Measurements are never exact. If you measure a can's radius and height, each a little off, how far off is the volume you compute? The tangent plane answers this: for small changes, the surface is nearly flat, and the change along the plane is easy to compute.
Key idea
The differential of $z = f(x, y)$ is
$$dz = f_x\,dx + f_y\,dy.$$
For small changes $dx$, $dy$, the true change $\Delta z$ is approximately $dz$. To bound the error in a computed quantity, use the largest possible measurement errors and add the sizes of the terms: $|dz| \le |f_x||dx| + |f_y||dy|$.
Why it works
Moving from $(a, b)$ to $(a + dx, b + dy)$ along the tangent plane changes the height by $f_x\,dx + f_y\,dy$: the $x$-slope times the $x$-step plus the $y$-slope times the $y$-step. Since the plane hugs the surface near the point, the real change differs from $dz$ only by a much smaller amount (roughly the square of the step). Dividing by $z$ gives relative error: for $Q = kx^ay^b$, $\frac{dQ}{Q} = a\frac{dx}{x} + b\frac{dy}{y}$, so percent errors add, each multiplied by its power.
Worked examples
Worked example: A differential
For $z = x^2y$, compare $dz$ with $\Delta z$ at $(2, 3)$ when $dx = 0.1$ and $dy = -0.05$.
- $f_x = 2xy = 12$ and $f_y = x^2 = 4$ at $(2, 3)$.
- $$dz = 12(0.1) + 4(-0.05) = 1.2 - 0.2 = 1.0.$$
- Exact: $$(2.1)^2(2.95) - 12 = 13.0095 - 12 = 1.0095.$$ The differential is off by less than ${}0.01$.
Answer: $dz = 1.0$, $\Delta z = 1.0095$
Worked example: Error in a volume
A can measures $r = 5$ cm and $h = 10$ cm, each to within ${}0.1$ cm and ${}0.2$ cm. Estimate the largest error in $V = \pi r^2h$.
- $V_r = 2\pi rh = 100\pi$ and $V_h = \pi r^2 = 25\pi$.
- Worst case: $$|dV| \le 100\pi(0.1) + 25\pi(0.2) = 10\pi + 5\pi = 15\pi \approx 47$$ cm$^3$.
- Relative: $V = 250\pi$, so the error is about $\frac{15\pi}{250\pi} = 6\%$, which is ${}2(2\%) + 2\%$, as the power rule predicts.
Answer: about ${}15\pi \approx 47$ cm$^3$
Worked example: Percent errors
The period of a pendulum is computed from $T = 2\pi\sqrt{L/g}$. If $L$ is off by at most ${}1\%$ and $g$ by at most ${}0.5\%$, how far off can $T$ be?
- $T = 2\pi L^{1/2}g^{-1/2}$, so $\frac{dT}{T} = \frac{1}{2}\frac{dL}{L} - \frac{1}{2}\frac{dg}{g}$.
- Worst case: $\frac{1}{2}(1\%) + \frac{1}{2}(0.5\%) = 0.75\%$.
Answer: about ${}0.75\%$
Common mistake
Letting errors cancel. In $dV = V_r\,dr + V_h\,dh$, the measurement errors could have either sign. To find the largest possible error, both must push the same way, so add the absolute values. Using the signs as given is right only when you know the actual changes.