Linear approximation in single-variable calculus

We've already learned how single-variable calculus — specifically, calculating the slope of a curve at a point, $x_o$ in its domain — can help us to quickly approximate the value of the function in the neighborhood of $x_o$. The graph shows such a scenario.

The key is to calculate the slope of the function at a point $x_o$. Then in the neighborhood (like inside of the blue-dashed circle) of that point, our linear approximation of the value of our function at some new point, $x$, is $f(x_o)$ plus that slope multiplied by the distance from $x$ to $x_o$.

$$f(x) \approx f(x_o) + f'(x_o)(x - x_o)$$

Any curved function $f(x)$, when viewed closely enough, is approximately linear, so this approximation works better the closer $x$ is to $x_o$.

In this section we'll generalize this concept to finding the equation of a plane tangent to 3D function $f(x,y)$. The ideas are similar, and they involve partial derivatives — rates of change of 3D functions in the $x$- and $y$-dimensions.

Simple example — tangent to a sphere

As a first example, let's look at the equation of a plane tangent to a sphere. Let's take an easy sphere, the sphere of radius 3 centered at the origin:

$$x^2 + y^2 + z^2 = 9$$

Tangent planes

In order to find an equation of a plane tangent to a function $f(x,y)$, we need

• A point on the surface of $f(x,y)$ which will also be on our plane


• A normal vector $\vec n$

The equation of our plane must obey the dot product $\vec n \cdot \vec{P_o P} = 0$, where $\vec{P_oP}$ is a vector in the plane. The coordinates of $P_o$ and $P$, respectively, are $(x_o, y_o, z_o)$ and $(x,y,z)$. The equation of our plane has the form

$$a(x - x_o) + b(y - y_o) + c(z - z_o) = 0,$$

where $\vec n = (a, b, c)$. Because there are an infinite number of equations for a plane, we can let $c$ be any nonzero number. Let's use $c = -1$, which will turn out to be a convenient choice. Then our equation changes to

$$a(x - x_o) + b(y - y_o) + z_o = z$$

Example 1

Find the equation of the plane tangent to the function $f(x,y) = -x^1 - y^2$ at the point $(x,y) = (1, 1)$.

Solution: First, the value of our function, $f(1, 1) = -1^2 - 1^2 = -2$, so our point of tangency is $(1, 1, -2)$.

The partial derivatives are

$$ \begin{align} f_x = -2x \; \rightarrow \; f_x(1,1) &= -2 \\[5pt] f_y = -2y \; \rightarrow \; f_y(1,1) &= -2 \end{align}$$

So our plane equation is

$$ \begin{align} z &= -2(x-1) - 2(y-1) -2 \\[5pt] &= -2x + 2 -2y + 2 -2 \\[5pt] &= -2x - 2y - 2 \end{align}$$

or $ \phantom{000}2x + 2y + z = 2$.

We can read the normal vector straight from this equation, $\vec n = (2, 2, 1)$. Here are 3D plots of our function and its tangent plane, one from the side so you can see the point of tangency (red). The second is just another view of $f(x,y)$ and the plane.

Example 2

Find the equation of the plane tangent to $f(x,y) = x^3 + xy^2 - y + 4$ at $(x,y) = (1,2)$.

Solution: First, the value of our function, $f(1, 2) = 1+4-2+4 = 7$, so our point of tangency is $(1, 2, 7)$.

The partial derivatives are

$$ \begin{align} f_x = 3x^2 + y^2 \; \rightarrow \; f_x(1,2) &= 3+4=7 \\[5pt] f_y = 2xy - 1 \; \rightarrow \; f_y(1,2) &= 4-1 = 3 \end{align}$$

So our plane equation is

$$ \begin{align} z &= 7(x-1) + 3(y-2) + 7 \\[5pt] &= 7x - 7 + 3y - 6 + 7 \\[5pt] &= 7x + 3y - 6 \end{align}$$

or $ \phantom{000}7x + 3y - z = -6$.

We can read the normal vector straight from this equation, $\vec n = (7,3,-1)$. Here is a 3D plot of our function and its tangent plane. This is a trick graph because the tangent plane intersects with the curved surface near our chosen point of tangency.

Another way to think about it: The gradient

Here's a schematic of a made-up function in $\mathbb{R}^3$. The function might be written as $z = f(x,y)$, but we could also write it as $F(x,y,z) = 0$. Just as a quick example, consider the function $f(x) = x^2 - y^2 + 3$. That's really $z = x^2 - y^2 + 3$, which we could also write as $z - x^2 + y^2 - 3 = 0$.

Now consider the point $P_o$ on the plane that's meant to be tangent to $f(x,y)$ at that point. If that plane is really tangent to the 3D curve, then it consists of all possible tangents to the curve at that point. Let's focus on one of those, the tangent to the curve $C$, one of many that traces the 3D graph and passes through $P_o$. We're going to claim that

$$\vec N = \nabla \vec F = (F_x, F_y, F_z),$$

where $\vec N$ is normal to the plane and any line in it, and $\nabla$ is the gradient "operator". Is our claim true?

Example 3

Let's revisit our first example, the sphere $x^2 + y^2 + z^2 = 9$, and find the plane tangent to it at $(1, 2, 2)$ using the gradient method.

Solution: first, we have $F = x^2 + y^2 + z^2 - 9 = 0$. Then the relevant first partial derivatives are

$$ \begin{align} F_x &= 2x \phantom{00} \rightarrow \phantom{00} F_x(1,2,2) = 2 \\[5pt] F_y &= 2y \phantom{00} \rightarrow \phantom{00} F_y(1,2,2) = 4 \\[5pt] F_z &= 2z \phantom{00} \rightarrow \phantom{00} F_z(1,2,2) = 4 \\[5pt] \end{align}$$

So $\vec n = \nabla F(1, 2, 2) = (2, 4, 4)$

Practice problems

Find an equation of the plane tangent to the surface at the given point.