Tangent Planes and Linear Approximations: Fresh Take

  • 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

The Main Idea 

Think of a tangent plane as the 3D version of a tangent line. In single-variable calculus, you found the line that just barely touches a curve at one point. Now you’re doing the same thing with surfaces—finding the plane that grazes the surface at a specific point.

Here’s the key insight: at any point on a smooth surface, there are infinitely many tangent lines. You get one in every direction you could move along the surface. If all these lines lie flat in the same plane, that’s your tangent plane.

The Formula 

For a surface [latex]z = f(x, y)[/latex] at point [latex](x_0, y_0)[/latex], the tangent plane equation is:

[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]

The formula comes from finding two tangent lines. One goes in the [latex]x[/latex]-direction, the other in the [latex]y[/latex]-direction. Then you determine the plane that contains both lines. The partial derivatives [latex]f_x(x_0, y_0)[/latex] and [latex]f_y(x_0, y_0)[/latex] tell you how steeply the surface is tilting in each direction at that point.

Problem-Solving Strategy

  1. Evaluate the function at your point to find [latex]f(x_0, y_0)[/latex]—this gives you the z-coordinate
  2. Find both partial derivatives [latex]f_x[/latex] and [latex]f_y[/latex]
  3. Evaluate the partials at [latex](x_0, y_0)[/latex]
  4. Plug everything into the formula and simplify

Not every surface has a tangent plane at every point. A surface might have a sharp corner or crease. Approaching from different directions might give conflicting information about the slope. In these cases, a tangent plane won’t exist. The function needs to be differentiable at the point—something you’ll explore more carefully later.

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

Linear Approximations

The Main Idea 

Linear approximation is about using something simple (a flat plane) to estimate something more complex (a curved surface). You’ve done this before with tangent lines in single-variable calculus. Now you’re extending it to functions of two variables using tangent planes.

The Core Idea

When you need to approximate [latex]f(x, y)[/latex] near a point [latex](x_0, y_0)[/latex] where you know the exact value, use the tangent plane at that point. The tangent plane equation becomes your approximation formula:

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

This is called the linear approximation of [latex]f[/latex] at [latex](x_0, y_0)[/latex].

Why This Works

Close to [latex](x_0, y_0)[/latex], the surface and its tangent plane are nearly touching. The farther you move from [latex](x_0, y_0)[/latex], the more the surface and plane separate. That’s why this method works best for points near your base point.

Problem-Solving Strategy

  1. Evaluate [latex]f(x_0, y_0)[/latex] to get the exact function value at your base point
  2. Find and evaluate both partial derivatives: [latex]f_x(x_0, y_0)[/latex] and [latex]f_y(x_0, y_0)[/latex]
  3. Plug into the formula to build your linear approximation [latex]L(x, y)[/latex]
  4. Substitute the point you want to approximate into [latex]L(x, y)[/latex]

Choose your base point [latex](x_0, y_0)[/latex] strategically. Pick a point where calculations are easy and that’s close to where you need the approximation. The closer your target point is to [latex](x_0, y_0)[/latex], the better your approximation will be.

Given the function [latex]f\,(x,\ y)=e^{5-2x+3y}[/latex] approximate [latex]f\,(4.1,\ 0.9)[/latex] using point [latex](4,\ 1)[/latex] for [latex](x_0,\ y_0)[/latex]. What is the approximate value of [latex]f\,(4.1,\ 0.9)[/latex] to four decimal places?

Watch the following video to see the worked solution to the above example.

You can view the transcript for “CP 4.20” here (opens in new window).

Differentiability

The Main Idea 

In single-variable calculus, a function is differentiable at a point if its derivative exists there. For functions of two variables, differentiability means something more: the surface must be genuinely smooth at that point, with a well-defined tangent plane.

The Key Issue

Just having both partial derivatives [latex]f_x(x_0, y_0)[/latex] and [latex]f_y(x_0, y_0)[/latex] exist isn’t enough. A function can have partial derivatives at a point but still have a corner or crease there. True differentiability requires the tangent plane to approximate the surface well in all directions near the point.

The Formal Definition

A function [latex]f(x, y)[/latex] is differentiable at [latex](x_0, y_0)[/latex] if you can write:

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

where the error term [latex]E(x, y)[/latex] shrinks faster than the distance to [latex](x_0, y_0)[/latex]:

[latex]\displaystyle\lim_{(x, y)\to(x_0, y_0)}\dfrac{E(x, y)}{\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}=0[/latex]

The Practical Test

Checking the formal definition is tedious. Here’s an easier way: if [latex]f[/latex], [latex]f_x[/latex], and [latex]f_y[/latex] are all continuous at [latex](x_0, y_0)[/latex], then [latex]f[/latex] is automatically differentiable there. Most functions you’ll work with pass this test.

Important Connections

  • Differentiability implies continuity: If [latex]f[/latex] is differentiable at a point, it’s continuous there
  • The reverse isn’t true: A function can be continuous at a point without being differentiable there
  • Watch for trouble spots: Points where partial derivatives don’t exist or aren’t continuous are where differentiability typically fails
Show that the function [latex]f\,(x,\ y)=3x-4y^{2}[/latex] is differentiable at point [latex](-1,\ 2)[/latex].

Watch the following video to see the worked solution to the above example.

You can view the transcript for “CP 4.21” here (opens in new window).

Differentials

The Main Idea 

The total differential gives you a way to estimate how much a function’s output changes when you make small adjustments to its inputs. It’s the multivariable version of the differential [latex]dy = f'(x)dx[/latex] from single-variable calculus.

For a function [latex]z = f(x, y)[/latex], the total differential at [latex](x_0, y_0)[/latex] is:

[latex]dz = f_x(x_0, y_0)dx + f_y(x_0, y_0)dy[/latex]

where [latex]dx = \Delta x[/latex] and [latex]dy = \Delta y[/latex] are small changes in [latex]x[/latex] and [latex]y[/latex].

This approximates the actual change [latex]\Delta z = f(x + \Delta x, y + \Delta y) - f(x, y)[/latex].

What This Means

The total differential tells you: if [latex]x[/latex] changes by a small amount [latex]dx[/latex] and [latex]y[/latex] changes by [latex]dy[/latex], then [latex]z[/latex] changes by approximately [latex]dz[/latex]. Each partial derivative shows how sensitive [latex]z[/latex] is to changes in that particular variable.

Problem-Solving Strategy

  1. Find the partial derivatives [latex]f_x[/latex] and [latex]f_y[/latex]
  2. Evaluate them at your base point [latex](x_0, y_0)[/latex]
  3. Identify the changes [latex]dx[/latex] and [latex]dy[/latex] in your input variables
  4. Calculate [latex]dz = f_x(x_0, y_0)dx + f_y(x_0, y_0)dy[/latex]
  5. Use this to approximate the new function value: [latex]f(x + dx, y + dy) \approx f(x, y) + dz[/latex]

Practical Application

Total differentials are especially useful for error propagation. If your measurements have small errors [latex]dx[/latex] and [latex]dy[/latex], the differential tells you approximately how much error to expect in your calculated result.

For Three Variables

The same idea extends naturally: [latex]dw = f_x dx + f_y dy + f_z dz[/latex] for functions [latex]w = f(x, y, z)[/latex].

Find the differential [latex]dz[/latex] of the function [latex]f(x, y)=4y^{2}+x^{2}y-2xy[/latex] and use it to approximate [latex]\Delta z[/latex] at point [latex](1, -1)[/latex]. Use [latex]\Delta x=0.03[/latex] and [latex]\Delta y=-0.02[/latex]. What is the exact value of [latex]\Delta z[/latex]?

Watch the following video to see the worked solution to the above example.