4.4 Tangent Planes and Linear Approximations

Big idea

The central move of one-variable calculus was replacing a curve near a point by its tangent line. Zoom in far enough on a smooth graph and it looks straight, and the straight thing is easier to compute with. This section carries that move up a dimension: zoom in on a smooth surface and it looks flat, and the flat thing is a plane.

Building that plane takes nothing new. You already have two tangent lines through the point, one along each trace, with slopes $f_x(a,b)$ and $f_y(a,b)$. Two intersecting lines determine a plane, so the tangent plane is forced. Writing it out gives a formula that should look familiar, because it is the tangent line formula with one extra term added for the second input.

Something does change, though, and it is worth being honest about. In one variable, having a derivative at a point was exactly the same as having a good linear approximation there. In two variables, having both partials is weaker than being locally flat. A surface can have a well-defined east-west slope and a well-defined north-south slope at a point and still be creased or torn in the diagonal directions, in which case the plane through those two lines does not hug the surface at all. The word differentiable is reserved for the honest situation where the plane really does approximate. The usable criterion is simple: if both partials exist and are continuous near the point, the function is differentiable there, and every formula built from standard functions passes that test wherever it is defined.

Once you trust the plane, you use it as a calculator. Replacing $f$ by its linearization turns awkward evaluations into arithmetic, and reading the same formula in terms of small changes gives the total differential, which is how small measurement errors in the inputs get propagated to an error in the output.

Decoder

If $f$ has continuous partial derivatives near $(a,b)$, the plane through the point $(a,b,f(a,b))$ with slope $f_x(a,b)$ in the $x$-direction and slope $f_y(a,b)$ in the $y$-direction approximates the surface well enough that the error shrinks faster than the distance you moved.

Everything hangs on the last clause. Any plane through a point on the surface has error going to zero as you approach, simply by continuity. The tangent plane is the one whose error goes to zero faster than the step size itself, so that the ratio of error to distance tends to zero. That is the precise sense in which the plane is the best linear fit and not merely one of many.

Reading the formula in slope terms explains its shape. Starting from the height $f(a,b)$, moving $x - a$ to the east adds $f_x(a,b)(x-a)$, and moving $y - b$ to the north adds $f_y(a,b)(y-b)$. The two contributions add, which is exactly what “flat” means.

The mistake to name here is evaluating the partials at the wrong place. The coefficients in a tangent plane are numbers, frozen at the point of tangency. If $f_x(a,b)$ still has an $x$ in it when you write the plane, you have written a surface, not a plane, and nothing that follows will work.

Definitions and results

Tangent plane. If $f$ is differentiable at $(a,b)$, the tangent plane to $z = f(x,y)$ there is

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

Normal vector. Rearranged, the plane has equation $f_x(a,b)(x-a) + f_y(a,b)(y-b) - (z - f(a,b)) = 0$, so the vector $\langle f_x(a,b),\, f_y(a,b),\, -1 \rangle$ is perpendicular to the surface at that point.

Differentiability. $f$ is differentiable at $(a,b)$ when the change in output can be written as

$$ \Delta z = f_x(a,b)\,\Delta x + f_y(a,b)\,\Delta y + \varepsilon_1 \Delta x + \varepsilon_2 \Delta y $$

where $\varepsilon_1$ and $\varepsilon_2$ both tend to $0$ as $(\Delta x, \Delta y) \to (0,0)$. The two error coefficients are what force the plane to fit better than any other.

Sufficient condition. If $f_x$ and $f_y$ exist and are continuous on a disk around $(a,b)$, then $f$ is differentiable at $(a,b)$. This is the criterion you use in practice.

Differentiable implies continuous. A function differentiable at a point is continuous there. The converse fails, and so does the weaker hope that existing partials alone imply continuity.

Linearization. The right-hand side of the tangent plane equation, viewed as a function, is the linearization

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

and $f(x,y) \approx L(x,y)$ for $(x,y)$ near $(a,b)$. Choose $(a,b)$ to be a nearby point where the exact value is easy.

Total differential. With $dx$ and $dy$ standing for small changes in the inputs, the resulting estimated change in output is

$$ dz = f_x(a,b)\,dx + f_y(a,b)\,dy $$

This is the linearization with the base value subtracted off, and it is the standard tool for propagating measurement error.

Relative error. Dividing $dz$ by $z$ converts absolute error to fractional error. For a product of powers the result is clean: each factor contributes its exponent times its own relative error.

Worked examples

A tangent plane to a paraboloid

Find the tangent plane to $z = x^2 + 2y^2$ at the point where $x = 1$ and $y = 1$.

The height is $f(1,1) = 1 + 2 = 3$. The partials are $f_x = 2x$ and $f_y = 4y$, so at the point the numbers are $f_x(1,1) = 2$ and $f_y(1,1) = 4$. The plane is

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

Check that the plane passes through the point of tangency: $2(1) + 4(1) - 3 = 3$, matching the height. Check the slopes: the plane’s coefficient of $x$ is $2$ and of $y$ is $4$, which are the partials, as required. A normal vector to the surface here is $\langle 2, 4, -1\rangle$.

Estimating a value with the linearization

Estimate $\sqrt{(3.1)^2 + (3.9)^2}$.

Let $f(x,y) = \sqrt{x^2 + y^2}$ and base the approximation at $(3,4)$, where $f(3,4) = 5$ exactly. The partials are

$$ f_x = \frac{x}{\sqrt{x^2+y^2}}, \qquad f_y = \frac{y}{\sqrt{x^2+y^2}} $$

so $f_x(3,4) = 3/5 = 0.6$ and $f_y(3,4) = 4/5 = 0.8$. With $\Delta x = 0.1$ and $\Delta y = -0.1$,

$$ f(3.1, 3.9) \approx 5 + 0.6(0.1) + 0.8(-0.1) = 5 + 0.06 - 0.08 = 4.98 $$

Check against the exact value: $(3.1)^2 = 9.61$ and $(3.9)^2 = 15.21$, summing to $24.82$, and $\sqrt{24.82} = 4.98196\ldots$. The estimate is off by about $0.002$, which is small compared with the step of size $0.1\sqrt{2} \approx 0.14$ that was taken. That ratio is the statement about error shrinking faster than distance, seen numerically.

Propagating measurement error

A cylindrical tank is measured as radius $5$ meters and height $10$ meters, each measurement good to within $0.1$ meter. Estimate the possible error in the computed volume $V = \pi r^2 h$.

The partials are $V_r = 2\pi r h$ and $V_h = \pi r^2$, so at $r = 5$, $h = 10$ they equal $100\pi$ and $25\pi$. With $|dr| \le 0.1$ and $|dh| \le 0.1$,

$$ |dV| \le 100\pi(0.1) + 25\pi(0.1) = 12.5\pi \approx 39.3 $$

cubic meters, against a computed volume of $250\pi \approx 785.4$.

Check this by relative error instead. Dividing $dV = 2\pi r h\,dr + \pi r^2 dh$ by $V = \pi r^2 h$ gives

$$ \frac{dV}{V} = \frac{2\,dr}{r} + \frac{dh}{h} = 2(0.02) + 0.01 = 0.05 $$

a five percent error, and five percent of $250\pi$ is $12.5\pi$. The two routes agree. Notice the factor of $2$ on the radius term: because the radius enters squared, an error there costs twice as much as the same fractional error in the height.

A linearization with an exponential

Approximate $f(x,y) = x e^{y}$ near $(2,0)$ and use it at $(2.05, -0.02)$.

Here $f(2,0) = 2$, $f_x = e^y$ gives $f_x(2,0) = 1$, and $f_y = x e^y$ gives $f_y(2,0) = 2$. The linearization is

$$ L(x,y) = 2 + (x - 2) + 2y $$

At the requested point, $L = 2 + 0.05 + 2(-0.02) = 2.01$. The exact value is $2.05 e^{-0.02} = 2.0094\ldots$, so the linear estimate is high by about $0.0006$. Base points are chosen exactly this way: pick the nearest place where the exponential and the arithmetic are both trivial.

Practice

First the plane itself. These give you a surface and a point and ask for the tangent plane, or for a normal vector to the surface there.

Then the plane used as a tool. These ask for a linearization, for an approximate value, or for the change in a quantity predicted by the total differential.

Quiz

Five items on tangent planes, normal vectors, differentiability, linear approximation, and error estimates with the total differential.