Difficulty: Intermediate | Prerequisites: Single-variable derivatives, Calculus I chain rule
Linear approximation uses the value of a function and its two partial derivatives at a known point to estimate the function nearby. When the function is given as a formula, you compute partial derivatives directly. When it is a table, you use centered differences. When it is a contour plot, you read off values at nearby contours and divide by the distance.
Partial derivative (f_x, f_y)
The rate of change of f with respect to one variable while the others are held fixed. f_x(a, b) = lim_{h→0} [f(a+h, b) – f(a, b)] / h. In simple terms, it is the slope of the surface in one coordinate direction.
Linear approximation (tangent plane approximation)
The first-order estimate f(x, y) ≈ f(a, b) + f_x(a, b)(x – a) + f_y(a, b)(y – b). Think of it as using the tangent plane to the surface at (a, b) as a stand-in for the surface itself, valid when (x, y) is close to (a, b).
Center difference derivative
An estimate of a partial derivative from tabulated data: f_x(a, b) ≈ [f(a+h, b) – f(a–h, b)] / (2h). More accurate than a one-sided (forward or backward) difference because errors cancel.
Forward difference derivative
f_x(a, b) ≈ [f(a+h, b) – f(a, b)] / h. Used when data on one side is unavailable.
Backward difference derivative
f_x(a, b) ≈ [f(a, b) – f(a–h, b)] / h. The mirror image of the forward difference.
Contour plot (level curve map)
A plot showing curves where f(x, y) takes a constant value. The spacing between contours indicates how fast f is changing: closely spaced contours mean a steep slope.
For a function f(x, y) differentiable at (a, b):
f(x, y) ≈ f(a, b) + f_x(a, b)(x – a) + f_y(a, b)(y – b)
This is the multivariable analogue of the single-variable tangent line approximation f(x) ≈ f(a) + f'(a)(x – a).
Given f(x, y) = x e^{y–x}, estimate f(1.1, 1.95).
Compute the partials: f_x = (1 – x)e^{y–x}, f_y = x e^{y–x}
Evaluate at (1, 2): f(1, 2) = e, f_x(1, 2) = 0, f_y(1, 2) = e
Plug in: f(1.1, 1.95) ≈ e + 0(0.1) + e(−0.05) = 0.95e
The key step is recognising that f_x vanishes at (1, 2), so the x-displacement contributes nothing.
Given a table of values for g(x, y) with integer grid spacing, estimate g(1.1, 1.95).
Use the centered difference because data is available on both sides:
g_x(1, 2) ≈ [g(2, 2) – g(0, 2)] / 2 = [22 – 18] / 2 = ... wait, let me be precise from the exam: g(2, 2) = 22, g(0, 2) = 18. So g_x = (22 – 18)/2 = 2? No: checking the table again, g(0, 2) is at row y = 2, column x = 0, which is 26, and g(2, 2) is at row y = 2, column x = 2, which is 22. So g_x(1, 2) = (22 – 26)/2 = −2.
g_y(1, 2) ≈ [g(1, 3) – g(1, 1)] / 2 = [28 – 22] / 2 = 3
g(1, 2) = 25
g(1.1, 1.95) ≈ 25 + (−2)(0.1) + 3(−0.05) = 25 – 0.2 – 0.15 = 24.65
Always use the centered difference when the grid allows it. It is more accurate than a one-sided difference.
Given level curves for h(x, y), estimate h(1.1, 1.95).
The approach is to average forward and backward difference estimates using distances to the nearest contour lines:
Read h(1, 2) ≈ 8 from the contour map
Estimate h_x(1, 2): the contour h = 4 lies at roughly x = 1.125 (forward), and h = 12 is at roughly x = 0.75 (backward). Average the two slopes: (1/2)[(4 – 8)/0.125 + (8 – 12)/(−0.25)] = (1/2)[−32 + (−16)] = −24
Estimate h_y(1, 2) similarly: −10
h(1.1, 1.95) ≈ 8 + (−24)(0.1) + (−10)(−0.05) = 8 – 2.4 + 0.5 = 6.1
Reading contour plots requires care. Interpolate positions by eye and use whichever nearby contours give the cleanest reading.
f(x,y) \approx f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)f_x(a,b) \approx \frac{f(a+h,b) - f(a-h,b)}{2h} \quad \text{(center difference)}f_x(a,b) \approx \frac{f(a+h,b) - f(a,b)}{h} \quad \text{(forward difference)}f_x(a,b) \approx \frac{f(a,b) - f(a-h,b)}{h} \quad \text{(backward difference)}For contour plots, h is the distance from the point to the nearest contour in that direction, and the function values come from the contour labels.
Linear approximation is the mathematical backbone of error propagation in engineering and the sciences. When you measure temperature and pressure with small uncertainties, linear approximation tells you how those uncertainties combine to affect a calculated quantity. Weather forecasting models use linearisation around a current state to project conditions forward in time.
Students often use a forward or backward difference when a centered difference is available. The centered difference is more accurate and should always be your first choice when data on both sides exists.
When reading contour plots, students sometimes measure distances from the wrong contour or confuse the direction of increasing values. Always check which contours are closer and which are farther, and note whether values are increasing or decreasing as you move.
A frequent error is forgetting the sign of (y – b) when the target y-value is less than b. If you are estimating at y = 1.95 with b = 2, the increment is −0.05, not +0.05.
Students sometimes compute partial derivatives symbolically and then forget to evaluate them at the specific point before plugging into the linear approximation formula.
⚠️ Linear approximation appears on nearly every MATH 215 exam. Expect to see it with a formula, a data table, and a contour plot, sometimes all three on the same exam.
⚠️ The exam tests whether you choose the right estimation method for the data type. Using a forward difference when a centered difference is available will lose marks.
⚠️ Pay close attention to units and signs in the displacement terms (x – a) and (y – b). Sign errors are the most common way to lose points on an otherwise correct setup.
Fill in the blank: the linear approximation formula is f(x, y) ≈ f(a, b) + ______ + ______.
True or false: the centered difference uses f(a+h) and f(a–h) but not f(a) itself.
If f_x(2, 3) = 5 and f_y(2, 3) = −3, and f(2, 3) = 10, estimate f(2.01, 3.02).
True or false: linear approximation is exact for any linear function f(x, y) = ax + by + c.
Answers: 1. f_x(a, b)(x – a) + f_y(a, b)(y – b). 2. True. 3. 10 + 5(0.01) + (−3)(0.02) = 10 + 0.05 – 0.06 = 9.99. 4. True.
Q: Given f(x, y) = x² sin(y), estimate f(1.02, π/2 + 0.03) using linear approximation.
A: f_x = 2x sin(y), f_y = x² cos(y). At (1, π/2): f = 1, f_x = 2, f_y = 0. So f(1.02, π/2 + 0.03) ≈ 1 + 2(0.02) + 0(0.03) = 1.04.
Q: A table gives g(3, 5) = 20, g(4, 5) = 23, g(2, 5) = 18, g(3, 6) = 22, g(3, 4) = 17. Estimate g(3.1, 4.8).
A: g_x(3, 5) ≈ (23 – 18)/2 = 2.5. g_y(3, 5) ≈ (22 – 17)/2 = 2.5. g(3.1, 4.8) ≈ 20 + 2.5(0.1) + 2.5(−0.2) = 20 + 0.25 – 0.5 = 19.75.
Q: On a contour plot, h(2, 1) = 10. The contour h = 12 is 0.5 units to the right of (2, 1), and h = 8 is 0.25 units to the left. Estimate h_x(2, 1).
A: Forward slope: (12 – 10)/0.5 = 4. Backward slope: (10 – 8)/0.25 = 8. Average: (4 + 8)/2 = 6.
Linear approximation is the foundation of the chain rule for multivariable functions and of related rates problems (where you differentiate a function of several time-dependent variables). The tangent plane, which is the geometric object behind linear approximation, reappears in the study of surfaces, gradients, and directional derivatives.
Related Terms / Search Tags
linear approximation, tangent plane, partial derivative, centered difference, center difference, forward difference, backward difference, contour plot, level curves, estimation, MATH 215, Calculus III, University of Michigan, multivariable calculus, differentiable, local linearity