Level 3 — ProductionMultivariable Calculus

Multivariable Calculus

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Level: 3 — Production (from-scratch derivations, explain-out-loud, code-from-memory) Time limit: 45 minutes Total marks: 60

Show all work. Derivations must proceed from definitions where indicated. Explanations must justify each step ("the why"), not just state results.

Question 1 (10 marks) — Path-dependence and continuity

Consider $f(x,y)=\frac{x^2 y}{x^4+y^2},\quad (x,y)\neq(0,0),\qquad f(0,0)=0.$

(a) From first principles, show that the limit along every straight line $y=mx$ through the origin equals $0$ . (4)

(b) By choosing a suitable non-linear path, prove the limit as $(x,y)\to(0,0)$ does not exist, and hence state whether $f$ is continuous at the origin. Explain why linear paths alone are insufficient to decide continuity. (6)

Question 2 (10 marks) — Gradient, steepest ascent, tangent plane

Let $f(x,y)=x^2+3xy+y^3$ and let $P=(1,2)$ .

(a) Compute $\nabla f(P)$ and, from the definition, the directional derivative $D_{\mathbf u}f(P)$ in the direction of $\mathbf v=(3,-4)$ . (4)

(b) State and derive (from the dot-product formula) the direction of steepest ascent at $P$ and the maximum rate of increase. (3)

Question 3 (12 marks) — Critical points and the second-derivative test

For $f(x,y)=x^3+y^3-3xy$ :

(a) Find all critical points from $\nabla f=\mathbf 0$ . (4)

(b) Derive the Hessian and, using the discriminant $D=f_{xx}f_{yy}-f_{xy}^2$ , classify each critical point. State clearly the rule for each case and why $D$ governs classification. (6)

Question 4 (12 marks) — Change of order / polar integration

(a) Evaluate by reversing the order of integration: $\int_0^1\int_{\sqrt{y}}^{1} e^{x^3}\,dx\,dy.$ Show the region and justify the reversal. (6)

(b) Evaluate using polar coordinates, stating the Jacobian explicitly: $\iint_D (x^2+y^2)\,dA,\qquad D=\{x^2+y^2\le 4,\ y\ge0\}.$ (6)

Question 5 (10 marks) — Conservative fields and line integrals

Let $\mathbf F=(2xy+z^2,\,x^2,\,2xz)$ .

(a) Show $\mathbf F$ is conservative by computing $\nabla\times\mathbf F$ from the definition, and find a potential function $\varphi$ . (6)

(b) Using the Fundamental Theorem for line integrals, evaluate $\int_C\mathbf F\cdot d\mathbf r$ where $C$ runs from $(0,0,0)$ to $(1,1,1)$ . Explain why the specific path is irrelevant. (4)

Question 6 (6 marks) — Explain-out-loud: unification

State Green's, Stokes', and the Divergence theorem, and explain in your own words how all three are special cases of the generalized Stokes theorem $\int_M d\omega=\int_{\partial M}\omega$ . Identify what plays the role of $\omega$ , $d\omega$ , $M$ , and $\partial M$ in each. (6)

Answer keyMark scheme & solutions

Question 1 (10)

(a) Along $y=mx$ : $f(x,mx)=\frac{x^2(mx)}{x^4+m^2x^2}=\frac{mx^3}{x^2(x^2+m^2)}=\frac{mx}{x^2+m^2}.$ As $x\to0$ , numerator $\to0$ , denominator $\to m^2\neq0$ (for $m\neq0$ ), so limit $=0$ . For $m=0$ ( $y=0$ ): $f=0$ . (4: substitution 2, limit 2)

(b) Take path $y=x^2$ : $f(x,x^2)=\frac{x^2\cdot x^2}{x^4+x^4}=\frac{x^4}{2x^4}=\tfrac12.$ Along this path the limit is $\tfrac12\neq0$ . Since two paths give different limits, $\lim_{(x,y)\to0}f$ does not exist; hence $f$ is not continuous at $0$ (limit $\neq f(0,0)=0$ ). Linear paths are insufficient because the denominator has an $x^4$ term matching the numerator's degree only along parabolic paths — straight lines all "wash out" the $x^4$ term, masking the discontinuity. (6: path 2, evaluation 2, conclusion + why 2)

Question 2 (10)

(a) $f_x=2x+3y,\ f_y=3x+3y^2$ . At $P=(1,2)$ : $\nabla f=(2+6,\,3+12)=(8,15)$ . Unit vector: $|\mathbf v|=\sqrt{9+16}=5$ , $\mathbf u=(3/5,-4/5)$ . $D_{\mathbf u}f=\nabla f\cdot\mathbf u=\tfrac{8\cdot3+15\cdot(-4)}{5}=\tfrac{24-60}{5}=-\tfrac{36}{5}=-7.2.$ (4)

(b) Steepest ascent is along $\nabla f=(8,15)$ (since $D_{\mathbf u}f=|\nabla f|\cos\theta$ maximised at $\theta=0$ ). Max rate $=|\nabla f|=\sqrt{64+225}=\sqrt{289}=17$ . (3)

(c) $f(P)=1+6+8=15$ . Tangent plane: $z=15+8(x-1)+15(y-2)=8x+15y-23.$ (3)

Question 3 (12)

(a) $f_x=3x^2-3y=0\Rightarrow y=x^2$ ; $f_y=3y^2-3x=0\Rightarrow x=y^2$ . Sub: $x=x^4\Rightarrow x(x^3-1)=0\Rightarrow x=0,1$ . Critical points: $(0,0)$ and $(1,1)$ . (4)

(b) $f_{xx}=6x,\ f_{yy}=6y,\ f_{xy}=-3$ . $D=36xy-9$ .

At $(0,0)$ : $D=-9<0\Rightarrow$ saddle.
At $(1,1)$ : $D=36-9=27>0$ and $f_{xx}=6>0\Rightarrow$ local minimum.

Rule: $D>0,f_{xx}>0$ min; $D>0,f_{xx}<0$ max; $D<0$ saddle; $D=0$ inconclusive. $D$ governs because it is the Hessian determinant whose sign determines definiteness of the quadratic approximation. (6)

(c) $f_{xy}=\partial_y(3x^2-3y)=-3$ ; $f_{yx}=\partial_x(3y^2-3x)=-3$ . Equal, confirming Clairaut (partials continuous). (2)

Question 4 (12)

(a) Region: $0\le y\le1,\ \sqrt y\le x\le1$ , i.e. $y\le x^2$ , $0\le x\le1$ . Reversed: $\int_0^1\int_0^{x^2}e^{x^3}\,dy\,dx=\int_0^1 x^2 e^{x^3}\,dx.$ Let $u=x^3,\ du=3x^2dx$ : $=\tfrac13\int_0^1 e^u du=\tfrac13(e-1).$ Reversal valid since region is bounded and both descriptions cover the same triangle-like region under $y=x^2$ . (6)

(b) Polar: $x=r\cos\theta,y=r\sin\theta$ , $dA=r\,dr\,d\theta$ (Jacobian $=r$ ). Upper half disc: $0\le r\le2$ , $0\le\theta\le\pi$ . $\int_0^\pi\int_0^2 r^2\cdot r\,dr\,d\theta=\pi\cdot\Big[\tfrac{r^4}{4}\Big]_0^2=\pi\cdot4=4\pi.$ (6)

Question 5 (10)

(a) $\mathbf F=(P,Q,R)=(2xy+z^2,x^2,2xz)$ . $\nabla\times\mathbf F=(R_y-Q_z,\ P_z-R_x,\ Q_x-P_y)=(0-0,\ 2z-2z,\ 2x-2x)=(0,0,0).$ Conservative. Find $\varphi$ : $\varphi_x=2xy+z^2\Rightarrow\varphi=x^2y+xz^2+g(y,z)$ . $\varphi_y=x^2+g_y=x^2\Rightarrow g_y=0$ . $\varphi_z=2xz+g_z=2xz\Rightarrow g_z=0$ . So $g$ =const. $\varphi=x^2y+xz^2.$ (6)

(b) $\int_C\mathbf F\cdot d\mathbf r=\varphi(1,1,1)-\varphi(0,0,0)=(1+1)-0=2.$ Path-independent because $\mathbf F=\nabla\varphi$ ; the integral depends only on endpoints (FTLI). (4)

Question 6 (6)

Green: $\oint_{\partial D}(P\,dx+Q\,dy)=\iint_D(Q_x-P_y)\,dA$ . (1)
Stokes: $\oint_{\partial S}\mathbf F\cdot d\mathbf r=\iint_S(\nabla\times\mathbf F)\cdot d\mathbf S$ . (1)
Divergence: $\iint_{\partial V}\mathbf F\cdot d\mathbf S=\iiint_V(\nabla\cdot\mathbf F)\,dV$ . (1)

Unification (3): Generalized Stokes $\int_M d\omega=\int_{\partial M}\omega$ . $\omega$ is a differential form on manifold $M$ with boundary $\partial M$ ; $d\omega$ is the exterior derivative. Green: $M$ =2D region, $\omega=P\,dx+Q\,dy$ (1-form), $d\omega=(Q_x-P_y)dx\wedge dy$ . Stokes: $M$ =surface, $\omega=\mathbf F\cdot d\mathbf r$ , $d\omega=$ curl flux 2-form. Divergence: $M$ =3D solid, $\omega=$ flux 2-form, $d\omega=(\nabla\cdot\mathbf F)dV$ (3-form). In each, integrating the derivative over the region equals integrating the form over its boundary.

[
  {"claim":"Q1(b) parabolic path gives 1/2","code":"x=symbols('x'); val=limit((x**2*x**2)/(x**4+(x**2)**2),x,0); result = (val==Rational(1,2))"},
  {"claim":"Q2 gradient magnitude at P is 17","code":"result = (sqrt(8**2+15**2)==17)"},
  {"claim":"Q2 directional derivative equals -36/5","code":"result = (Rational(8*3+15*(-4),5)==Rational(-36,5))"},
  {"claim":"Q3 (1,1) is local min: D=27>0, fxx=6>0","code":"x,y=symbols('x y'); D=36*x*y-9; result = (D.subs({x:1,y:1})==27 and 6>0)"},
  {"claim":"Q4a integral equals (e-1)/3","code":"x=symbols('x'); val=integrate(x**2*exp(x**3),(x,0,1)); result = (simplify(val-(exp(1)-1)/3)==0)"},
  {"claim":"Q4b polar integral equals 4pi","code":"r,t=symbols('r t'); val=integrate(integrate(r**2*r,(r,0,2)),(t,0,pi)); result = (val==4*pi)"},
  {"claim":"Q5b potential difference equals 2","code":"x,y,z=symbols('x y z'); phi=x**2*y+x*z**2; result = (phi.subs({x:1,y:1,z:1})-phi.subs({x:0,y:0,z:0})==2)"}
]