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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:$({x}_{0},{y}_{0})$

## Want to join the conversation?

- Is finding a tangent plan similar/related to the use of a Jacobian to approximate nonlinear functions?(13 votes)
- The short answer is: yes. In our case, the Jacobian is the gradient vector.(12 votes)

- 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.(9 votes)

- someone please explain to me why the equation g(x,y)=ax+by+c satisfying the slope constant(2 votes)
- 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.(6 votes)

- 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 :)(2 votes) - Find an equation of the tangent plane to the surface

z =

√

xy

at the point (1, 1, 1).(1 vote) - 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(2 votes)

- 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))

please help as fast as possible

thanks in advance(1 vote)- Your question might be in a wrong page, an equation for f(x,y) and a specific coordinate are needed to calculate the tangent plane.(1 vote)

- I have a surface equation of: z(x,y) = √(12(〖sec〗^2 x/y-1))+ln(9/10 (x^2/π^2 +y^2/4)).

I need to find the tangent plane to the surface at the point P(π/3, 2).

I can get halfway through this problem to find z_0 = 2 but cannot find the constants f_x or f_y.

Any help would be greatly appreciated.(1 vote) - so this the equation of the plane with two input varibles. kind of a headache . when already we define a plane to be represented as an input of three variables.... is there a convenient way to see it as of the form ax + by +cz= d or should we not attempt to do that.(1 vote)
- It's perfectly fine to put it in the form ax+by+cz=d.(1 vote)