MATH 126 Calculus III (Multivariable Calculus), University of Washington
Difficulty: Intermediate | Prerequisites: MATH 124/125 (single-variable calculus, partial derivatives basics)
This set of topics sits at the heart of multivariable calculus: extending the familiar ideas of domains, tangent lines, and local linear approximation from single-variable calculus into two (or more) dimensions. In single-variable calculus you found where a function was defined, drew a tangent line at a point, and used that line to estimate nearby values. Here you do the same, but the tangent "line" becomes a tangent plane, and the linear approximation uses partial derivatives in two directions instead of one ordinary derivative. You should already be comfortable computing partial derivatives before working through this material.
The domain of a multivariable function is the set of (x, y) inputs where the expression is defined, which for square roots means keeping the inside non-negative. The tangent plane is the flat surface that best approximates a surface z = f(x, y) at a given point, built from the two partial derivatives. Linear approximation uses that tangent plane to estimate f at a nearby point without computing the exact value.
Domain of a multivariable function
The set of all input pairs (x, y) for which f(x, y) produces a real number. For square roots, the expression under the radical must be greater than or equal to zero. For fractions, the denominator cannot be zero. In simple terms, it is the collection of (x, y) points where the function "works."
Partial derivative (fx and fy)
The rate of change of f with respect to one variable while holding the other constant. fx means differentiate with respect to x, treating y as a constant, and fy means differentiate with respect to y, treating x as a constant. Think of it as: how does the output change if you nudge just one input?
Tangent plane
The flat plane that best approximates the surface z = f(x, y) at a particular point (a, b, f(a, b)). It is the two-variable analogue of the tangent line in single-variable calculus. The equation is z = f(a, b) + fx(a, b)(x - a) + fy(a, b)(y - b).
Linear approximation (linearisation)
Using the tangent plane at a known point (a, b) to estimate f at a nearby point (x, y). The formula is L(x, y) = f(a, b) + fx(a, b)(x - a) + fy(a, b)(y - b). In simple terms, you use a flat surface to guess the value of a curved one, and it works well when you are close to the base point.
Differentiable (at a point)
A function f(x, y) is differentiable at (a, b) if its partial derivatives exist and the linear approximation error shrinks faster than the distance to (a, b). If both partial derivatives are continuous near (a, b), then f is differentiable there. Think of it as: the tangent plane is a good fit, not just something you can write down.
The domain of f(x, y) is determined by whatever restrictions the formula imposes. The most common restrictions in MATH 126 are:
Square roots require the radicand to be >= 0.
Denominators cannot equal zero.
Logarithms require the argument to be > 0.
Given f(x, y) = 1 + sqrt(x^2 - 2xy), the domain requires:
x^2 - 2xy >= 0
Factor: x(x - 2y) >= 0
This product is non-negative when both factors share the same sign (or either is zero).
Case 1: x >= 0 and x - 2y >= 0, which gives y <= x/2 with x >= 0.
Case 2: x <= 0 and x - 2y <= 0, which gives y >= x/2 with x <= 0.
The boundary is the pair of lines y = x/2 and x = 0 (the y-axis). The domain is the region at or below y = x/2 when x >= 0, and at or above y = x/2 when x <= 0. Graphically, this is the region on and outside the line y = x/2 measured from the origin, forming a wedge that includes both the positive-x side (below the line) and the negative-x side (above the line).
Find the boundary curve(s) by setting the inequality to equality.
Test a sample point in each region to determine which side satisfies the inequality.
Shade the valid region. Use a solid boundary line if the inequality is >= (closed), or dashed if > (open).
The tangent plane to z = f(x, y) at the point (a, b) is:
z = f(a,b) + f_x(a,b)(x - a) + f_y(a,b)(y - b)This is the direct generalisation of the tangent line formula z = f(a) + f'(a)(x - a) from single-variable calculus, extended to two directions.
Compute f(a, b) to get the z-value of the point of tangency.
Compute the partial derivative fx, then evaluate fx(a, b).
Compute the partial derivative fy, then evaluate fy(a, b).
Substitute everything into the tangent plane equation.
Simplify fully.
Given f(x, y) = 1 + sqrt(x^2 - 2xy), find the tangent plane at (1, 0).
f(1, 0) = 1 + sqrt(1 - 0) = 2
For the partial derivatives, write g(x, y) = x^2 - 2xy so that f = 1 + sqrt(g).
fx = (2x - 2y) / (2 sqrt(x^2 - 2xy))
At (1, 0): fx = 2 / (2 * 1) = 1
fy = (-2x) / (2 sqrt(x^2 - 2xy))
At (1, 0): fy = -2 / (2 * 1) = -1
Tangent plane: z = 2 + 1(x - 1) + (-1)(y - 0) = x - y + 1
Final answer: z = x - y + 1
Linear approximation uses the tangent plane at a "nice" base point (a, b) to estimate f at a nearby point. The formula is:
f(x,y) \approx L(x,y) = f(a,b) + f_x(a,b)(x - a) + f_y(a,b)(y - b)The key is choosing a base point where f, fx, and fy are easy to compute exactly, then letting the tangent plane do the work for the nearby, messier point.
Identify the base point (a, b) where everything is easy to compute.
Compute f(a, b), fx(a, b), and fy(a, b) exactly.
Set dx = x - a and dy = y - b (the small displacements).
Plug into L(x, y) = f(a, b) + fx(a, b) * dx + fy(a, b) * dy.
Given f(x, y) = sqrt(x^2 + 3xy), approximate f(11, 12) using the linear approximation at (10, 10).
f(10, 10) = sqrt(100 + 300) = sqrt(400) = 20
fx = (2x + 3y) / (2 sqrt(x^2 + 3xy))
fx(10, 10) = (20 + 30) / (2 * 20) = 50 / 40 = 5/4
fy = 3x / (2 sqrt(x^2 + 3xy))
fy(10, 10) = 30 / 40 = 3/4
Displacements: dx = 11 - 10 = 1, dy = 12 - 10 = 2.
L(11, 12) = 20 + (5/4)(1) + (3/4)(2) = 20 + 5/4 + 6/4 = 20 + 11/4 = 91/4 = 22.75
The linear approximation gives f(11, 12) approximately equal to 91/4 (or 22.75).
\text{Tangent plane: } z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)\text{Linear approximation: } L(x,y) = f(a,b) + f_x(a,b)\,\Delta x + f_y(a,b)\,\Delta y\text{Chain rule for } f(x,y) = h(g(x,y))\text{: } f_x = h'(g) \cdot g_x,\quad f_y = h'(g) \cdot g_yFor f = sqrt(g(x,y)), the partial derivatives follow the pattern fx = gx / (2 sqrt(g)) and fy = gy / (2 sqrt(g)).
Linear approximation is used in engineering whenever you need a quick estimate of a complicated function near a known operating point, for example estimating how a small change in temperature and pressure affects gas volume. Tangent planes underpin the idea of a "best local flat approximation," which is foundational in computer graphics for rendering smooth surfaces from polygonal meshes.
Students often forget to check that the base point (a, b) is in the domain of f before computing partial derivatives. If (a, b) is on the boundary or outside the domain, the tangent plane does not exist there.
A common error is writing the tangent plane as z = fx(x - a) + fy(y - b), forgetting to add f(a, b). The tangent plane passes through the point (a, b, f(a, b)), not through the origin.
Students sometimes compute fx and fy correctly as general expressions but then forget to evaluate them at the specific point (a, b) before plugging in.
When finding domains involving inequalities like x(x - 2y) >= 0, students often only consider one case (both positive) and miss the other (both negative).
Domain problems are a reliable opener on MATH 126 midterms. They test whether you can translate algebraic constraints into a region in the xy-plane and sketch it. Always show the factoring and case analysis.
Tangent plane questions require you to show all partial derivative work. A correct final answer with no intermediate steps earns no credit on this exam.
Linear approximation problems will specify that you must use the approximation method. Computing the exact value with a calculator earns zero marks.
Results must be fully simplified: the exam instructions state that (x - 1)^2 + y = x^2 must be reduced to -2x + y + 1 = 0.
True or false: The domain of f(x, y) = sqrt(4 - x^2 - y^2) is the disk x^2 + y^2 <= 4, including the boundary circle.
Answer: True.
Fill in the blank: The tangent plane equation requires three ingredients: f(a, b), ______, and fy(a, b).
Answer: fx(a, b).
True or false: If fx(a, b) = 0 and fy(a, b) = 0, the tangent plane at (a, b) is horizontal (parallel to the xy-plane).
Answer: True.
Fill in the blank: In linear approximation, the closer (x, y) is to ______, the more accurate the estimate.
Answer: the base point (a, b).
Q: Find the domain of g(x, y) = ln(x + y) + sqrt(x - y).
A: Two restrictions apply. The logarithm requires x + y > 0, and the square root requires x - y >= 0. The domain is the set of all (x, y) such that x + y > 0 and y <= x. Graphically, this is the region above the line x + y = 0 (but not including it) and at or below the line y = x.
Q: Find the equation of the tangent plane to f(x, y) = x^2 + y^2 at the point (1, 2).
A: f(1, 2) = 1 + 4 = 5. Then fx = 2x, so fx(1, 2) = 2. And fy = 2y, so fy(1, 2) = 4. The tangent plane is z = 5 + 2(x - 1) + 4(y - 2) = 2x + 4y - 5.
Q: Use the linear approximation of f(x, y) = sqrt(x + 3y) at (1, 5) to estimate f(1.1, 4.9).
A: f(1, 5) = sqrt(1 + 15) = 4. Then fx = 1/(2 sqrt(x + 3y)), so fx(1, 5) = 1/8. And fy = 3/(2 sqrt(x + 3y)), so fy(1, 5) = 3/8. Displacements: dx = 0.1, dy = -0.1. L = 4 + (1/8)(0.1) + (3/8)(-0.1) = 4 + 0.0125 - 0.0375 = 3.975.
Q: True or false: the tangent plane to f(x, y) = x^2 + y^2 at the origin is z = 0.
A: True. f(0, 0) = 0, fx(0, 0) = 0, fy(0, 0) = 0, so z = 0 + 0 + 0 = 0. The tangent plane is the xy-plane itself.
Tangent planes connect directly to the gradient vector: the gradient at a point gives the direction of steepest ascent on the surface, and its components are exactly the partial derivatives that build the tangent plane. This material also feeds into the second derivative test and optimisation (covered in the companion study notes), because critical points are where the tangent plane is horizontal. Linear approximation generalises to higher-order Taylor expansions in two variables.
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