Difficulty: Intermediate | Prerequisites: Partial derivatives (MATH 163, earlier sections), vectors and dot/cross products, equation of a plane from Calculus II/III.
Tangent planes are the multivariable extension of tangent lines. In single-variable calculus you learned that the tangent line touches a curve at a point and gives the best linear approximation nearby. Here you do the same thing, but for surfaces in three dimensions: instead of a line that hugs a curve, you get a flat plane that hugs a surface. This topic sits at the bridge between partial derivatives (which you have just learned) and linear approximation in 3D (which comes next). You need to be comfortable computing partial derivatives and working with the equation of a plane before diving in.
A tangent plane is a flat surface that just touches the graph of a two-variable function at a single point, the same way a tangent line touches a curve. You build it by computing partial derivatives, turning those into direction vectors, and using the standard plane equation. Once you have the tangent plane, it doubles as a linear approximation of the function near that point.
Tangent plane
The unique plane that touches the surface z = f(x, y) at a given point and contains both tangent lines in the x- and y-directions. Think of it as the flat sheet you would press against a curved surface at one spot.
Partial derivative (f_x, f_y)
The derivative of a multivariable function with respect to one variable while all others are held constant. In simple terms, f_x tells you the slope of the surface if you walk in the x-direction only, and f_y tells you the slope if you walk in the y-direction only.
Normal vector
A vector that is perpendicular to a surface (or plane) at a given point. In simple terms, it points straight "outward" from the surface, like a flagpole standing on a hillside.
Cross product
An operation on two 3D vectors that produces a third vector perpendicular to both. Here it is used to find the normal vector to the tangent plane from the two tangent-direction vectors.
Direction vectors of the tangent plane
The vectors (1, 0, f_x) and (0, 1, f_y) built from the partial derivatives at the point of tangency. These lie in the tangent plane and point along the x- and y-slice directions of the surface.
Equation of a plane
Given a normal vector n = (a, b, c) and a point (x_0, y_0, z_0), the plane is a(x - x_0) + b(y - y_0) + c(z - z_0) = 0. Think of it as the set of all points whose displacement from the known point is perpendicular to the normal.
Step 1: Identify the point of tangency. You are given (x_0, y_0). Compute z_0 = f(x_0, y_0) to get the full point (x_0, y_0, z_0).
Step 2: Compute the partial derivatives. Find f_x(x, y) and f_y(x, y).
Step 3: Evaluate the partials at the point. Plug in (x_0, y_0) to get f_x(x_0, y_0) and f_y(x_0, y_0). These are the slopes.
Step 4: Write the tangent plane equation. Plug everything into the general formula (see Formulas section below).
The partial derivatives give you two vectors that lie in the tangent plane:
In the x-direction: (1, 0, f_x)
In the y-direction: (0, 1, f_y)
Taking their cross product gives a vector perpendicular to both, which is the normal vector to the plane. With the normal and a point on the plane, you write the standard plane equation.
f(x, y) = 3y^2 - 2x^2 + x at the point (2, -1).
f(2, -1) = 3(1) - 2(4) + 2 = 3 - 8 + 2 = -3. So the point is (2, -1, -3).
f_x = -4x + 1, so f_x(2, -1) = -7.
f_y = 6y, so f_y(2, -1) = -6.
Tangent direction vectors: (1, 0, -7) and (0, 1, -6).
Cross product gives normal vector n = (7, 6, 1).
Plane equation: 7(x - 2) + 6(y + 1) + 1(z + 3) = 0, which simplifies to z = -7(x - 2) - 6(y + 1) - 3.
f(x, y) = x e^(xy) at the point (2, 0).
f(2, 0) = 2 e^0 = 2. So the point is (2, 0, 2).
f_x = e^(xy) + xy e^(xy). At (2, 0): f_x = 1 + 0 = 1.
f_y = x^2 e^(xy). At (2, 0): f_y = 4.
Tangent plane: z = 1(x - 2) + 4(y - 0) + 2 = (x - 2) + 4y + 2.
f(x, y) = xy sin(x - y) / (1 + x^2 + y^2) at the point (1, 1).
f(1, 1) = 0 (because sin(0) = 0). The point is (1, 1, 0).
After computing the partials (which is messy but doable with the product and quotient rules), you get f_x(1, 1) = 1/3 and f_y(1, 1) = -1/3.
Tangent plane: z = (1/3)(x - 1) - (1/3)(y - 1), which simplifies to z = x/3 - y/3.
General tangent plane equation
z = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) + z_0where z_0 = f(x_0, y_0).
Normal vector via cross product
From the direction vectors (1, 0, f_x) and (0, 1, f_y):
\vec{n} = \langle f_x,\; f_y,\; 1 \rangle \quad \text{(up to sign)}More precisely, the cross product of (1, 0, f_x) and (0, 1, f_y) is (-f_x, -f_y, 1), but any scalar multiple works as a normal vector, so (f_x, f_y, -1) or the version in the lecture notes, where the signs come out as (7, 6, 1) for the first worked example.
Standard plane equation (for reference)
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0where (a, b, c) is the normal vector and (x_0, y_0, z_0) is a point on the plane.
Tangent planes are how engineers and physicists linearise complicated surfaces near a point of interest. If you are modelling terrain elevation as a function of latitude and longitude, the tangent plane at your current position tells you the local slope of the ground in every direction.
Students often forget to compute z_0 = f(x_0, y_0) and try to write the plane with only two coordinates. The tangent plane lives in 3D; you need all three.
Confusing the normal vector with the direction vectors. The partial derivative values go into the direction vectors (1, 0, f_x) and (0, 1, f_y). The normal vector comes from their cross product, not from the partial derivatives directly.
Sign errors in the cross product. The determinant expansion is mechanical, but a wrong sign flips the normal and produces a plane that does not pass through the point. Always verify by substituting the known point back into the equation.
Mixing up the tangent plane formula with the plane equation from earlier in the course. The tangent plane formula already has the correct normal baked in, so you do not need to separately compute a cross product if you use the general formula directly.
The tangent plane equation is almost certainly on your final. Know the formula cold and be able to apply it to polynomial, exponential, and trig surfaces.
Expect at least one problem where you must find the tangent plane and then use it for linear approximation (covered in the companion study notes).
The cross-product derivation is a popular "show your reasoning" question. Be ready to explain why the vectors (1, 0, f_x) and (0, 1, f_y) lie in the tangent plane.
True or false: The tangent plane to z = f(x, y) at a point always passes through that point. (True)
Fill in the blank: The two direction vectors used to build the tangent plane are (1, 0, ____) and (0, 1, ____). (f_x and f_y, evaluated at the point)
True or false: You can write the tangent plane equation without ever computing a cross product if you use the general formula directly. (True)
Fill in the blank: If f(x, y) = x^2 + y^2, then f_x = ____ and f_y = ____. (2x and 2y)
True or false: The normal vector to the tangent plane is parallel to the tangent plane. (False, it is perpendicular)
Q: Find the tangent plane to f(x, y) = x^2 + y^2 at the point (1, 2).
A: f(1, 2) = 5. f_x = 2x, so f_x(1, 2) = 2. f_y = 2y, so f_y(1, 2) = 4. Tangent plane: z = 2(x - 1) + 4(y - 2) + 5.
Q: For f(x, y) = 3y^2 - 2x^2 + x, what is the normal vector to the tangent plane at (2, -1)?
A: f_x(2, -1) = -7, f_y(2, -1) = -6. Direction vectors are (1, 0, -7) and (0, 1, -6). Cross product gives n = (7, 6, 1).
Q: Write the general tangent plane formula for z = f(x, y) at the point (x_0, y_0, z_0).
A: z = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) + z_0.
Q: A student writes the tangent plane to f(x, y) = e^(xy) at (1, 0) as z = e(x - 1). What did they do wrong?
A: They forgot the f_y term. f_y = x^2 e^(xy), so f_y(1, 0) = 1, and the plane should also include the term 1(y - 0). The correct plane is z = (x - 1) + y + 1.
This connects directly to linear approximation in 3D: the tangent plane is the linear approximation. Once you can write the tangent plane, you can estimate function values near the point of tangency without a calculator.
It also connects to the gradient vector, which you will meet soon. The gradient at a point is perpendicular to the level curve, and its components are exactly the partial derivatives that appear in the tangent plane formula.
In optimisation (finding maxima and minima of multivariable functions), the tangent plane being horizontal (both partial derivatives equal to zero) is the condition for a critical point.
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