MATH 126 Calculus III (Multivariable Calculus), University of Washington
Difficulty: Intermediate | Prerequisites: Partial derivatives, tangent planes (see companion Study Notes)
Once you can compute partial derivatives and tangent planes, the next question is: where does a surface reach its highest or lowest points? This set of topics covers how to find and classify critical points of a function of two variables (local maxima, local minima, saddle points), how to find global extrema on a closed bounded region, and how to work with parametric space curves, including tangent vectors and curvature. These are the core optimisation and curve-geometry tools in MATH 126 and tend to appear on every midterm. You should be confident with partial derivatives before tackling this material.
Critical points are where both partial derivatives are zero; you classify them using the second derivative test (the discriminant D). To find global extrema on a closed region, check interior critical points and then parametrise the boundary. For space curves, the tangent vector is the derivative of the position vector, and curvature measures how sharply the curve bends at a point.
Critical point
A point (a, b) where both fx(a, b) = 0 and fy(a, b) = 0 (or where a partial derivative does not exist). These are the candidates for local maxima, local minima, or saddle points. Think of it as: the tangent plane is horizontal here.
Second derivative test (discriminant D)
A classification method for critical points. Compute D = fxx * fyy - (fxy)^2 at the critical point. If D > 0 and fxx > 0, the point is a local minimum. If D > 0 and fxx < 0, it is a local maximum. If D < 0, it is a saddle point. If D = 0, the test is inconclusive.
Saddle point
A critical point that is neither a local max nor a local min. The surface curves upward in one direction and downward in another, like the centre of a horse saddle or a Pringle crisp. In simple terms, it looks like a maximum along one cross-section and a minimum along another.
Global (absolute) extrema
The largest and smallest values of f on an entire region, including its boundary. On a closed, bounded set, the Extreme Value Theorem guarantees these exist. You find them by comparing the values at interior critical points and along the boundary.
Lagrange multipliers
A method for optimising f(x, y) subject to a constraint g(x, y) = c, by solving the system grad f = lambda * grad g together with the constraint. This technique is used to handle the boundary of a region. In simple terms, the optimal point on a constraint curve is where the level curves of f are tangent to the constraint.
Parametric curve (vector-valued function)
A curve in space described by r(t) = (x(t), y(t), z(t)). The variable t is the parameter. Think of it as: the curve is the path traced out as t varies.
Tangent vector
The derivative r'(t) = (x'(t), y'(t), z'(t)) gives the direction of the curve at parameter value t. Two lines are parallel if and only if their direction vectors are scalar multiples of each other.
Curvature (kappa)
A measure of how sharply a curve bends at a given point. The formula uses the cross product of r'(t) and r''(t). A straight line has curvature zero; a tight circle has high curvature. In simple terms, curvature is the reciprocal of the radius of the best-fitting circle at that point.
Compute fx and fy.
Set fx = 0 and fy = 0 simultaneously. Solve the system to find all critical points.
Compute the second partial derivatives: fxx, fyy, fxy.
For each critical point, evaluate D = fxx * fyy - (fxy)^2.
Classify: D > 0 and fxx > 0 means local min. D > 0 and fxx < 0 means local max. D < 0 means saddle point. D = 0 means inconclusive.
Given f(x, y) = x^3 + y^3 - 3xy.
fx = 3x^2 - 3y = 0, so y = x^2
fy = 3y^2 - 3x = 0, so x = y^2
Substitute y = x^2 into x = y^2: x = (x^2)^2 = x^4, so x^4 - x = 0, giving x(x^3 - 1) = 0. This yields x = 0 or x = 1.
x = 0 gives y = 0. Critical point: (0, 0).
x = 1 gives y = 1. Critical point: (1, 1).
Second derivatives: fxx = 6x, fyy = 6y, fxy = -3.
At (0, 0): fxx = 0, fyy = 0, fxy = -3. D = (0)(0) - (-3)^2 = -9 < 0. Saddle point.
At (1, 1): fxx = 6, fyy = 6, fxy = -3. D = (6)(6) - (-3)^2 = 36 - 9 = 27 > 0, and fxx = 6 > 0. Local minimum.
Final answer: (0, 0) is a saddle point, (1, 1) is a local minimum.
When the region S is closed and bounded, the Extreme Value Theorem guarantees that f attains both a global maximum and a global minimum somewhere on S. The strategy has two phases: check the interior, then check the boundary.
Find all interior critical points (fx = 0 and fy = 0) that lie inside S.
Parametrise the boundary of S and reduce f to a function of one variable along each boundary piece.
Find the critical points of each single-variable function and evaluate f at those, as well as at the endpoints where boundary pieces meet.
Compare all values from steps 1 through 3. The largest is the global maximum, the smallest is the global minimum.
Given f(x, y) = x^2 + 2y^2 + 8y on S = {(x, y) | x^2 + y^2 <= 16}.
Interior critical points:
fx = 2x = 0, so x = 0
fy = 4y + 8 = 0, so y = -2
Critical point: (0, -2). Check it is inside: 0 + 4 = 4 <= 16. Yes.
f(0, -2) = 0 + 8 - 16 = -8
Boundary: x^2 + y^2 = 16. Parametrise as x = 4 cos(t), y = 4 sin(t), with t in [0, 2 pi].
Substitute into f:
f = 16 cos^2(t) + 2(16 sin^2(t)) + 8(4 sin(t))
f = 16 cos^2(t) + 32 sin^2(t) + 32 sin(t)
f = 16 + 16 sin^2(t) + 32 sin(t)
Let u = sin(t). Then h(u) = 16 + 16u^2 + 32u for u in [-1, 1].
h'(u) = 32u + 32 = 0 gives u = -1, so sin(t) = -1, meaning t = 3 pi/2, which gives (x, y) = (0, -4).
f(0, -4) = 0 + 32 - 32 = 0
Endpoints of u in [-1, 1]:
u = -1: h(-1) = 16 + 16 - 32 = 0, corresponding to (0, -4).
u = 1: h(1) = 16 + 16 + 32 = 64, corresponding to (0, 4).
f(0, 4) = 0 + 32 + 32 = 64
Comparison:
f(0, -2) = -8 (interior)
f(0, -4) = 0 (boundary)
f(0, 4) = 64 (boundary)
Global minimum: -8 at (0, -2). Global maximum: 64 at (0, 4).
A parametric curve r(t) = (x(t), y(t), z(t)) traces a path in three-dimensional space. The tangent vector at parameter t is r'(t) = (x'(t), y'(t), z'(t)). Two lines are parallel when their direction vectors are scalar multiples of one another.
Given r(t) = (4t, t^2, t^3). Find a point on this curve whose tangent line is parallel to the line x = 4t, y = -2t, z = 3t.
The given line has direction vector (4, -2, 3).
The tangent vector to the curve is r'(t) = (4, 2t, 3t^2).
For parallelism, r'(t) must be a scalar multiple of (4, -2, 3). So there exists some constant k such that:
4 = 4k, giving k = 1
2t = -2k = -2, giving t = -1
3t^2 = 3k = 3, giving t^2 = 1, so t = 1 or t = -1
The value t = -1 satisfies all three equations. At t = -1:
r(-1) = (4(-1), (-1)^2, (-1)^3) = (-4, 1, -1)
Final answer: the point is (-4, 1, -1).
Curvature measures how quickly the direction of a curve changes. For a parametric curve r(t), the curvature at a point is:
\kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}Compute r'(t) and r''(t).
Compute the cross product r'(t) x r''(t).
Find the magnitude of the cross product.
Find |r'(t)|^3.
Divide.
Given r(t) = (4t, t^2, t^3). Find the curvature at B(0, 0, 0).
First, find the parameter value: r(t) = (0, 0, 0) when t = 0.
r'(t) = (4, 2t, 3t^2). At t = 0: r'(0) = (4, 0, 0).
r''(t) = (0, 2, 6t). At t = 0: r''(0) = (0, 2, 0).
Cross product r'(0) x r''(0):
Using the determinant formula with i, j, k:
i component: (0)(0) - (0)(2) = 0
j component: (0)(0) - (4)(0) = 0, but with the negative sign: -(0 - 0) = 0
k component: (4)(2) - (0)(0) = 8
So r'(0) x r''(0) = (0, 0, 8).
|r'(0) x r''(0)| = 8.
|r'(0)| = sqrt(16 + 0 + 0) = 4. So |r'(0)|^3 = 64.
Curvature: kappa = 8 / 64 = 1/8.
Final answer: kappa = 1/8.
\text{Second derivative test: } D = f_{xx} f_{yy} - (f_{xy})^2\text{Curvature: } \kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}\text{Tangent vector: } \mathbf{r}'(t) = \langle x'(t),\, y'(t),\, z'(t) \rangle\text{Cross product: } \mathbf{a} \times \mathbf{b} = \langle a_2 b_3 - a_3 b_2,\; a_3 b_1 - a_1 b_3,\; a_1 b_2 - a_2 b_1 \rangleFor global extrema on a disk x^2 + y^2 <= R^2, parametrise the boundary circle as x = R cos(t), y = R sin(t).
Students often assume D > 0 alone is enough to classify a critical point. You must also check the sign of fxx to distinguish between a local min (fxx > 0) and a local max (fxx < 0).
A common error on global extrema problems is forgetting to check the boundary. The interior critical point might be neither the global max nor the global min.
Students sometimes confuse the tangent vector r'(t) with the unit tangent vector T(t) = r'(t) / |r'(t)|. The curvature formula uses r'(t) and r''(t) directly, not the unit tangent.
When computing cross products, sign errors on the j-component are extremely common. The j-component carries a negative sign in the cofactor expansion: -(a1b3 - a3b1).
The second derivative test problem is a near-certainty on MATH 126 midterms. Make sure you can solve a system of two equations in two unknowns cleanly, and that you evaluate D at each critical point separately.
Global extrema problems always require showing work on both the interior and the boundary. Stating only the boundary analysis, or only the interior, will cost marks.
Curvature problems test your cross product mechanics. Practise the determinant expansion until it is automatic, paying special attention to the sign of the middle component.
The exam requires full simplification. Leave no unsimplified square roots, fractions, or unexpanded expressions.
True or false: If D = 0 at a critical point, the point must be a saddle point.
Answer: False. D = 0 means the test is inconclusive; you cannot classify the point with this test alone.
Fill in the blank: To find global extrema on a closed region, you must check both ______ critical points and the ______.
Answer: interior; boundary.
True or false: The curvature of a straight line is always zero.
Answer: True.
Fill in the blank: Two lines are parallel if their direction vectors are ______ of each other.
Answer: scalar multiples.
True or false: A local minimum found by the second derivative test is automatically a global minimum.
Answer: False. A local minimum may not be the global minimum, especially on a bounded region where the boundary must also be checked.
Q: Find and classify the critical points of f(x, y) = x^2 + xy + y^2 + 3x - 3y + 4.
A: fx = 2x + y + 3 = 0 and fy = x + 2y - 3 = 0. From the first equation, y = -2x - 3. Substituting: x + 2(-2x - 3) - 3 = 0 gives x - 4x - 6 - 3 = 0, so -3x = 9, x = -3, y = 3. Critical point: (-3, 3). Second derivatives: fxx = 2, fyy = 2, fxy = 1. D = 4 - 1 = 3 > 0 and fxx > 0, so (-3, 3) is a local minimum.
Q: Find the global maximum and minimum of f(x, y) = x + 2y on the disk x^2 + y^2 <= 1.
A: Interior: fx = 1 and fy = 2 are never zero, so there are no interior critical points. Boundary: x = cos(t), y = sin(t). Then f = cos(t) + 2 sin(t). f'(t) = -sin(t) + 2 cos(t) = 0 gives tan(t) = 2. At this t, f = sqrt(1 + 4) = sqrt(5) (using the identity). The global max is sqrt(5) and the global min is -sqrt(5).
Q: Find the curvature of r(t) = (t, t^2, 0) at t = 0.
A: r'(t) = (1, 2t, 0), r''(t) = (0, 2, 0). At t = 0: r'(0) = (1, 0, 0), r''(0) = (0, 2, 0). Cross product: (0, 0, 2). |cross product| = 2. |r'(0)| = 1, so |r'(0)|^3 = 1. Curvature kappa = 2/1 = 2.
Q: A curve has r(t) = (cos(t), sin(t), t). Is its tangent at t = 0 parallel to the vector (0, 1, 1)?
A: r'(t) = (-sin(t), cos(t), 1). At t = 0: r'(0) = (0, 1, 1). This is exactly (0, 1, 1), so yes, the tangent line at t = 0 is parallel (in fact, it has the same direction vector).
The second derivative test connects back to tangent planes: a critical point is where the tangent plane is horizontal, and the second derivatives tell you whether the surface curves up (bowl) or down (dome) from there. Global extrema on boundaries connect to Lagrange multipliers, which appear later in the course for constrained optimisation on more complex regions. Curvature feeds into the study of the TNB frame (tangent, normal, binormal vectors) and applications in physics, such as centripetal acceleration of a particle moving along a curve.
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