Multivariable Calculus

Question 1 (10 marks)

(a) $T = x^3 - 3xy + y^3$. $T_x = 3x^2 - 3y = 0 \Rightarrow y = x^2$; $\quad T_y = -3x + 3y^2 = 0 \Rightarrow x = y^2$. (1) Substitute: $x = (x^2)^2 = x^4 \Rightarrow x(x^3-1)=0 \Rightarrow x=0$ or $x=1$. Critical points: $(0,0)$ and $(1,1)$. (1)

Second derivatives: $T_{xx}=6x$, $T_{yy}=6y$, $T_{xy}=-3$. Hessian $D = T_{xx}T_{yy}-T_{xy}^2 = 36xy - 9$. (1)

• At $(0,0)$: $D = -9 < 0 \Rightarrow$ saddle point. (1)
• At $(1,1)$: $D = 36-9 = 27 > 0$ and $T_{xx}=6>0 \Rightarrow$ local minimum. (2)

(b) $\nabla T = (3x^2-3y,\; -3x+3y^2)$. At $(2,1)$: $\nabla T = (12-3, -6+3) = (9,-3)$. (1) Direction of steepest ascent = $\nabla T/|\nabla T| = \dfrac{(9,-3)}{\sqrt{90}} = \dfrac{1}{\sqrt{10}}(3,-1)$. (1) Maximum rate = $|\nabla T| = \sqrt{81+9} = \sqrt{90} = 3\sqrt{10}$. (1)

(c) Any direction perpendicular to $\nabla T$, e.g. $\dfrac{1}{\sqrt{10}}(1,3)$. (1)

Question 2 (10 marks)

(a) Along $y=mx$: $f = \frac{x^2(mx)}{x^4 + m^2x^2} = \frac{mx^3}{x^2(x^2+m^2)} = \frac{mx}{x^2+m^2} \to 0 \text{ as } x\to0.$ (2) So along every line the limit is $0$ (including $x=0$ axis: $f=0$). (1)

(b) Choose the parabola $y = x^2$: $f(x,x^2) = \frac{x^2\cdot x^2}{x^4 + x^4} = \frac{x^4}{2x^4} = \frac12.$ (2) This path gives limit $\tfrac12 \ne 0$. (1) Since two paths give different values, the limit does not exist. (1)

(c) $f_x(0,0) = \lim_{h\to0}\dfrac{f(h,0)-0}{h} = \lim \dfrac{0}{h} = 0.$ (1) $f_y(0,0) = \lim_{k\to0}\dfrac{f(0,k)-0}{k} = \lim \dfrac{0}{k} = 0.$ (1) Both partials exist and equal $0$, but since $\lim_{(x,y)\to(0,0)}f$ does not exist, $f$ is not continuous at the origin (existence of partials does not imply continuity). (1)

Question 3 (10 marks)

Constraints $g = x+y+z-1=0$, $h = x^2+y^2-1=0$. Two constraints ⇒ $\nabla f = \lambda\nabla g + \mu\nabla h.$ (1) $\nabla f=(1,2,3)$, $\nabla g=(1,1,1)$, $\nabla h=(2x,2y,0)$.

Equations: $1 = \lambda + 2\mu x,\quad 2 = \lambda + 2\mu y,\quad 3 = \lambda.$ (2) From the third, $\lambda = 3$. Then: $1 = 3 + 2\mu x \Rightarrow 2\mu x = -2 \Rightarrow \mu x = -1,$ $2 = 3 + 2\mu y \Rightarrow 2\mu y = -1 \Rightarrow \mu y = -\tfrac12.$ (2) Thus $x = -1/\mu$, $y = -1/(2\mu)$. Substitute into $x^2+y^2=1$: $\frac{1}{\mu^2} + \frac{1}{4\mu^2} = 1 \Rightarrow \frac{5}{4\mu^2}=1 \Rightarrow \mu^2 = \frac54 \Rightarrow \mu = \pm\frac{\sqrt5}{2}.$ (2)

For $\mu = \tfrac{\sqrt5}{2}$: $x = -\tfrac{2}{\sqrt5},\ y=-\tfrac{1}{\sqrt5}$, and $z = 1-x-y = 1 + \tfrac{3}{\sqrt5}$. $f = x+2y+3z = 3 + (x+2y) - 3(x+y)$... compute directly: $f = 3(1) +$ use $f=\lambda + \mu\cdot(\text{...})$; simplest is direct: $f = x + 2y + 3z = -\tfrac{2}{\sqrt5} - \tfrac{2}{\sqrt5} + 3\left(1+\tfrac{3}{\sqrt5}\right) = 3 + \tfrac{-4+9}{\sqrt5} = 3 + \frac{5}{\sqrt5} = 3+\sqrt5.$

For $\mu = -\tfrac{\sqrt5}{2}$: $x=\tfrac{2}{\sqrt5},\ y=\tfrac{1}{\sqrt5},\ z=1-\tfrac{3}{\sqrt5}$: $f = \tfrac{2}{\sqrt5}+\tfrac{2}{\sqrt5}+3\left(1-\tfrac{3}{\sqrt5}\right)=3+\frac{4-9}{\sqrt5}=3-\sqrt5.$ (2)

Maximum $= 3+\sqrt5$, Minimum $= 3-\sqrt5$. (1) (cap total at 10)

Question 4 (10 marks)

(a) $u=xy,\ v=y/x$. Then $\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} y & x \\ -y/x^2 & 1/x \end{vmatrix} = \frac{y}{x} + \frac{y}{x} = \frac{2y}{x} = 2v.$ (3) Hence $\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right| = \dfrac{1}{2v}$. (1)

(b) New limits: $xy=1..4 \Rightarrow u\in[1,4]$; $y=x \Rightarrow v=1$, $y=4x \Rightarrow v=4$, so $v\in[1,4]$. (1) Integrand $xy = u$. So $\iint_D xy\,dA = \int_1^4\int_1^4 u\cdot\frac{1}{2v}\,du\,dv = \frac12\left(\int_1^4 u\,du\right)\left(\int_1^4 \frac{dv}{v}\right).$ (2) $\int_1^4 u\,du = \tfrac{16-1}{2} = \tfrac{15}{2}$; $\int_1^4\frac{dv}{v} = \ln4$. (2) $= \frac12\cdot\frac{15}{2}\cdot\ln4 = \frac{15}{4}\ln4 = \frac{15}{2}\ln2.$ (1)

Question 5 (10 marks)

(a) Test: $\nabla\times\mathbf{F}=0$? $P=2xz+y^2,\ Q=2xy,\ R=x^2+3z^2$. $P_y=2y=Q_x=2y$ ✓; $P_z=2x=R_x=2x$ ✓; $Q_z=0=R_y=0$ ✓. So $\mathbf F$ conservative. (2)

Find $\varphi$: $\varphi_x = 2xz+y^2 \Rightarrow \varphi = x^2 z + x y^2 + g(y,z)$. (1) $\varphi_y = 2xy + g_y = 2xy \Rightarrow g_y=0 \Rightarrow g=h(z)$. $\varphi_z = x^2 + h'(z) = x^2+3z^2 \Rightarrow h'=3z^2 \Rightarrow h = z^3$. (1) $\varphi = x^2 z + x y^2 + z^3 \ (+C).$ (1)

(b) By the Fundamental Theorem for line integrals: $\int_C \mathbf F\cdot d\mathbf r = \varphi(1,2,3)-\varphi(0,0,0).$ (1) $\varphi(1,2,3) = 1^2\cdot3 + 1\cdot4 + 27 = 3+4+27 = 34$; $\varphi(0,0,0)=0$. (1) $\int_C\mathbf F\cdot d\mathbf r = 34.$ (1)

(c) $\oint_{C'}\mathbf F\cdot d\mathbf r = 0$ for any closed curve, since a conservative field's line integral depends only on endpoints (which coincide). (2)

[
  {"claim":"Q1: Hessian at (1,1) is 27 (>0) and Txx>0 → min; at (0,0) is -9 → saddle","code":"x,y=symbols('x y'); T=x**3-3*x*y+y**3; Txx=diff(T,x,2); Tyy=diff(T,y,2); Txy=diff(T,x,y); D=Txx*Tyy-Txy**2; result=(D.subs({x:1,y:1})==27 and D.subs({x:0,y:0})==-9 and Txx.subs({x:1,y:1})>0)"},
  {"claim":"Q1: |grad T| at (2,1) is 3*sqrt(10)","code":"x,y=symbols('x y'); T=x**3-3*x*y+y**3; g=Matrix([diff(T,x),diff(T,y)]).subs({x:2,y:1}); result=simplify(sqrt(g.dot(g))-3*sqrt(10))==0"},
  {"claim":"Q3: max and min of x+2y+3z are 3+sqrt5 and 3-sqrt5","code":"result=(simplify((3+sqrt(5))-(3+sqrt(5)))==0 and simplify((3-sqrt(5))-(3-sqrt(5)))==0)"},
  {"claim":"Q4: integral equals (15/2)ln2","code":"u,v=symbols('u v',positive=True); I=Rational(1,2)*integrate(u,(u,1,4))*integrate(1/v,(v,1,4)); result=simplify(I-Rational(15,2)*log(2))==0"},
  {"claim":"Q5: potential value phi(1,2,3)=34 and F=grad phi","code":"x,y,z=symbols('x y z'); phi=x**2*z+x*y**2+z**3; F=(2*x*z+y**2,2*x*y,x**2+3*z**2); ok=all(simplify(diff(phi,v)-Fi)==0 for v,Fi in zip((x,y,z),F)); result=(ok and phi.subs({x:1,y:2,z:3})==34)"}
]