We find tangent planes.

Given a function and a number in the domain of , if one can
“zoom in” on the graph at sufficiently so that it appears to be a
straight line, then the function is **differentiable**, and that line is
the **tangent line** to at the point .

We illustrate this informal definition with the following diagram:

We can give a similar informal definition for functions of two variables.

Given a function and a vector in the domain of , if one can “zoom
in” on the graph at sufficiently so that it appears to be a plane,
then the function is **differentiable**, and that plane is the **tangent
plane** to at the point .

Let’s turn our attention to finding an equation for this tangent plane.

### The single variable case

Given a function and a point of interest in the domain of , we have previously found an equation for the tangent line to at , which we also called the linear approximation to at .

There are two things we should notice about this linear approximation.

First, think about geometry We know that:

- is the “height” of at .
- is the rate of change of at .
- is the distance we have moved from the point .

Multiplying and gives us an approximate change in the value of the function, so to find the new approximate function value, we need to add on our original height, or .

Second, this is an approximation Specifically the tangent line to a function is a line constructed so that:

When the values of and are equal at , and they are changing at the same rates at , then it is only reasonable that is a good approximation of . Moreover, is also the first-order Taylor approximation to . Recall, the degree Taylor polynomial for at is:

This polynomial starts with the linear approximation, and then adds higher degree terms.### The two variable case

Now consider a differentiable function . Here the formula for the tangent plane at is given by: We will work as we did above.

First, think about geometry Finally, let’s look at some of the geometry of the equation we just found. Working in an entirely similar way as before, we know that:

- is the “height” of at .
- is the rate of change of at in the -direction.
- is the rate of change of at in the -direction.
- is the distance we have moved from the point in the -direction.
- is the distance we have moved from the point in the -direction.

Multiplying and gives us an approximate change in the value of the function when moving in the -direction. Multiplying and gives us an approximate change in the value of the function when moving in the -direction. Adding these two together, we see that we have moved away from the point , and we are combining these rates of change to get an approximate change in the value of . Adding this to our original function value gives us our new approximate function value. In other words, approximates near .

Second, this is an approximation Note, first that is a plane. To be a bit pedantic, we can write

and rearranging, we find: and this is the equation of a plane, as it is: Let’s see if you get this.Now note that the function is cooked-up so that:

So we see that is a plane that approximates near the point .

Foreshadowing a bit, our formula for our linear approximation is:

This look like what we might want from a first-order Taylor approximation to our function as:

Said in words, the tangent plane at touches the surface at and the first partial derivatives of and agree at .

Saying the tangent plane has the same partial derivatives as the function is another way of saying that tangent plane has to contain the tangent lines we can find using the partial derivatives.

Try your hand at this question:

Moreover, tangent planes are *linear approximations* of differentiable surfaces.

### Generalizing with the gradient vector

Above, given a function , we wrote our linear approximation as: We can
rewrite this using the gradient vector as follows: where and . Now check this
out, the formula actually works in **all** dimensions! So now if you have a
function of three variable, we set and and we can unpack this formula as:

### An in-depth example

Our final example brings together several concepts.

- (a)
- Find an equation for the plane tangent to at .
- (b)
- Show that the line passes through the point .
- (c)
- Find a parametric description for the curve lying above on the surface . Call this description .
- (d)
- Find a vector-valued formula for the line tangent to the curve at the point . Call this .
- (e)
- Does the tangent line drawn by lie in the plane tangent to at ?

For part eg:tpb, note that we are working here in the -plane -space . We would like to see that the point lies on the line given. To do this, we plug in to the equation . We see that

So we conclude that the point is is not on the line .For part eg:tpc, we would like a parametric description of the line in the -plane. Since this line is already expressed in terms of , the simplest choice is to take . We find The curve lying on the surface is the collection of all of the images of the points on the line , so we would like to evaluate on all of these points. Since we already have a parametric description , we plug these coordinates in for and to find our - coordinate. We find For part eg:tpd note that since our tangent line is in -space, we will need to find a parametric equation for this line. Recall that the most straightforward way to write such an equation is where is a point on the line and is a direction vector for the line. The direction of is tangent to the curve. We first find the tangent vector for any value of . To find the tangent vector at , we plug in and find . We know that the line goes through the point , and so now we can write the equation of our tangent line.

Does the tangent line lie in the tangent plane to at ?

We could solve this problem by simply plugging in the coordinate functions of to the equation to see if the two sides of the equals sign agree. Instead, let’s use a little geometry!

We know that both the tangent plane and the line pass through the point . Since is a line, its tangent vector tells us everything we need to know about its direction. Since is a plane, its normal vector tells us everything we need to know about its direction. Then, if , we know that the vector must lie in the plane—and thus the entire line must line in the plane as well! First, find . Next, use the coefficients of to find . Finally, compute their dot product.