Tangent Planes

A tangent line, one dimension down

The curve $y=x^2$y = x² has slope 2x, which is 4 at the point (2, 4). The tangent line there is y = 4 + 4(x − 2), that is y = 4x − 4. Near x = 2 the line and the curve are hard to tell apart: at x = 2.1 the curve is at 4.41 and the line at 4.4, a gap of 0.01.

So near the point, the line can stand in for the curve. A surface has the same thing one dimension up: a flat plane that rests on it at one point.

The curve $y=x^2$y = x² and its tangent line y = 4x − 4 at the marked point (2, 4). The line meets the curve there with the same slope, 4, and stays close to it on either side.

A flat plane resting on a surface

Take the dome $f\left(x,y\right)=16\left(1-x^2-y^2\right)$f(x, y) = 16(1 − x² − y²) and the point above (0.54, 0.42), at height 16(1 − 0.2916 − 0.1764) = 8.512. Close to that point the dome curves only a little, and a flat plane can be laid against it so that it touches there and tilts the same way the dome does. That plane is the tangent plane at the point.

At (0.54, 0.42) the partial derivatives are $\frac{\partial f}{\partial x}=-32x=-17.28$∂f/∂x = −32x = −17.28 and $\frac{\partial f}{\partial y}=-32y=-13.44$∂f/∂y = −32y = −13.44: the dome falls in both the x and the y directions, steeply, and the plane falls with it.

The dome $z=16\left(1-x^2-y^2\right)$z = 16(1 − x² − y²) and a gold patch of its tangent plane at the marked point above (0.54, 0.42). The patch is built from the two partial derivatives there: its height changes by −17.28 per unit in x and by −13.44 per unit in y.

Two slopes fix the plane

A plane through the point $\left(x_0,y_0,z_0\right)$(x₀, y₀, z₀) that is not vertical can be written $z-z_0=a\left(x-x_0\right)+b\left(y-y_0\right)$z − z₀ = a(x − x₀) + b(y − y₀). Moving in x alone, it climbs a for each unit; moving in y alone, it climbs b. To rest on the surface, its slope in x must equal $\frac{\partial f}{\partial x}$∂f/∂x there and its slope in y must equal $\frac{\partial f}{\partial y}$∂f/∂y.

So the tangent plane at $\left(x_0,y_0\right)$(x₀, y₀) is $z-z_0=\left(\frac{\partial f}{\partial x}\right)\left(x-x_0\right)+\left(\frac{\partial f}{\partial y}\right)\left(y-y_0\right)$z − z₀ = (∂f/∂x)(x − x₀) + (∂f/∂y)(y − y₀), with both partial derivatives evaluated at $\left(x_0,y_0\right)$(x₀, y₀) and $z_0=f\left(x_0,y_0\right)$z₀ = f(x₀, y₀). One slope alone is not enough: it fixes a line in the plane, and the plane could still swing about that line.

The bowl at (3, 4)

For $f\left(x,y\right)=x^2+y^2$f(x, y) = x² + y² at (3, 4): $z_0=9+16=25$z₀ = 9 + 16 = 25, $\frac{\partial f}{\partial x}=2x=6$∂f/∂x = 2x = 6 and $\frac{\partial f}{\partial y}=2y=8$∂f/∂y = 2y = 8. The tangent plane is z = 25 + 6(x − 3) + 8(y − 4), which multiplies out to z = 6x + 8y − 25. At (3, 4) it gives 18 + 32 − 25 = 25, so the plane passes through the point on the surface.

Cut both with the vertical plane y = 4. The surface becomes the curve $z=x^2+16$z = x² + 16 and the tangent plane becomes the line z = 25 + 6(x − 3), that is z = 6x + 7: the tangent line to that curve at x = 3. Cut with x = 3 instead, and the curve $z=9+y^2$z = 9 + y² meets the line z = 8y − 7 at y = 4. The tangent plane holds the tangent line of every slice through the point.

The bowl $z=x^2+y^2$z = x² + y² and its tangent plane at (3, 4), both cut along y = 4. The plain curve is the slice $z=x^2+16$z = x² + 16; the gold line is the plane’s slice, z = 6x + 7, touching the curve at (3, 25) with slope 6, the value of $\frac{\partial f}{\partial x}$∂f/∂x there.

How close it stays

Step from (3, 4) to (3.1, 4.2). The surface is at $3.1^2+4.2^2=27.25$3.1² + 4.2² = 27.25 and the plane at 25 + 0.6 + 1.6 = 27.2, a gap of 0.05. Step a tenth as far, to (3.01, 4.02): the surface is at 25.2205 and the plane at 25.22, a gap of 0.0005, a hundredth of the first.

For this bowl the gap is exactly $h^2+k^2$h² + k² for a step of h in x and k in y, so it shrinks with the square of the step while the step itself shrinks only in proportion. That is what makes the plane the best flat approximation: near the point, $f\left(x,y\right)\approx z_0+\left(\frac{\partial f}{\partial x}\right)\left(x-x_0\right)+\left(\frac{\partial f}{\partial y}\right)\left(y-y_0\right)$f(x, y) ≈ z₀ + (∂f/∂x)(x − x₀) + (∂f/∂y)(y − y₀), and the estimate 27.2 for f(3.1, 4.2) is out by 0.05.

A plane built at one point

The partial derivatives change from point to point, so each point has its own tangent plane, and one plane cannot fit a curved surface everywhere.

Nor need the surface stay on one side of it. For the saddle $z=x^2-y^2$z = x² − y² at the origin, both partial derivatives are 0 and the tangent plane is z = 0. The saddle lies above that plane along the x-axis, where $z=x^2$z = x², and below it along the y-axis, where $z=-y^2$z = −y², crossing it along the lines y = x and y = −x.

The usual mistakes

Pairing a slope with the wrong variable. $\frac{\partial f}{\partial x}$∂f/∂x multiplies $\left(x-x_0\right)$(x − x₀) and $\frac{\partial f}{\partial y}$∂f/∂y multiplies $\left(y-y_0\right)$(y − y₀): for the bowl at (3, 4), z = 25 + 8(x − 3) + 6(y − 4) is a different plane, and it misses the surface’s slope in both directions.

Leaving the partial derivatives as functions. In z = 25 + 2x(x − 3) + 2y(y − 4) the terms are not linear, so it is not a plane; evaluate 2x and 2y at (3, 4) first.

Dropping $z_0$z₀. z = 6(x − 3) + 8(y − 4) has the right tilt but passes through (3, 4, 0), 25 units below the surface.

A tree measured with a tape and a clinometer

In the application below, a tree’s height is $H=d\tan \theta$H = d tan θ, a function of the distance d and the angle $\theta$θ. Its tangent plane at the readings estimates the height from a second pair of readings, and adding the sizes of the two partial-derivative terms gives the largest error each instrument can cause.