}_{y}\left({x}_{0},{y}_{0}\right)\left(y-{y}_{0}\right)$

## Want to join the conversation?

• Is finding a tangent plan similar/related to the use of a Jacobian to approximate nonlinear functions?
• The short answer is: yes. In our case, the Jacobian is the gradient vector.
• How can you guarantee that the graph of a linear function passes through a particular point (x0,y0,z0) in space?

One clean way to do this is to write our linear function as:
g(x,y)=a(x−x0​)+b(y−y0​)+z0​
-------------------------------------------
Can someone please explain this part to me? May be this obvious but I do not understand why it is always true. An illustrative example will be really helpful
(1 vote)
• That's because if you plug in x_0 and y_0 into the equation, the equation becomes:

g(x,y)=a(x_0 - x_0​)+b(y_0 − y_0​)+z_0​
=z_0

In other words, if you plug in x_o and y_0, the first two terms go to zero and so the whole thing evaluates to z_0, which is exactly just a way of saying that the function passes through x_0, y_0 and z_0.

It's a bit like if you wanted to write a one-variable linear function that passes through, say, the point (1, 2). What you would do is you would use the point-slope form and write:

y - 2 = m(x - 1)

Just by looking at the equation, you know that this line would pass through (1, 2). But to make it look more like the two-variable case, you could write it as:

y = m(x - 1) + 2

If x = 1, then the equation becomes y = 2, which is equivalent to saying that the line passes though the point (1, 2). Just like what I said earlier about the two-variable case.
• someone please explain to me why the equation g(x,y)=ax+by+c satisfying the slope constant
• Because the partial derivatives with respect to X and Y of this particular function are respectively a and b (two constant), therefore they are independent from the specific point in which they are calculated (i.e. they don’t depend on the two variable x and y). Given that the two partial derivatives express the slope of the function in the direction of x axis and y axis respectively, as long as they are constant in any point of the function so is the slope. Hope it helps.
• hey guys.. i actually understood how to get the tangent function.. at the end, how does should we interpret the T(x,y)?

should be the entire equation as it is, i mean what x and y need for, should make a new equation of value with x and y being the position of the tangent plane?
thanks gian :)
• Find an equation of the tangent plane to the surface

z =

xy

at the point (1, 1, 1).
(1 vote)
• z=(x+y)/2
• here we are finding tangent planes but the equation used is of linearization function please answer me fastly
(1 vote)
• The tangent plane is a linearization of a three dimensional surface
• how do you find a tangent line from a triangle?
(1 vote)
• One is a line that has a coincident point with a curve and that is parallel to the curve at that point... the other is a three-sided shape (or more accurately, a 3-angled shape: a trigon is a 3-sided shape).
If you would like a more specific answer, please make the question more specific.
(1 vote)
• how do i find a tangent plane for a function thats defined as follows: R^2 ---> R^2
more specifically:
(x,y)--> (sqrt(x^2+y^2 , arctan(y/x))