Difficulty: Intermediate | Prerequisites: All Midterm 2 topics (double/triple integrals, gradients, optimisation, coordinate systems)
True/false questions on MATH 215 exams test conceptual understanding rather than computation. They probe whether you can spot the difference between a theorem's actual statement and a plausible-sounding variant, whether properties of individual functions carry over to combinations, and whether you can quickly verify or disprove claims about integrals, optimisation, and coordinate geometry using symmetry, counterexamples, or known results.
(a) FALSE: If f and g each have a saddle point at (0,0), so must f + g.
Counterexample: f = x² - y², g = -x² + 2y². Both have saddle points at the origin (D < 0 for each). But f + g = y², which has D = 0 · 2 - 0² = 0 at the origin with f_xx = 0, f_yy = 2 > 0. In fact y² has a local (degenerate) minimum along the x-axis, not a saddle.
(b) TRUE: The change-of-variables integral identity for D = {(x,y) | -1 ≤ x+y ≤ 1, -1 ≤ x-y ≤ 1}.
Let u = x + y, v = x - y. Then D in (u, v) coordinates is the square [-1, 1] × [-1, 1]. The Jacobian of the transformation from (u, v) to (x, y) is |∂(x,y)/∂(u,v)| = 1/2. But the right-hand side of the given equation has no Jacobian factor of 1/2, so the statement requires checking carefully. The integrand on the right is f(x+y, x-y), and the integral is over [-1,1]×[-1,1] in (x, y), not (u, v). Actually, the substitution u = x+y, v = x-y maps the square [-1,1]×[-1,1] in (x,y) to the diamond D in (u,v), with Jacobian |∂(u,v)/∂(x,y)| = 2. So ∬_D f(u,v) du dv = ∫∫ f(x+y, x-y) · 2 dx dy over the square. This means the statement as written (without the factor of 2) is actually FALSE.
(c) TRUE: The level surface of f = x² + y² + z² through the maximum point P on the ellipse C is perpendicular to C at P.
At a constrained maximum of f on C (where C is the intersection of two surfaces), the gradient ∇f is a linear combination of the normals to the two constraint surfaces (Lagrange multipliers). The level surface f = c passes through P with normal ∇f. Since ∇f lies in the plane spanned by the constraint normals, and C is tangent to the intersection of the two constraint surfaces, ∇f is perpendicular to C at P.
(d) TRUE: ∫₀¹ ∫₀¹ f(x) dy dx = ∫₀¹ f(x) dx.
Since f(x) does not depend on y, the inner integral ∫₀¹ f(x) dy = f(x) · 1 = f(x). So the double integral collapses to ∫₀¹ f(x) dx.
(e) TRUE: There exists a unit vector u such that D_u f(1,2,3) = -∂f/∂x(1,2,3).
Choose u = (-1, 0, 0). Then D_u f = ∇f · (-1,0,0) = -f_x. This is a unit vector in R³, so it always works.
(a) FALSE: The surface area of z = 1/(1 + x²⁰²² + y²⁰²²) over the unit disc is π/2.
The surface area formula is ∬_D √(1 + f_x² + f_y²) dA. Since the integrand √(1 + f_x² + f_y²) ≥ 1, the surface area is at least the area of the disc, which is π. So the surface area cannot be π/2.
(b) FALSE: The max of xe^(-(x²+y²)³) on x² + y² = 1 is 1.
On the circle x² + y² = 1, the function becomes xe^(-1). The maximum of x on the unit circle is 1 (at (1,0)), so the maximum is 1 · e^(-1) = 1/e, not 1.
(c) TRUE: The mass of a liquid filling the unit ball with density ρ = 3 + xy³z⁵ is 4π.
The term xy³z⁵ is odd in x (and also odd in z), and the unit ball is symmetric about x = 0. So ∭ xy³z⁵ dV = 0 by symmetry. The mass is therefore ∭ 3 dV = 3 · (4π/3) = 4π.
(d) Requires computation: Midpoint rule comparison.
With m = n = 1 (one rectangle), the midpoint rule uses the function value at the centre of each rectangle.
For ∫₁³ ∫₋₂⁰ f(x,y) dy dx: the rectangle is [1,3] × [-2,0], midpoint = (2, -1). f(2, -1) = 9. Area = 2 × 2 = 4. Estimate = 9 × 4 = 36.
For ∫₀² ∫₃⁵ f(x,y) dx dy: the rectangle is [0,2] × [3,5], midpoint = (1, 4). We need f(1, 4): from the table, f(1, ...) has x = 1 as a row, but y = 4 is not directly in the table columns. The y-columns are -2, -1, 0, 1, 2. The x-rows are 1, 2, 3, 4, 5. So (x,y) = (1,4) is not in the table. We need the midpoint of [0,2] × [3,5], which is x = 1, y = 4. Reading from the table with x\y layout: row x = 4, column y = 1 gives f(4, 1) = 5. Wait: the midpoint for the second integral ∫₀² ∫₃⁵ f(x,y) dx dy has the rectangle [0,2] in y and [3,5] in x, midpoint x = 4, y = 1. f(4, 1) = 5. Estimate = 5 × 4 = 20.
36 > 20, so the statement is TRUE.
(e) FALSE: ∫₀¹ ∫₀^(x²) f(x,y) dy dx = ∫₀¹ ∫_(y²)¹ f(x,y) dx dy.
The region is 0 ≤ y ≤ x², 0 ≤ x ≤ 1. This is the same as y ≤ x², x ≤ 1, y ≥ 0. Solving for x: x ≥ √y. So the correct reversed integral is ∫₀¹ ∫_(√y)¹ f dx dy. The proposed form has y² in place of √y, which is incorrect.
If f is odd in one variable and the region is symmetric in that variable, the integral is zero. This principle settles Q5(c) instantly: xy³z⁵ is odd in x over the unit ball.
Surface area ≥ area of the base (because √(1 + f_x² + f_y²) ≥ 1). This kills Q5(a) without any computation.
Directional derivatives satisfy |D_u f| ≤ |∇f|, which settles Q2(d) and is the key inequality for all "does such a u exist" questions.
On x² + y² = 1, any expression involving x² + y² simplifies to a constant. In Q5(b), e^(-(x²+y²)³) becomes e^(-1), reducing the problem to finding the max of x on the circle.
Q3(b) tests whether students recognise that a change of variables requires a Jacobian correction. The substitution u = x+y, v = x-y has |∂(u,v)/∂(x,y)| = 2, so dropping the Jacobian makes the identity false.
Q3(e) and Q5's directional derivative questions test whether students remember that e₁ = (1,0,0), e₂ = (0,1,0), e₃ = (0,0,1) and their negatives are all valid unit vectors, and dotting ∇f with -e₁ gives -f_x.
Q5(e) tests whether students can correctly reverse the order of integration. The region y ≤ x² becomes x ≥ √y, not x ≥ y².
Students assume properties are preserved under addition. If f and g each have property P at a point, f + g need not. Test this by constructing a counterexample where f + g cancels the problematic behaviour.
Students forget bounding arguments. Before attempting a computation, check whether the claim is even possible. "Surface area = π/2 over a region of area π" is immediately suspicious because surface area ≥ projected area.
Students confuse inverse functions when reversing integration order. The curve y = x² inverts to x = √y, not x = y². Draw the region before writing the reversed limits.
Students forget to substitute constraints before optimising. On x² + y² = 1, the function xe^(-(x²+y²)³) simplifies to xe^(-1). The max of x on the unit circle is 1, so the answer is e^(-1), not 1.
Students assume the midpoint rule requires complicated computation. With m = n = 1, there is only one rectangle, and you evaluate f at exactly one point: the centre.
Q: True or false: If f(x, y) has a local max at (a, b), then every directional derivative D_u f(a, b) = 0.
A: True. At a local max, ∇f(a,b) = 0, so D_u f = ∇f · u = 0 for every u.
Q: True or false: ∬_D 1 dA ≥ 0 for any region D.
A: True. The integrand is non-negative and we integrate over a region with non-negative area.
Q: True or false: If ∇f(P) = 0 and ∇g(P) = 0, then ∇(fg)(P) = 0.
A: True. By the product rule, ∇(fg) = f∇g + g∇f. At P, both ∇f and ∇g are zero, so ∇(fg)(P) = 0.
Q: True or false: The integral ∫₀¹ ∫₀¹ (x - 1/2) dy dx = 0.
A: True. The function x - 1/2 is odd about x = 1/2 on [0,1], so ∫₀¹ (x - 1/2) dx = 0.
Q: True or false: If f(x,y) ≤ 5 on D and area(D) = 3, then ∬_D f dA ≤ 15.
A: True. By the comparison property of double integrals, ∬_D f dA ≤ ∬_D 5 dA = 5 · area(D) = 15.
Q: True or false: Switching to polar coordinates always simplifies the computation of a double integral.
A: False. Polar coordinates help when the region or integrand has circular symmetry. For a rectangular region with a polynomial integrand, Cartesian is typically simpler.
True/false calculus, conceptual review, saddle point counterexample, surface area lower bound, odd function symmetry, midpoint rule, Fubini's theorem, order of integration reversal, directional derivative bound, level surface perpendicularity, Lagrange multipliers, change of variables Jacobian, MATH 215 Midterm 2, Calculus III, UMich Math 215, multivariable calculus conceptual questions