Multivariable Calculus

Q1. (3 marks) Level curve $x^2+y^2=4$ is a circle centred at the origin of radius $2$.

• Recognising circle: 1
• Equation $x^2+y^2=4$: 1
• Radius $2$: 1

Why: Level curves are the sets where $f$ is constant; $x^2+y^2=c$ describes concentric circles for $c>0$.

Q2. (4 marks) $f_x = 3x^2 y - 4y^2 + y\cos(xy)$ $f_y = x^3 - 8xy + x\cos(xy)$

• $f_x$ correct (each term, ~2): 2
• $f_y$ correct: 2

Why: Hold the other variable constant; chain rule on $\sin(xy)$ gives $\cos(xy)\cdot(\partial/\partial x\, xy)$.

Q3. (5 marks) Clairaut's theorem: If $f_{xy}$ and $f_{yx}$ are continuous on an open region containing a point, then $f_{xy}=f_{yx}$ there. (2) Verification for $f=x^2y^3$: $f_x = 2xy^3,\quad f_{xy}=6xy^2.$ $f_y = 3x^2y^2,\quad f_{yx}=6xy^2.$ So $f_{xy}=f_{yx}=6xy^2$. ✓

• Statement: 2, $f_{xy}$: 1.5, $f_{yx}$: 1.5

Q4. (5 marks) $z_x = 2x \Rightarrow$ at $(1,1)$: $2$. $z_y = 6y \Rightarrow 6$. Tangent plane: $z - 4 = 2(x-1) + 6(y-1)$, i.e. $z = 2x + 6y - 4.$

• $z_x, z_y$: 2, plug in: 1, plane equation: 2

Why: Tangent plane $z=z_0+f_x(x-x_0)+f_y(y-y_0)$.

Q5. (4 marks) $\dfrac{dz}{dt} = \dfrac{\partial z}{\partial x}\dfrac{dx}{dt} + \dfrac{\partial z}{\partial y}\dfrac{dy}{dt}$ $= (2xy)(2t) + (x^2)(3t^2)$. Substitute $x=t^2$, $y=t^3$: $= 2t^5\cdot 2t + t^4\cdot 3t^2 = 4t^6 + 3t^6 = 7t^6.$

• Chain rule setup: 2, substitution: 1, answer $7t^6$: 1

Q6. (5 marks) $\nabla f = \langle 2x, 2y\rangle$; at $(1,2)$: $\langle 2, 4\rangle$. Unit vector: $|\mathbf{v}|=5$, $\hat{\mathbf{u}}=\langle 3/5, 4/5\rangle$. $D_{\mathbf u}f = \nabla f\cdot\hat{\mathbf u} = 2\cdot\tfrac35 + 4\cdot\tfrac45 = \tfrac{6}{5}+\tfrac{16}{5} = \tfrac{22}{5}.$

• Gradient: 2, unit vector: 1, dot product & answer $22/5$: 2

Q7. (5 marks) $f_x = 2x-4=0\Rightarrow x=2$; $f_y=2y+6=0\Rightarrow y=-3$. Critical point $(2,-3)$. $f_{xx}=2$, $f_{yy}=2$, $f_{xy}=0$. Hessian $D = f_{xx}f_{yy}-f_{xy}^2 = 4>0$ and $f_{xx}=2>0$. $\Rightarrow$ local minimum at $(2,-3)$.

• Critical point: 2, second derivatives: 1, $D=4$: 1, classification: 1

Q8. (4 marks) $\int_0^1\int_0^2 (x+y)\,dy\,dx = \int_0^1\left[xy+\tfrac{y^2}{2}\right]_0^2 dx = \int_0^1 (2x+2)\,dx = [x^2+2x]_0^1 = 3.$

• Inner integral: 2, outer & answer $3$: 2

Q9. (3 marks) $\operatorname{div}\mathbf F = \nabla\cdot\mathbf F = P_x + Q_y + R_z$. (2) For $\mathbf F=\langle x^2,y^2,z^2\rangle$: $\operatorname{div}\mathbf F = 2x+2y+2z$. (1)

Q10. (2 marks) Circulation form: $\displaystyle\oint_C P\,dx+Q\,dy = \iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$ (1) Flux form: $\displaystyle\oint_C \mathbf F\cdot\mathbf n\,ds = \iint_D\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\right)dA.$ (1)

[
  {"claim":"Chain rule Q5 gives 7t**6","code":"t=symbols('t'); x=t**2; y=t**3; z=x**2*y; result = simplify(diff(z,t) - 7*t**6)==0"},
  {"claim":"Directional derivative Q6 equals 22/5","code":"x,y=symbols('x y'); f=x**2+y**2; grad=[diff(f,x),diff(f,y)]; g=[gr.subs({x:1,y:2}) for gr in grad]; u=[Rational(3,5),Rational(4,5)]; D=sum(a*b for a,b in zip(g,u)); result = D==Rational(22,5)"},
  {"claim":"Double integral Q8 equals 3","code":"x,y=symbols('x y'); result = integrate(integrate(x+y,(y,0,2)),(x,0,1))==3"},
  {"claim":"Hessian determinant Q7 positive (local min)","code":"x,y=symbols('x y'); f=x**2+y**2-4*x+6*y+1; D=diff(f,x,2)*diff(f,y,2)-diff(f,x,y)**2; result = (D==4) and (diff(f,x,2)>0)"}
]