
We find tangent planes.

As we are hoping is becoming a habit, let’s start by looking again at the case of functions of one variable, and use what we know from our previous calculus courses to answer our questions in higher dimensions. As usual with this strategy, we proceed with care. In the past you may have learned:

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

We illustrate this informal definition with the following diagram:

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

Given a function $F:\R ^2\to \R$ and a vector $\vec {a}$ in the domain of $F$, if one can “zoom in” on the graph at $(\vec {a}, F(\vec {a}))$ sufficiently so that it appears to be a plane, then the function is differentiable, and that plane is the tangent plane to $F$ at the point $(\vec {a},F(\vec {a}))$.

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

### The single variable case

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

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

First, think about geometry We know that:

• $f(a)$ is the “height” of $f$ at $x=a$.
• $f'(a)$ is the rate of change of $f$ at $a$.
• $(x-a)$ is the distance we have moved from the point $a$.

Multiplying $f'(a)$ and $(x-a)$ 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 $f(a)$.

So we see that:

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

• $\l (a) = f(a)$
• $\l '(a) = f'(a)$

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

This polynomial starts with the linear approximation, and then adds higher degree terms.

### The two variable case

Now consider a differentiable function $F : \R ^2 \to \R$. Here the formula for the tangent plane at $\vec {a} =\vector {a,b}$ 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:

• $F(\vec {a})$ is the “height” of $F$ at $\vector {x,y} = \vec {a}$.
• $F^{(1,0)}(\vec {a})$ is the rate of change of $F$ at $\vector {x,y} = \vec {a}$ in the $x$-direction.
• $F^{(0,1)}(\vec {a})$ is the rate of change of $F$ at $\vector {x,y} = \vec {a}$ in the $y$-direction.
• $(x-a)$ is the distance we have moved from the point $a$ in the $x$-direction.
• $(y-b)$ is the distance we have moved from the point $a$ in the $y$-direction.

Multiplying $F^{(1,0)}(\vec {a})$ and $(x-a)$ gives us an approximate change in the value of the function when moving in the $x$-direction. Multiplying $F^{(0,1)}(\vec {a})$ and $(y-b)$ gives us an approximate change in the value of the function when moving in the $y$-direction. Adding these two together, we see that we have moved away from the point $\vec {a}$, and we are combining these rates of change to get an approximate change in the value of $F$. Adding this to our original function value $F(\vec {a})$ gives us our new approximate function value. In other words, $L$ approximates $F$ near $\vector {x,y} = \vec {a}$.

So we see that:

Second, this is an approximation Note, first that $z=L (x,y)$ 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.

Given a differentiable function $F:\R ^2\to \R$, find a vector normal to the plane $z= L (x,y)$.

Now note that the function $L$ is cooked-up so that:

• $L(a,b) = F(a,b)$
• $L^{(1,0)} (a,b) = F^{(1,0)}(a,b)$
• $L^{(0,1)} (a,b) = F^{(0,1)}(a,b)$

So we see that $z=L(x,y)$ is a plane that approximates $z=F(x,y)$ near the point $(a,b,F(a,b))$.

Let $F(x,y) = 5-2(x-3)^2 -3(y-2)^2$. Find a tangent plane at $(x,y) = (2,1)$.
Find a vector normal to the plane you found above.
From above you see: So now write:

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 $F$ as:

• $L(a,b) = F(a,b)$
• $L^{(1,0)} (a,b) = F^{(1,0)}(a,b)$
• $L^{(0,1)} (a,b) = F^{(0,1)}(a,b)$

Said in words, the tangent plane at $\vec {a}$ touches the surface at $\vec {a}$ and the first partial derivatives of $L$ and $F$ agree at $\vec {a}$.

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:

Find an equation for the tangent plane to $F(x,y) = 3\cos (x)\sin (y)$ at $(x,y) = (\pi /3,\pi /6)$.

Moreover, tangent planes are linear approximations of differentiable surfaces.

Let $F:\R ^2\to \R$ be a differentiable function where you know that $F(1,-3) = 5$ and that $F^{(1,0)}(1,-3) = -4$ and $F^{(0,1)}(1,-3) = 2$. Give a plane tangent to $z =F(x,y)$ at the point $(1,-3)$ and use this tangent plane to approximate the value of $F(2,-4)$. First we write the linear approximation, meaning the tangent plane: So we believe that $F(2,-4) \approx L\left (\answer {2},\answer {-4}\right ) = \answer {-1}$.

### Generalizing with the gradient vector

Above, given a function $F:\R ^2\to \R$, we wrote our linear approximation as: We can rewrite this using the gradient vector as follows: where $\vec {x} = \vector {x,y}$ and $\vec {a} = \vector {a,b}$. Now check this out, the formula actually works in all dimensions! So now if you have a function of three variable, we set $\vec {x} = \vector {x,y,z}$ and $\vec {a} = \vector {a,b,c}$ and we can unpack this formula as:

Moreover, if we are working with functions of a single variable, then

where $\vec {a} = \vector {a}$ and $\vec {x} = \vector {x}$. I like to say that the formula is “one formula to rule them all,” since when you know this formula, you know how to compute linear approximations in all dimensions.

### An in-depth example

Our final example brings together several concepts.