Difficulty: Intermediate | Prerequisites: Tangent planes study notes (companion doc), partial derivatives, MATH 151 linear approximation concept.
Linear approximation is one of the most practical tools in calculus: it lets you estimate function values that would otherwise require a calculator, by using the tangent line (in 1D) or the tangent plane (in 2D) as a stand-in. This lecture extends the idea you first met in MATH 151 into three and even four dimensions. If you are comfortable finding tangent planes (see the companion study notes), this is a short step further. The payoff is that you can approximate messy expressions like ln(0.72) or 4(-1.9)^2 - 4.1e^(0.3) by hand, to reasonable accuracy.
Linear approximation replaces a complicated function with its tangent plane near a chosen point, giving you a quick estimate without a calculator. You pick a nearby point where the function is easy to evaluate, build the tangent plane there, and plug in your target values. The technique extends naturally from 2D to any number of dimensions by adding one partial-derivative term per extra variable.
Linear approximation (linearisation)
An estimate of a function's value near a known point, obtained by replacing the function with its tangent line (1D) or tangent plane (2D+). Think of it as zooming in so close that the curve looks flat, then reading the value off the flat surface.
Centre point (a, b) or (a, b, c)
The point around which you build the approximation. You choose it because the function is easy to evaluate there and it is close to the point you want to estimate. In simple terms, it is your "base camp" for the approximation.
Tangent plane as approximation (T(x, y))
The tangent plane equation, re-read as an approximating function: f(x, y) is approximately equal to T(x, y) for (x, y) near (a, b). This is the same equation from the tangent planes topic, used for a different purpose.
Differentiable
A function is differentiable at a point if it is smooth enough there for the linear approximation to work. Roughly, the surface has no corners, tears, or vertical jumps at that point. For the functions in this course, differentiability is usually given or obvious.
Error of approximation
The difference between the true function value and the linear approximation. The error is small when (x, y) is close to (a, b) and grows as you move further away.
For a single-variable function f(x), the linear approximation centred at x = a is:
f(x) is approximately L(x) = f(a) + f'(a)(x - a).
This is just the tangent line at x = a, used to estimate f(x) for x near a. The key insight is that locally, a smooth curve behaves like a straight line.
For f(x, y), the linear approximation centred at (a, b) is:
f(x, y) is approximately T(x, y) = f_x(a, b)(x - a) + f_y(a, b)(y - b) + f(a, b).
This is identically the tangent plane equation. The difference is the intent: here you are using the plane to estimate function values, not just to describe the geometry of the surface.
Pick a point (a, b) that satisfies two conditions:
It is close to the point you want to estimate.
f(a, b) is easy to compute (preferably by hand, no calculator).
Good candidates are round numbers, values that zero out exponents, or arguments that make logarithms or trig functions land on known values.
Estimate f(6.9, 2.06) where f(x, y) = ln(x - 3y).
Choose the centre (7, 2). It is close to (6.9, 2.06), and f(7, 2) = ln(7 - 6) = ln(1) = 0.
f_x = 1/(x - 3y). At (7, 2): f_x = 1/(7 - 6) = 1.
f_y = -3/(x - 3y). At (7, 2): f_y = -3/(7 - 6) = -3.
T(x, y) = 1(x - 7) - 3(y - 2) + 0 = (x - 7) - 3(y - 2).
Estimate: T(6.9, 2.06) = (6.9 - 7) - 3(2.06 - 2) = -0.1 - 0.18 = -0.28.
Calculator check: ln(6.9 - 3(2.06)) = ln(0.72) is approximately -0.33. The approximation is in the right ballpark.
Estimate f(-1.9, 4.1) where f(x, y) = 4x^2 - ye^(2x + y).
Choose the centre (-2, 4). Then 2x + y = 2(-2) + 4 = 0, so e^0 = 1, making things clean.
f(-2, 4) = 4(4) - 4e^0 = 16 - 4 = 12.
f_x = 8x - 2ye^(2x + y). At (-2, 4): f_x = -16 - 8(1) = -24.
f_y = -e^(2x + y) - ye^(2x + y). At (-2, 4): f_y = -1 - 4 = -5.
T(x, y) = -24(x + 2) - 5(y - 4) + 12.
Estimate: T(-1.9, 4.1) = -24(0.1) - 5(0.1) + 12 = -2.4 - 0.5 + 12 = 9.1.
Calculator check: f(-1.9, 4.1) is approximately 8.91. Close.
For f(x, y, z), the linear approximation centred at (x_0, y_0, z_0) is:
q(x, y, z) = f_x(x_0, y_0, z_0)(x - x_0) + f_y(x_0, y_0, z_0)(y - y_0) + f_z(x_0, y_0, z_0)(z - z_0) + f(x_0, y_0, z_0).
The pattern is the same: one term per variable, each weighted by the corresponding partial derivative evaluated at the centre.
f(x, y, z) = 3xy^2 - 2xyz + z^3 - 4xz, centred at (2, 1, 3).
f(2, 1, 3) = 6 - 12 + 27 - 24 = -3.
f_x = 3y^2 - 2yz - 4z. At (2, 1, 3): f_x = 3 - 6 - 12 = -15.
f_y = 6xy - 2xz. At (2, 1, 3): f_y = 12 - 12 = 0.
f_z = -2xy + 3z^2 - 4x. At (2, 1, 3): f_z = -4 + 27 - 8 = 15.
q(x, y, z) = -15(x - 2) + 0(y - 1) + 15(z - 3) - 3.
1D linear approximation (MATH 151 recap)
f(x) \approx L(x) = f(a) + f'(a)(x - a)2D linear approximation (tangent plane)
f(x,y) \approx T(x,y) = f_x(a,b)(x - a) + f_y(a,b)(y - b) + f(a,b)3D linear approximation (extending to 4D)
f(x,y,z) \approx q(x,y,z) = f_x(x_0,y_0,z_0)(x - x_0) + f_y(x_0,y_0,z_0)(y - y_0) + f_z(x_0,y_0,z_0)(z - z_0) + f(x_0,y_0,z_0)The general pattern for n variables: one partial-derivative term per variable, each multiplied by the displacement from the centre, plus the function value at the centre.
Linear approximation is the mathematical backbone of any situation where you need a quick, rough answer from a complicated model. Engineers use it constantly: if a structural simulation gives stress as a function of load and temperature, the linear approximation tells you how stress changes for a small shift in either input without re-running the whole model.
Students sometimes choose a centre point that is close to the target but hard to evaluate. The whole point of the method is that f(a, b) should be trivial to compute. If it requires a calculator, you have chosen the wrong centre.
Forgetting to add f(a, b) at the end. The linear approximation is not just the partial-derivative terms; you must include the function value at the centre. Without it, the approximation is centred at the origin instead of at your chosen point.
Assuming the approximation is exact. It is an estimate, and the error grows as you move away from the centre. The lecture notes compare the approximation to the calculator value to make this point.
In the 4D extension, students sometimes leave out one of the partial-derivative terms. The rule is one term per input variable, no exceptions.
Linear approximation is a near-certain exam topic. You will likely be given a function and a target point and asked to estimate the value without a calculator.
The key exam skill is choosing a good centre point. Practise spotting round numbers that zero out exponentials or land logarithms on ln(1) = 0.
Some problems ask you to both approximate and compute the actual value, then compare. Be ready to show both and comment on the accuracy.
The 4D extension is a favourite "stretch" question. If it appears, it is simply the 3D formula with one extra term.
True or false: The linear approximation of f(x, y) at (a, b) is the same equation as the tangent plane at (a, b). (True)
Fill in the blank: A good centre point for estimating f(6.9, 2.06) where f(x, y) = ln(x - 3y) is (__, __). (7, 2)
True or false: The linear approximation always gives the exact value of the function. (False, it gives an estimate that improves as you get closer to the centre)
Fill in the blank: To extend linear approximation from 2 input variables to 3, you add one more ____ term. (partial derivative)
True or false: If the centre point requires a calculator to evaluate, you should pick a different one. (True)
Q: Use linear approximation to estimate ln(1.05) without a calculator. (Hint: this is the 1D version, centred at a = 1.)
A: f(x) = ln(x), f(1) = 0, f'(x) = 1/x, f'(1) = 1. So L(x) = 0 + 1(x - 1) = x - 1. L(1.05) = 0.05. (Calculator gives approximately 0.0488.)
Q: For f(x, y) = ln(x - 3y), write the linear approximation centred at (7, 2) and use it to estimate f(7.1, 1.95).
A: T(x, y) = (x - 7) - 3(y - 2). T(7.1, 1.95) = 0.1 - 3(-0.05) = 0.1 + 0.15 = 0.25.
Q: For f(x, y) = 4x^2 - ye^(2x+y), why is (-2, 4) a better centre point than (-1.9, 4.1) for estimating f(-1.9, 4.1)?
A: At (-2, 4), the exponent 2x + y = 0, so e^0 = 1, making f(-2, 4) trivial to compute by hand. At (-1.9, 4.1), the exponent is 0.3, and you would need a calculator for e^(0.3).
Q: Write the general linear approximation formula for a function of three variables f(x, y, z) centred at (x_0, y_0, z_0).
A: q(x,y,z) = f_x(x_0,y_0,z_0)(x - x_0) + f_y(x_0,y_0,z_0)(y - y_0) + f_z(x_0,y_0,z_0)(z - z_0) + f(x_0,y_0,z_0).
Q: True or false: The error of a linear approximation tends to be smaller when the target point is closer to the centre point.
A: True. The tangent plane hugs the surface tightly near the centre and drifts away further out.
Linear approximation is the direct application of tangent planes (companion study notes). If you can find a tangent plane, you can do a linear approximation, and vice versa.
This also connects to differentials (dz = f_x dx + f_y dy), which are the formal language for "how much does the output change for a small change in the inputs." Differentials and linear approximation are two ways of saying the same thing.
Later in the course, the idea of replacing a function with a simpler one near a point appears again in Taylor series (second-order and higher approximations), where you add quadratic and cubic terms for better accuracy.
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