Let f(x, y) = xy/(x² + y²) for (x, y) ≠ (0, 0) and f(0, 0) = 0. At the origin,

both partial derivatives exist but f is not continuous

f is continuous but neither partial derivative exists

f is continuous and both partial derivatives exist

Let g(x, y) = x²y/(x⁴ + y²) for (x, y) ≠ (0, 0) and g(0, 0) = 0. Which statements are true at the origin?

the directional derivative D_{u}g exists for every unit vector u

D_{u}g = ∇g·u for every unit vector u

For f(x, y) = x³ + y³ − 3xy, the critical points are

(0, 0), a saddle point, and (1, 1), a local minimum

(0, 0), a local maximum, and (1, 1), a local minimum

(0, 0), a saddle point, and (1, 1), a local maximum

Let F = (−y/(x² + y²), x/(x² + y²)) on ℝ² ∖ {(0, 0)}. Which statement is correct?

curl F = 0 but F is not conservative on ℝ² ∖ {(0, 0)}

curl F = 0, so F is conservative on ℝ² ∖ {(0, 0)}

curl F ≠ 0, which is why the circulation is non-zero

the circulation of F round the unit circle is 0

Calculus of Several Variables and Vector Calculus

Section 1 of the GATE Mathematics (MA) paper is calculus of several variables and then vector calculus, and it is examined differently from the engineering papers’ calculus: MA asks what the definitions imply and what they do not. A function of two variables can have both partial derivatives at a point and still be discontinuous there; it can have a directional derivative in every direction and still fail to be differentiable. So this chapter puts the chain of implications first — continuous partials ⇒ differentiable ⇒ continuous, with every converse broken by a named counterexample — and only then turns to the computations the paper also sets: Taylor’s theorem and the second-derivative test with its saddle points, Lagrange’s multipliers, double and triple integrals with the Jacobian of a change of variables, area, volume and surface area, and finally gradient, divergence and curl, line and surface integrals and the theorems of Green, Stokes and Gauss, each with the hypotheses that make it true. The field −y/(x² + y²), x/(x² + y²) is here on purpose: its curl is zero and its circulation round the unit circle is 2π, and that one example answers a whole family of “which of these is true” questions.

1. Continuity, partial, directional and total derivatives

Let f: ℝⁿ → ℝ and a ∈ ℝⁿ. The partial derivative ∂f/∂xᵢ(a) differentiates along one coordinate axis; the directional derivative D_uf(a) = lim_t→0 [f(a + tu) − f(a)]/t does the same along a unit vector u. f is differentiable at a (has a total derivative) if there is a linear map L with f(a + h) = f(a) + L(h) + o(‖h‖). When it exists, L is given by the gradient, L(h) = ∇f(a)·h, and then D_uf(a) = ∇f(a)·u for every u, so the greatest rate of increase is ‖∇f(a)‖, in the direction of ∇f(a).

The implications, and the counterexample that breaks each converse
Implication	Converse fails for
partials exist near a and are continuous at a ⇒ f differentiable at a	f = (x² + y²) sin(1/√(x² + y²)), f(0) = 0: differentiable at 0, partials not continuous there
differentiable ⇒ continuous, and every D_uf exists and equals ∇f·u	g = x²y/(x⁴ + y²), g(0) = 0: every directional derivative exists at 0, yet g → 1/2 along y = x², so g is not even continuous
differentiable ⇒ both partials exist	f = xy/(x² + y²), f(0) = 0: fₓ(0) = f_y(0) = 0, but f = m/(1 + m²) on y = mx, so f is discontinuous at 0

⚠️ The gradient formula needs differentiability

For g = x²y/(x⁴ + y²) both partials at the origin are 0, because g vanishes on both axes, so ∇g(0) = 0. But D_ug(0) = u₁²/u₂ for u₂ ≠ 0, which is not ∇g(0)·u = 0. The formula D_uf = ∇f·u is a consequence of differentiability, not of the partials existing; an option that computes a directional derivative from the gradient of a non-differentiable function is the trap.

Mixed partials commute — f_xy = f_yx — when both are continuous (Schwarz/Clairaut). Without continuity they can differ: f = xy(x² − y²)/(x² + y²), f(0) = 0, has f_xy(0) = −1 and f_yx(0) = 1.

2. Taylor’s theorem, maxima and minima, saddle points and Lagrange’s multipliers

For f of class C² near a, Taylor’s theorem gives f(a + h) = f(a) + ∇f(a)·h + ½ hᵀH(a + θh)h for some θ ∈ (0, 1), where H is the Hessian matrix of second partials. At a critical point (∇f(a) = 0) the sign of the quadratic form hᵀH(a)h decides: positive definite gives a strict local minimum, negative definite a strict local maximum, indefinite a saddle point, and semidefinite leaves the test inconclusive.

The two-variable test, with D = f_xxf_yy − f_xy² at a critical point
Condition	Conclusion	Example at the origin
D > 0, f_xx > 0	local minimum	x² + y²
D > 0, f_xx < 0	local maximum	−x² − y²
D < 0	saddle point	x² − y², xy
D = 0	no conclusion	x⁴ + y⁴ (minimum), x⁴ − y⁴ (saddle), x³ (neither)

Worked example. f = x³ + y³ − 3xy has ∇f = (3x² − 3y, 3y² − 3x) = 0 at (0, 0) and (1, 1). With f_xx = 6x, f_yy = 6y, f_xy = −3: at (0, 0), D = −9 < 0, a saddle; at (1, 1), D = 36 − 9 = 27 > 0 and f_xx = 6 > 0, a local minimum with value −1. It is not a global minimum, since f → −∞ along the negative x-axis.

🎯 Why Lagrange’s condition is ∇f = λ∇g

To extremise f on the level set g = c, where ∇g ≠ 0, move only along the surface: any tangent direction v satisfies ∇g·v = 0, and at an extremum D_vf = ∇f·v must vanish for all such v. So ∇f is orthogonal to the whole tangent space, which forces ∇f = λ∇g. The hypothesis ∇g ≠ 0 is essential: minimise f = x on g = y² − x³ = 0; the minimum is at (0, 0), where ∇g = 0 and no λ exists, since ∇f = (1, 0). With two constraints the condition is ∇f = λ∇g + μ∇h.

On a sphere the multiplier method has a one-line shortcut: the extreme values of a linear function c·x on ‖x‖ = r are ±r‖c‖, attained at x = ±rc/‖c‖. So x + 2y + 2z on x² + y² + z² = 9 ranges over [−9, 9].

3. Double and triple integrals, Jacobians and change of variables

For f continuous on a region that is simultaneously of type I and type II, Fubini’s theorem lets the order of integration be exchanged, and this is how an impossible inner integral becomes an easy one: ∫₀¹∫ₓ¹ e^y² dy dx has no elementary inner integral, but the region is 0 ≤ x ≤ y ≤ 1, so it equals ∫₀¹∫₀ʸ e^y² dx dy = ∫₀¹ y e^y² dy = (e − 1)/2.

Change of variables. If T: (u, v) ↦ (x, y) is a C¹ bijection with non-vanishing Jacobian J = ∂(x, y)/∂(u, v), then ∬_T(D) f dx dy = ∬_D f(T(u, v)) |J| du dv. The absolute value matters; the Jacobian of the inverse map is 1/J.

The Jacobians used most
Coordinates	Map	\|J\|
polar	x = r cos θ, y = r sin θ	r
cylindrical	x = r cos θ, y = r sin θ, z = z	r
spherical	x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ	ρ² sin φ
linear	(x, y) = A(u, v)	\|det A\|

Area of D is ∬_D 1 dA; volume under z = f ≥ 0 over D is ∬_D f dA, and the volume of a solid V is ∭_V 1 dV (a ball of radius a: ∫₀^2π∫₀^π∫₀^a ρ² sin φ dρ dφ dθ = 4πa³/3).
Surface area of the graph z = f(x, y) over D is ∬_D √(1 + fₓ² + f_y²) dA; for a parametrised surface r(u, v) it is ∬ ‖r_u × r_v‖ du dv. The paraboloid z = x² + y² below z = 2 has area ∫₀^2π∫₀^√2 √(1 + 4r²) r dr dθ = 2π(27 − 1)/12 = 13π/3.

4. Gradient, divergence, curl, line and surface integrals

For a scalar φ and a vector field F = (P, Q, R): ∇φ = (φₓ, φ_y, φ_z); div F = Pₓ + Q_y + R_z; curl F = (R_y − Q_z, P_z − Rₓ, Qₓ − P_y). For C² fields, curl ∇φ = 0 and div curl F = 0, and div ∇φ = ∇²φ. A line integral ∫_C F·dr = ∫ₐᵇ F(r(t))·r′(t) dt depends on the orientation of C but not on its parametrisation; a surface integral ∬_S F·n dS is the flux of F through S.

On an open connected set, F is conservative (F = ∇φ, equivalently ∮F·dr = 0 round every closed curve, equivalently path independence) exactly when line integrals depend only on end points; then ∫_C ∇φ·dr = φ(end) − φ(start). On a simply connected open set, curl F = 0 is also sufficient.

⚠️ Zero curl is not enough on a punctured plane

F = (−y/(x² + y²), x/(x² + y²)) on ℝ² ∖ {0} has Qₓ − P_y = 0 everywhere, yet on the unit circle r = (cos t, sin t), F·r′ = sin²t + cos²t = 1 and ∮F·dr = 2π ≠ 0. So F is not conservative on the punctured plane; it is the gradient of the polar angle θ only on a slit plane, which is simply connected. The hole is what breaks “curl zero ⇒ gradient”.

5. The theorems of Green, Stokes and Gauss

Statements with their hypotheses
Theorem	Statement	Hypotheses
Green	∮_C (P dx + Q dy) = ∬_D (Qₓ − P_y) dA	C a positively oriented, piecewise smooth, simple closed curve bounding D; P, Q of class C¹ on an open set containing D
Stokes	∮_∂S F·dr = ∬_S (curl F)·n dS	S an oriented, piecewise smooth surface with boundary ∂S oriented by the right-hand rule; F of class C¹ near S
Gauss (divergence)	∯_∂V F·n dS = ∭_V div F dV	V a bounded solid whose boundary is a closed piecewise smooth surface with outward normal n; F of class C¹ on V

Area by Green. Taking (P, Q) = (−y/2, x/2) gives area(D) = ½∮_C(x dy − y dx); for the ellipse x = a cos t, y = b sin t this is ½∫₀^2π ab dt = πab.
Gauss in one line. F = (x, y, z) has div F = 3, so its outward flux through the sphere of radius a is 3 · (4/3)πa³ = 4πa³, which the direct computation F·n = a on the sphere, times area 4πa², confirms.
Stokes lets you pick the surface. Every surface with the same oriented boundary gives the same flux of curl F, so replace a hemisphere by the flat disc it caps.

⚠️ Green on a region with a hole

Green’s theorem needs P and Q to be C¹ on the whole of D. For the field of Section 4 and the unit disc, the right side would be ∬0 dA = 0 while the left side is 2π; there is no contradiction, because the field is undefined at the origin. For an annulus the boundary has two pieces, the outer counter-clockwise and the inner clockwise, and then the theorem holds.

Key takeaways

Continuous partials ⇒ differentiable ⇒ continuous with D_uf = ∇f·u; each converse fails, and xy/(x² + y²) and x²y/(x⁴ + y²) are the counterexamples to keep ready.
At a critical point D = f_xxf_yy − f_xy² decides: D > 0 is an extremum by the sign of f_xx, D < 0 a saddle, D = 0 nothing.
Lagrange’s condition ∇f = λ∇g holds at a constrained extremum only where ∇g ≠ 0.
Change variables with |J|: r for polar and cylindrical, ρ² sin φ for spherical; reverse the order of integration when the inner integral has no primitive.
Green, Stokes and Gauss need C¹ fields on the whole region; curl F = 0 gives a potential only on a simply connected domain.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Let f(x, y) = xy/(x² + y²) for (x, y) ≠ (0, 0) and f(0, 0) = 0. At the origin,
1. both partial derivatives exist but f is not continuous
2. f is continuous but neither partial derivative exists
3. f is differentiable
4. f is continuous and both partial derivatives exist
Show answer
Answer: A — both partial derivatives exist but f is not continuous
f vanishes on both axes, so fₓ(0, 0) = lim [f(h, 0) − 0]/h = 0 and likewise f_y(0, 0) = 0. On the line y = mx, f = m/(1 + m²), which depends on m (1/2 on y = x, 0 on y = 0), so the limit at the origin does not exist and f is discontinuous; a discontinuous function cannot be differentiable.
Let g(x, y) = x²y/(x⁴ + y²) for (x, y) ≠ (0, 0) and g(0, 0) = 0. Which statements are true at the origin?
1. the directional derivative D_ug exists for every unit vector u
2. g is continuous
3. g is differentiable
4. D_ug = ∇g·u for every unit vector u
Show answer
Answer: A — the directional derivative D_{u}g exists for every unit vector u
(1) True: g(tu)/t = u₁²u₂/(t²u₁⁴ + u₂²) → u₁²/u₂ if u₂ ≠ 0, and g(tu) = 0 if u₂ = 0. (2) False: on y = x², g = x⁴/(2x⁴) = 1/2 ≠ 0. (3) False: differentiability implies continuity, which fails. (4) False: ∇g(0) = (0, 0) because g vanishes on both axes, but D_ug = u₁²/u₂ ≠ 0 for u = (1, 1)/√2.
The directional derivative of f(x, y, z) = x²y + yz³ at the point (1, 2, 1) in the direction of the vector 2i − j + 2k is ____.
Numerical answer — type the value.
Show answer
Answer: 6
∇f = (2xy, x² + z³, 3yz²) = (4, 2, 6) at (1, 2, 1). The unit vector is (2, −1, 2)/3, since ‖(2, −1, 2)‖ = 3. D_uf = (8 − 2 + 12)/3 = 18/3 = 6. Forgetting to normalise gives 18.
For f(x, y) = x³ + y³ − 3xy, the critical points are
1. (0, 0), a saddle point, and (1, 1), a local minimum
2. (0, 0), a local maximum, and (1, 1), a local minimum
3. (0, 0), a saddle point, and (1, 1), a local maximum
4. (1, 1) only, a global minimum
Show answer
Answer: A — (0, 0), a saddle point, and (1, 1), a local minimum
fₓ = 3x² − 3y = 0 and f_y = 3y² − 3x = 0 give y = x², x = x⁴, so (0, 0) and (1, 1). With f_xx = 6x, f_yy = 6y, f_xy = −3: D(0, 0) = −9 < 0, a saddle; D(1, 1) = 27 > 0 with f_xx = 6 > 0, a local minimum. It is not global: f(x, 0) = x³ → −∞.
The maximum value of x + 2y + 2z subject to x² + y² + z² = 9 is ____.
Numerical answer — type the value.
Show answer
Answer: 9
Lagrange: (1, 2, 2) = 2λ(x, y, z), so (x, y, z) is parallel to (1, 2, 2), of length 3; on the sphere of radius 3, (x, y, z) = (1, 2, 2) and the value is 1 + 4 + 4 = 9. Equivalently, Cauchy–Schwarz gives |c·x| ≤ ‖c‖‖x‖ = 3 × 3. The minimum is −9.
The value of ∫₀¹ ∫ₓ¹ e^y² dy dx, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.86
The inner integral has no elementary primitive, so reverse the order over 0 ≤ x ≤ y ≤ 1: ∫₀¹ (∫₀ʸ dx) e^y² dy = ∫₀¹ y e^y² dy = [e^y²/2]₀¹ = (e − 1)/2 = 1.71828/2 = 0.859, which is 0.86.
Under spherical coordinates x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ, the volume element dx dy dz equals
1. ρ² sin φ dρ dφ dθ
2. ρ sin φ dρ dφ dθ
3. ρ² cos φ dρ dφ dθ
4. ρ² dρ dφ dθ
Show answer
Answer: A — ρ² sin φ dρ dφ dθ
The Jacobian determinant ∂(x, y, z)/∂(ρ, φ, θ) equals ρ² sin φ, which is non-negative for 0 ≤ φ ≤ π. Integrating it over 0 ≤ ρ ≤ a, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π gives (a³/3)(2)(2π) = 4πa³/3, the volume of the ball, a quick check that rules out the other factors.
The surface area of the part of the paraboloid z = x² + y² that lies below the plane z = 2 is kπ. The value of k, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 4.33
Area = ∬ √(1 + 4x² + 4y²) dA over x² + y² ≤ 2 = ∫₀^2π ∫₀^√2 √(1 + 4r²) r dr dθ. With u = 1 + 4r², r dr = du/8 and u runs from 1 to 9: 2π · (1/8)(2/3)(27 − 1) = 2π · 26/12 = 13π/3. So k = 13/3 = 4.33.
Let F = (−y/(x² + y²), x/(x² + y²)) on ℝ² ∖ {(0, 0)}. Which statement is correct?
1. curl F = 0 but F is not conservative on ℝ² ∖ {(0, 0)}
2. curl F = 0, so F is conservative on ℝ² ∖ {(0, 0)}
3. curl F ≠ 0, which is why the circulation is non-zero
4. the circulation of F round the unit circle is 0
Show answer
Answer: A — curl F = 0 but F is not conservative on ℝ² ∖ {(0, 0)}
Qₓ = (y² − x²)/(x² + y²)² = P_y, so the (scalar) curl vanishes. On r = (cos t, sin t), F·r′ = sin²t + cos²t = 1, so ∮F·dr = 2π ≠ 0, and a conservative field has zero circulation. Curl zero implies a potential only on a simply connected domain; the punctured plane is not one.
Let C be the unit circle x² + y² = 1 traversed counter-clockwise. Then ∮_C (−y³ dx + x³ dy) = kπ, where k = ____.
Numerical answer — type the value.
Show answer
Answer: 1.5
By Green, the integral is ∬ (Qₓ − P_y) dA = ∬ (3x² + 3y²) dA = 3∫₀^2π∫₀¹ r² · r dr dθ = 3 · 2π · 1/4 = 3π/2, so k = 1.5. The hypotheses hold: P = −y³, Q = x³ are polynomials.
The outward flux of F = (x³, y³, z³) through the unit sphere x² + y² + z² = 1 is kπ, where k = ____.
Numerical answer — type the value.
Show answer
Answer: 2.4
div F = 3(x² + y² + z²) = 3ρ². By Gauss, flux = ∭ 3ρ² · ρ² sin φ dρ dφ dθ = 3 · (1/5) · 2 · 2π = 12π/5, so k = 2.4. Using the volume 4π/3 with div F treated as 3 gives the wrong 4.
Let F = (y, z, x) and let C be the circle x² + y² = 1, z = 0, oriented counter-clockwise when viewed from above. Then ∮_C F·dr = kπ, where k = ____.
Numerical answer — type the value.
Show answer
Answer: -1
curl F = (R_y − Q_z, P_z − Rₓ, Qₓ − P_y) = (0 − 1, 0 − 1, 0 − 1) = (−1, −1, −1). By Stokes over the unit disc with n = k, the integral is (−1) × π = −π, so k = −1. Directly: r = (cos t, sin t, 0), F·r′ = sin t · (−sin t) + 0 + cos t · 0 = −sin²t, integrating to −π. Typed, the answer is -1.
For a scalar field φ and a vector field F, both of class C² on ℝ³, which identities hold everywhere?
1. curl(∇φ) = 0
2. div(curl F) = 0
3. curl(curl F) = 0
4. div(∇φ) = ∇²φ
Show answer
Answer: A — curl(∇φ) = 0; B — div(curl F) = 0; D — div(∇φ) = ∇²φ
(1) and (2) hold because mixed second partials of C² functions commute. (4) is the definition of the Laplacian. (3) is false: curl curl F = ∇(div F) − ∇²F; for F = (0, 0, x²), curl F = (0, −2x, 0) and curl curl F = (0, 0, −2) ≠ 0.
Using area = ½∮(x dy − y dx), the area enclosed by the ellipse x²/9 + y²/4 = 1 is
1. 6π
2. 12π
3. 36π
4. 3π
Show answer
Answer: A — 6π
With x = 3 cos t, y = 2 sin t, x dy − y dx = (6 cos²t + 6 sin²t) dt = 6 dt, so the area is ½ · 6 · 2π = 6π = πab with a = 3, b = 2. 12π forgets the factor ½; 36π multiplies a² by b².