*Question 1* (a) Consider the following derivatives of $u(x,t)=t^{-\frac{1}{2}}e^{-\frac{x^{2}}{8}}$: $\begin{align} \frac{ \partial u }{ \partial t } &= -\frac{1}{2} t^{-3/2} e^{-x^{2}/8t} + t^{-1/2} \left( \frac{x^{2}}{8t^{2}} e^{-x^{2}/8t} \right) \\ &= e^{-x^{2}/8t}\left(\frac{x^{2}t^{-5/2}}{8}-\frac{t^{-3/2}}{2} \right). \\ \frac{ \partial u }{ \partial x } &=t^{-1/2} \left( \frac{-2x}{8t}e^{-x^{2}/8} \right)=-\frac{xt^{-\frac{3}{2}}}{4}e^{-x^{2}/8t} .\\ \frac{ \partial^{2} u }{ \partial x^{2} } &= -\frac{t^{-\frac{3}{2}}}{4}e^{-x^{2}/8t} +\left( \frac{xt^{-\frac{3}{2}}}{4} \right)\left( \frac{2x}{8t} \right)e^{-x^{2}/8t} \\ &= e^{-x^{2}/8t}\left( \frac{x^{2}t^{-5/2}}{16}-\frac{t^{-3/2}}{4} \right). \end{align}$Hence $\alpha=2$. (b) (i) Note that $\frac{ \partial^{2} u }{ \partial x^{2} } = - \sin x \cos \beta t$ and $\frac{ \partial^{2} u }{ \partial t^{2} }=-\beta^{2}\sin x\cos \beta t$. Substituting these in the wave equation gives $1=\frac{\beta^{2}}{c^{2}}\implies\beta= c \;\text{ since $c,\beta>0$} .$ (b) (ii) Rewrite the given equation as $u(r,s)=f(r)+g(s)$where $r(x,t)=x+ct$ and $s(x,t)=x-ct$. Using chain rule, $\begin{align} \frac{ \partial u }{ \partial x } &= \frac{ \partial u }{ \partial r } \frac{ \partial r }{ \partial x } + \frac{ \partial u }{ \partial s } \frac{ \partial s }{ \partial x }. \end{align}$Substituting $\frac{ \partial r }{ \partial x } =\frac{ \partial s }{ \partial x }=1$ then $\frac{ \partial g }{ \partial r } =\frac{ \partial f }{ \partial s }=0$ gives $\frac{ \partial u }{ \partial x } =\frac{ \partial u }{ \partial r } + \frac{ \partial u }{ \partial s } = \frac{ \partial f }{ \partial r } + \frac{ \partial g }{ \partial s } .$Differentiating again w.r.t $x$ gives $\begin{align} \frac{ \partial^{2} u }{ \partial x^{2} } &= \left( \frac{ \partial^{2} f }{ \partial r^{2} } \frac{ \partial r }{ \partial x } + \frac{ \partial^{2} f }{ \partial rs } \frac{ \partial s }{ \partial x } \right) + \left( \frac{ \partial^{2} g }{ \partial sr } \frac{ \partial s }{ \partial x } + \frac{ \partial^{2} g }{ \partial s^{2} } \frac{ \partial s }{ \partial x } \right) \\ &= \frac{ \partial^{2} f }{ \partial r^{2} } \frac{ \partial r }{ \partial x } + \frac{ \partial^{2} g }{ \partial s^{2} } \frac{ \partial s }{ \partial x } &\text{(since $\frac{ \partial^{2} g }{ \partial sr } = \frac{ \partial^{2} f }{ \partial rs }=0 $)} \\ &= \frac{ \partial^{2} f }{ \partial r^{2} } + \frac{ \partial^{2} g }{ \partial s^{2} } & \text{(substituting $\frac{ \partial r }{ \partial x } =\frac{ \partial s }{ \partial x }=1$)} \end{align} $ Similarly $\frac{ \partial^{2} u }{ \partial t^{2} } = \frac{ \partial^{2} f }{ \partial r^{2} } \left( \frac{ \partial r }{ \partial t } \right)^{2} + \frac{ \partial^{2} g }{ \partial s^{2} } \left( \frac{ \partial s }{ \partial t } \right)^{2} = c^{2} \left( \frac{ \partial^{2} f }{ \partial r^{2} } + \frac{ \partial^{2} g }{ \partial s^{2} } \right) .$ So indeed $\frac{ \partial^{2} u }{ \partial x^{2} } = \frac{1}{c^{2}} \frac{ \partial^{2} u }{ \partial t^{2} }$ and it satisfies the wave equation. *Question 2* (a) Its first derivatives are $f_{x}=e^{x}-2$ and $f_{y}=3y^{2}-3$. Equating them to zero gives that the critical points are $(\ln 2,1)$ and $(\ln 2,-1)$. Its second derivatives are by $f_{xx}=e^{x},\quad f_{xy} =f_{yx} = 0, \quad f_{yy} = 6y $hence the determinant of its Hessian matrix is given by $D=6e^{x}y$. At $(\ln2,1)$: $D=12>0$ & $f_{xx}=2>0$ so it is a minimum point. At $(\ln2,-1)$: $D=-12<0$ so it is a saddle point. (b) The linear approximation around $P$ is given by $\begin{align} L(x,y) &=f(0,2) + xf_{x}(0,2) + (y-2)f_{y}(0,2) \\ &=3-x+9(y-2) \\ &= -x + 9y -15. \end{align}$So $f(x,y)\approx L(x,y) =-0.01+9(2-0.01)-15=18-10\times 0.01-15=3-0.1=2.9$. (c) ![[Pasted image 20240206175343.png|350]] (d) ![[Pasted image 20240206213718.png]] ![[Pasted image 20240206213759.png]]