Tangent Planes and Linear Approximations: Learn It 1

  • Find equations of tangent planes to surfaces
  • Use tangent planes for approximation
  • Explain when a function of two variables is differentiable
  • Use the total differential to approximate the change in a function of two variables

Tangent Planes

You’ve already worked with tangent lines to curves when studying functions of one variable. For [latex]y=f(x)[/latex], the slope of the tangent line at [latex]x=a[/latex] is [latex]m=f^{\prime}(a)[/latex]. Now we’re extending this idea to three dimensions—finding tangent planes to surfaces.

In two dimensions, only one line can be tangent to a curve at a point. In three dimensions, things get more interesting—infinitely many lines can be tangent to a surface at a single point. If all these tangent lines lie in the same plane, that plane is the tangent plane.

A more intuitive way to think of a tangent plane is to assume the surface is smooth at that point (no corners). Then, a tangent line to the surface at that point in any direction does not have any abrupt changes in slope because the direction changes smoothly.

tangent plane

Let [latex]P_{0}=(x_0, y_0, z_0)[/latex] be a point on a surface [latex]S[/latex], and let [latex]C[/latex] be any curve passing through [latex]P_0[/latex] and lying entirely in [latex]S[/latex]. If the tangent lines to all such curves [latex]C[/latex] at [latex]P_0[/latex] lie in the same plane, then this plane is called the tangent plane to [latex]S[/latex] at [latex]P_0[/latex].

A surface S is shown with a point P0 = (x0, y0, z0). There are two intersecting curves shown on S that pass through P0. There are tangents drawn for each of these curves at P0, and these tangent lines create a plane, namely, the tangent plane at P0.
Figure 1. The tangent plane to a surface [latex]\small{S}[/latex] at a point [latex]\small{P_{0}}[/latex] contains all the tangent lines to curves in [latex]\small{S}[/latex] that pass through [latex]\small{P_{0}}[/latex].

For a tangent plane to exist at a point, the function defining the surface needs to be differentiable at that point. We’ll define differentiability more carefully later in this section. For now, here’s the formula for finding tangent planes.

equation of a tangent plane

Let [latex]S[/latex] be a surface defined by a differentiable function [latex]z=f\,(x,\ y)[/latex], and let [latex]P_0=(x_0,\ y_0)[/latex] be a point in the domain of [latex]f[/latex]. Then, the equation of the tangent plane to [latex]S[/latex] at [latex]P_0[/latex] is given by

[latex]z=f\,(x_0,\ y_0)+f_x\,(x_0,\ y_0)(x-x_0)+f_y\,(x_0,\ y_0)(y-y_0)[/latex].

Where Does This Formula Come From?

To understand why this works, let’s build it from the ground up using two tangent lines to the surface [latex]S[/latex]. First, recall what you learned about tangent lines in single-variable calculus.

Definition of a Tangent Line

If [latex]y = f(x)[/latex] is differentiable at [latex]x = x_0[/latex], then the equation of the tangent line to the curve at [latex]x=x_0[/latex] is:

[latex]y = f(x_0)+f'(x_0)(x-x_0)[/latex]

Now, consider two curves on our surface [latex]S[/latex]. The first curve is where [latex]S[/latex] intersects the vertical plane [latex]x=x_0[/latex]. The tangent line to this curve at our point is [latex]z=f(x_0, y_0)+f_y(x_0, y_0)(y-y_0)[/latex]. The second curve is where [latex]S[/latex] intersects the vertical plane [latex]y=y_0[/latex]. Its tangent line is [latex]z=f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)[/latex].

These two tangent lines determine our tangent plane. To find the equation of the plane, we need a normal vector. Let’s find vectors parallel to each tangent line:

  • Parallel to the first tangent line: [latex]{\bf{a}}={\bf{j}}+f_y(x_0, y_0){\bf{k}}[/latex]
  • Parallel to the second tangent line: [latex]{\bf{b}}={\bf{i}}+f_x(x_0, y_0){\bf{k}}[/latex]

The cross product of these vectors gives us a normal vector to the plane:

[latex]\begin{array}{ccc}\hfill {{\bf{a}}\times{\bf{b}}} & =\hfill & {({\bf{j}}+f_y\,(x_0,\ y_0)\,{\bf{k}})\times({\bf{i}}+f_x\,(x_0,\ y_0)\,{\bf{k}})} \hfill \\ \hfill & =\hfill & {\begin{vmatrix}{\bf{i}}&{\bf{j}}&{\bf{k}}\\0&1&f_y\,(x_0,\ y_0)\\1&0&f_x(x_0,\ y_0)\end{vmatrix}} \hfill \\ \hfill & =\hfill & {f_x\,(x_0,\ y_0)\,{\bf{i}}+f_y\,(x_0,\ y_0)\,{\bf{j}}-{\bf{k}}.} \hfill \\ \hfill \end{array}[/latex]

This vector is perpendicular to both tangent lines, so it’s perpendicular to the tangent plane. We can now use this normal vector [latex]{\bf{n}}[/latex] and the point [latex]P_0=(x_0, y_0, f(x_0, y_0))[/latex] to write the equation of the plane:

[latex]\begin{array}{ccc}\hfill {{\bf{n}}\cdot ((x-x_0)\,{\bf{i}}+(y-y_0)\,{\bf{j}}+(z-f\,(x_0,\ y_0))\,{\bf{k}})} & =\hfill & {0} \hfill \\ \hfill {(f_x\,(x_0,\ y_0)\,{\bf{i}}+f_y\,(x_0,\ y_0)\,{\bf{j}}-{\bf{k}})\cdot ((x-x_0)\,{\bf{i}}+(y-y_0)\,{\bf{j}}+(z-f\,(x_0,\ y_0))\,{\bf{k}})} & =\hfill & {0} \hfill \\ \hfill {f_x\,(x_0,\ y_0)(x-x_0)+f_y\,(x_0,\ y_0)(y-y_0)-(z-f\,(x_0,\ y_0))} & =\hfill & {0} \hfill \\ \hfill \end{array}[/latex]

 

Solving for [latex]z[/latex] gives us the tangent plane equation.

Find the equation of the tangent plane to the surface defined by the function [latex]f\,(x,\ y)=2x^{2}-3xy+8y^{2}+2x-4y+4[/latex] at point [latex](2,-1)[/latex].

Find the equation of the tangent plane to the surface defined by the function [latex]f\,(x,y)=\sin{(2x)}\cos{(3y)}[/latex] at point [latex](\pi /3,\pi /4)[/latex].

When Tangent Planes Don’t Exist

Not every surface has a tangent plane at every point. Consider this function:

[latex]f\,(x,\ y)=\begin{cases}\frac{xy}{\sqrt{x^{2}+y^{2}}}\ \ (x,\ y)\neq(0,\ 0) \\ 0\ \ \ \ \ \ \ \ \ \ \ \ (x,\ y)=(0,\ 0).\end{cases}[/latex]

Here’s what its graph looks like:

A curved surface that passes through (0, 0, 0) and that folds up on either side of the z axis.
Figure 3. Graph of a function that does not have a tangent plane at the origin.

When either [latex]x=0[/latex] or [latex]y=0[/latex], we have [latex]f(x, y)=0[/latex]. The function doesn’t change along either axis, so [latex]f_x(x, 0)=f_y(0, y)=0[/latex]. As [latex]x[/latex] or [latex]y[/latex] approach zero along the axes, these partial derivatives stay zero. Using our formula, we’d get [latex]z=0[/latex] as the equation of the tangent plane.

But watch what happens when we approach the origin from a different direction. Let’s approach along the line [latex]y=x[/latex]:

[latex]f\,(x,\ x)=\dfrac{x(x)}{\sqrt{x^{2}+(x)^{2}}}=\dfrac{x^{2}}{\sqrt{2x^{2}}}=\dfrac{|x|}{\sqrt{2}}.[/latex]

When [latex]x>0[/latex], this curve has slope [latex]\sqrt{2}/2[/latex]. When [latex]x<0[/latex], the slope is [latex]-\sqrt{2}/2[/latex].

This creates a problem. Our definition of tangent plane assumes all tangent lines through a point lie in the same plane. That’s clearly not happening here—we’re getting different slopes depending on which direction we approach from. When we discuss differentiability later, we’ll see that this function isn’t differentiable at the origin, which is why a tangent plane doesn’t exist there.