Tangent Planes

A tangent plane to the surface z = f(x, y) at (a, b) is the plane that touches the surface there and has the same slopes in the x and y directions. Its equation is z = f(a, b) + f_x(a, b)(x − a) + f_y(a, b)(y − b). Near the point, the plane approximates the function.

Before this lesson: Partial Derivatives, Equation of a Plane and Linear Approximation

What a tangent plane is

Zoom in on a smooth surface and it starts to look flat. The flat piece it looks like is the tangent plane. In one variable, a curve looks like its tangent line up close; this is the same idea one dimension higher.

The tangent plane has to match the surface’s slope in the x direction, which is f_x(a, b), and its slope in the y direction, which is f_y(a, b). It also has to pass through the point (a, b, f(a, b)). Those three facts pin down exactly one plane.

Tangent plane equation

z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)

• Linearization: the right side is called the linearization L(x, y). For (x, y) near (a, b), f(x, y) ≈ L(x, y). This is linear approximation in two variables.
• Total differential: a small change in the inputs changes z by about dz = f_x \, dx + f_y \, dy.
• Surfaces F(x, y, z) = k: the tangent plane at (x_0, y_0, z_0) is F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0, with the partials evaluated at the point.
• Normal vector: for z = f(x, y), the vector \langle f_x, f_y, -1 \rangle is perpendicular to the tangent plane. For F(x, y, z) = k, use \langle F_x, F_y, F_z \rangle.
• When it works: if f_x and f_y are continuous near (a, b), then f is differentiable there and the tangent plane is a good fit.

Tangent plane examples

Example 1: Tangent plane to a paraboloid

Find the tangent plane to z = x² + 3y² at the point (2, 1, 7).

f(2, 1) = 2^2 + 3(1)^2 = 7

f_x = 2x

f_x(2, 1) = 2(2) = 4

f_y = 6y

f_y(2, 1) = 6(1) = 6

z = 7 + 4(x - 2) + 6(y - 1)

z = 7 + 4x - 8 + 6y - 6

z = 4x + 6y - 7

Answer: the tangent plane is z = 4x + 6y − 7.

Example 2: Linear approximation of a square root

Use a linear approximation to estimate √(3.02² + 3.97²).

f(x, y) = \sqrt{x^2 + y^2}, \quad (a, b) = (3, 4)

f(3, 4) = \sqrt{9 + 16} = 5

f_x = \dfrac{x}{\sqrt{x^2 + y^2}}

f_x(3, 4) = \dfrac{3}{5} = 0.6

f_y = \dfrac{y}{\sqrt{x^2 + y^2}}

f_y(3, 4) = \dfrac{4}{5} = 0.8

L = 5 + 0.6(3.02 - 3) + 0.8(3.97 - 4)

= 5 + 0.6(0.02) + 0.8(-0.03)

= 5 + 0.012 - 0.024

= 4.988

Answer: √(3.02² + 3.97²) ≈ 4.988. A calculator gives 4.98812…, so the estimate is off by about 0.0001.

Example 3: A logarithm surface

Find the tangent plane to z = ln(x − 2y) at the point where x = 3 and y = 1.

f(3, 1) = \ln(3 - 2) = \ln 1 = 0

f_x = \dfrac{1}{x - 2y}

f_x(3, 1) = \dfrac{1}{1} = 1

f_y = \dfrac{-2}{x - 2y}

f_y(3, 1) = \dfrac{-2}{1} = -2

z = 0 + 1(x - 3) - 2(y - 1)

z = x - 3 - 2y + 2

z = x - 2y - 1

Answer: the tangent plane at (3, 1, 0) is z = x − 2y − 1.

Example 4: A surface given by an equation

Find the tangent plane and the normal line to the ellipsoid x² + 2y² + 3z² = 21 at (1, 2, 2).

1^2 + 2(2)^2 + 3(2)^2 = 1 + 8 + 12 = 21

F_x = 2x = 2

F_y = 4y = 8

F_z = 6z = 12

2(x - 1) + 8(y - 2) + 12(z - 2) = 0

(x - 1) + 4(y - 2) + 6(z - 2) = 0

x - 1 + 4y - 8 + 6z - 12 = 0

x + 4y + 6z = 21

x = 1 + t, \quad y = 2 + 4t, \quad z = 2 + 6t

Answer: the tangent plane is x + 4y + 6z = 21, and the normal line is x = 1 + t, y = 2 + 4t, z = 2 + 6t.

Example 5: Estimating a change in volume (word problem)

A cone has radius 5 cm and height 12 cm, so its volume is V = (1/3)πr²h. The radius grows to 5.1 cm while the height shrinks to 11.8 cm. Estimate the change in volume.

V_r = \dfrac{2}{3}\pi rh

V_r(5, 12) = \dfrac{2}{3}\pi(5)(12) = 40\pi

V_h = \dfrac{1}{3}\pi r^2

V_h(5, 12) = \dfrac{1}{3}\pi(25) = \dfrac{25\pi}{3}

dr = 0.1, \quad dh = -0.2

dV = 40\pi(0.1) + \dfrac{25\pi}{3}(-0.2)

= 4\pi - \dfrac{5\pi}{3}

= \dfrac{7\pi}{3}

\approx 7.33

Answer: the volume grows by about 7.33 cm³. The exact change is about 7.24 cm³, so the estimate is close. The growing radius adds more volume than the shrinking height takes away.

Common mistakes

Related methods

One-variable linear approximation

With only one input, the tangent plane formula becomes the tangent line y = f(a) + f′(a)(x − a) from linear approximation in Calculus 1. The two-variable version just adds a second slope term.

Total differential and error estimates

The total differential dz = f_x \, dx + f_y \, dy estimates how measurement errors in x and y affect z. It is the same calculation as Example 5, which is the two-variable version of differentials.

Normal lines

The line through the point in the direction of the normal vector is the normal line. It is perpendicular to the surface, and you write it with the parametric equations of a line.

Practice problems

1. Find the tangent plane to z = 3x² − xy + y² at (1, 2, 5).
2. Find the tangent plane to z = xy² at (2, −1, 2).
3. Find the tangent plane to z = eˣ cos y at (0, 0, 1).
4. Find the tangent plane to z = √(25 − x² − y²) at (3, 0, 4).
5. Find the linearization of f(x, y) = x/y at (6, 3), and use it to estimate f(6.3, 2.9).
6. Use a linear approximation to estimate (1.98)³(3.01)².
7. Find the tangent plane to x² + y² − z² = 4 at (2, 1, 1).
8. Find the tangent plane to xyz = 6 at (1, 2, 3).
9. Find the normal line to z = x² + y² at (1, 2, 5).
10. A metal plate is 20 cm long and 15 cm wide. Heating stretches the length by 0.04 cm and the width by 0.03 cm. Use the total differential to estimate the change in area.

f_x = 6x - y

f_x(1, 2) = 6 - 2 = 4

f_y = -x + 2y

f_y(1, 2) = -1 + 4 = 3

z = 5 + 4(x - 1) + 3(y - 2)

z = 5 + 4x - 4 + 3y - 6

z = 4x + 3y - 5

f_x = y^2, \quad f_x(2, -1) = 1

f_y = 2xy, \quad f_y(2, -1) = -4

z = 2 + 1(x - 2) - 4(y + 1)

z = 2 + x - 2 - 4y - 4

z = x - 4y - 4

f_x = e^x\cos y, \quad f_x(0, 0) = 1

f_y = -e^x\sin y, \quad f_y(0, 0) = 0

z = 1 + 1(x - 0) + 0(y - 0)

z = 1 + x

f_x = \dfrac{-x}{\sqrt{25 - x^2 - y^2}}, \quad f_x(3, 0) = -\dfrac{3}{4}

f_y = \dfrac{-y}{\sqrt{25 - x^2 - y^2}}, \quad f_y(3, 0) = 0

z = 4 - \dfrac{3}{4}(x - 3)

z = 4 - \dfrac{3}{4}x + \dfrac{9}{4}

z = -\dfrac{3}{4}x + \dfrac{25}{4}

f(6, 3) = \dfrac{6}{3} = 2

f_x = \dfrac{1}{y} = \dfrac{1}{3}

f_y = -\dfrac{x}{y^2} = -\dfrac{6}{9} = -\dfrac{2}{3}

L(x, y) = 2 + \dfrac{1}{3}(x - 6) - \dfrac{2}{3}(y - 3)

L(6.3, 2.9) = 2 + \dfrac{1}{3}(0.3) - \dfrac{2}{3}(-0.1)

= 2 + 0.1 + 0.0667

\approx 2.167

f = x^3y^2, \quad f(2, 3) = 8(9) = 72

f_x = 3x^2y^2 = 3(4)(9) = 108

f_y = 2x^3y = 2(8)(3) = 48

L = 72 + 108(-0.02) + 48(0.01)

= 72 - 2.16 + 0.48

= 70.32

F_x = 2x = 4

F_y = 2y = 2

F_z = -2z = -2

4(x - 2) + 2(y - 1) - 2(z - 1) = 0

2(x - 2) + (y - 1) - (z - 1) = 0

2x - 4 + y - 1 - z + 1 = 0

2x + y - z = 4

F_x = yz = 6

F_y = xz = 3

F_z = xy = 2

6(x - 1) + 3(y - 2) + 2(z - 3) = 0

6x - 6 + 3y - 6 + 2z - 6 = 0

6x + 3y + 2z = 18

f_x = 2x = 2, \quad f_y = 2y = 4

\langle 2, 4, -1 \rangle

x = 1 + 2t, \quad y = 2 + 4t, \quad z = 5 - t

A = lw, \quad A_l = w = 15, \quad A_w = l = 20

dA = 15(0.04) + 20(0.03)

= 0.6 + 0.6

= 1.2

Tangent planes FAQ

What is the equation of a tangent plane?

For z = f(x, y) at (a, b), it is z = f(a, b) + f_x(a, b)(x − a) + f_y(a, b)(y − b). It uses the value of f and both partial derivatives at the point.

How is a tangent plane used for linear approximation?

Near (a, b), the surface and its tangent plane are very close, so the plane’s height L(x, y) estimates f(x, y). The closer (x, y) is to (a, b), the better the estimate.

How do you find the tangent plane to a surface like x² + y² + z² = 9?

Write the surface as F(x, y, z) = k. The gradient ⟨F_x, F_y, F_z⟩ at the point is a normal vector, so the plane is F_x(x − x₀) + F_y(y − y₀) + F_z(z − z₀) = 0.

When does a surface not have a tangent plane?

At points where the surface has a sharp edge or point, like the tip of the cone z = √(x² + y²). There the partial derivatives do not exist, so no single plane fits.

Summary

• The tangent plane to z = f(x, y) at (a, b) is z = f(a, b) + f_x(a, b)(x − a) + f_y(a, b)(y − b).
• Near (a, b), f(x, y) ≈ L(x, y), the linearization.
• The total differential dz = f_x \, dx + f_y \, dy estimates small changes.
• For F(x, y, z) = k, use F_x(x − x₀) + F_y(y − y₀) + F_z(z − z₀) = 0.
• The normal vector is ⟨f_x, f_y, −1⟩, or ⟨F_x, F_y, F_z⟩.

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