Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / Section B

Substituting $x=r\cos \theta$, $y=r\sin \theta$ into the circle equation gives $r^2-2r\cos \theta =0$. Thus its nonzero radial intersection is $r_Q=2\cos \theta$, with the origin also on the circle. The tangent to a circle at the rightmost point is $x=2$, or $r_R\cos \theta =2$.

Take rays with $-\pi /2<\theta <\pi /2$. The difference of the two radial distances gives the cissoid of Diocles:Multiplying by $\cos \theta$ gives $x=2{\sin }^2\theta =2y^2/(x^2+y^2)$, hence the Cartesian equation $y^2(2-x)=x^3$. A convenient parametric curve, with $t=\tan \theta$, isThe real curve has $0\le x<2$, a cusp at the origin, symmetry about the $x$ axis, and the vertical asymptote $x=2$. Near the origin, $x\sim 2t^2$ and $y\sim 2t^3$, so its cusp has horizontal tangent direction. As $|t|\to \infty$, $x\to 2^{-}$ while $|y|\to \infty$. The circle and curve meet away from the origin at $(1,\pm 1)$.

The tangent half-angle substitution $t=\tan (x/2)$ gives $\sin x=2t/(1+t^2)$ and $dx=2dt/(1+t^2)$. The endpoints $x=0,\pi /2$ become $t=0,1$, respectively. HenceCompleting the square, $t^2+t+1=(t+1/2{)}^2+3/4$, then gives

Normalization of a probability distribution and the geometric series give $1=A/(1-\rho )$, so $A=1-\rho$. For a nonnegative integer $n$, the strict tail starts at $n+1$:For nonnegative integers $n>m$, the event $\{N>n\}$ is contained in $\{N>m\}$. ThereforeThis is the memorylessness of the geometric distribution with the zero-based counting convention. Keeping strict versus non-strict inequalities explicit is essential: $N-m-1$, conditional on $N>m$, has the original distribution of $N$.

Write $\lambda =\ln ({\rho }^{-1})>0$. The probability density function normalizes when $1=B{\int }_0^{\infty }{e}^{-\lambda x}dx=B/\lambda$, so $B=\lambda$. Its survival function is $e^{-\lambda x}$ for $x\ge 0$; hence, for $x>y\ge 0$,Differentiating this conditional tail givesThe value assigned at the single endpoint does not affect any probability. The memorylessness of the exponential distribution says that $X-y$, conditional on $X>y$, is again an exponential distribution of rate $\lambda$. Its first two moments are $1/\lambda$ and $2/{\lambda }^2$, found by integration by parts, so translation leaves the variance unchanged:

The characteristic equation for the homogeneous solution is $(r+2{)}^2=0$, giving $(C_1+C_{2}x)e^{-2x}$. For a particular solution, substitute $A{e}^{2x}$: the left side becomes $16A{e}^{2x}$, so $A=1/16$. The general solution is

The characteristic equation has roots $-3\pm 4i$, giving the homogeneous solution $e^{-3x}(A\cos 4x+B\sin 4x)$. For the sinusoidal forcing term, $y_p=\sin 5x$ is a particular solution, since its second derivative cancels $25y_p$ and $6y_p^{\prime }=30\cos 5x$. The zero initial displacement gives $A=0$; zero initial derivative gives $4B+5=0$. ThusThe transient response decays as $e^{-3x}$, leaving the steady-state response $\sin 5x$. Near zero, $y(0)=y^{\prime }(0)=0$ and $y^{\prime \prime }(0)=30$, so the sketch starts with $y\sim 15x^2$ and a horizontal tangent line.

If $udx+vdy=dF$ with continuous second partial derivatives of $F$, then $u=F_x$ and $v=F_y$. The symmetry of second derivatives gives the necessary condition for an exact differential form:This condition is also locally sufficient on a small rectangle for continuously differentiable coefficients; global sufficiency additionally needs suitable topology of the domain.

Holding $y$ fixed, the chain rule giveswhere the local change of variable from $x$ to $u$ is valid.

For the internal energy $U(S,V)$, its exact differential identifies $T=U_S$ and $p=-U_V$. The Maxwell relation from equality of mixed partial derivatives is $T_V{|}_S=-p_S{|}_V$. At fixed volume, the constant-volume heat capacity satisfiesso $T_S{|}_V=T/C_V$, assuming nonzero $C_V$ and a locally invertible relation. Applying the chain rule to $p(T,V)$ gives the isentropic temperature-volume derivativeFor the stated one-mole ideal gas, the equation of state gives $p_T{|}_V=R/V$ and $C_V=3R/2$, so the right side is $-2T/(3V)$. Along an isentropic process, $p\propto V^{-5/3}$ and $T=pV/R\propto V^{-2/3}$, giving the same left side:

Taking the curl component by component gives

Integrating ${\psi }_x=-(2xy+z)$ with respect to $x$ gives $\psi =-x^{2}y-xz+h(y,z)$. Matching ${\psi }_y=-(x^2+2y)$ requires $h=-y^2+j(z)$, and matching ${\psi }_z=-x$ requires $j$ constant. A scalar potential with the requested sign isThus the force field is a conservative vector field on all of ${\mathbb{R}}^3$.

Parametrize the straight line by $\mathbf{x}(t)=(t,t,t)$, $0\le t\le 1$. The line integral isThe change in scalar potential is $\psi (1,1,1)-\psi (0,0,0)=-3$, so the answer is $-\Delta \psi$, as required by the fundamental theorem for line integrals for $\mathbf{F}=-\nabla \psi$.

For positive semiaxes $a,b$, the cross product of the two parametric tangent vectors gives the outward oriented surface element of an ellipsoid of revolution:Substituting the parametrization into the vector field and taking its dot product with this element givesThe middle term has zero integral over $0\le \varphi \le 2\pi$. In the other terms, ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$, ${\int }_0^{\pi }{\sin }^3\theta {\cos }^2\theta d\theta =4/15$, and ${\int }_0^{\pi }\sin \theta {\cos }^4\theta d\theta =2/5$. The directly evaluated flux is therefore

True. Multiplicativity of the determinant and the identity $\det (B^{-1})=1/\det B$ give $\det (B^{-1}AB)=\det (B^{-1})\det A\det B=\det A$. This is invariance under matrix similarity.

False. The cyclic property of the trace permits $ABC\mapsto BCA$, but not an arbitrary interchange. Even with $A$ and $B$ both invertible matrices, takeThen $\mathrm{Tr}(ABC)=1$ and $\mathrm{Tr}(BAC)=2$, providing a counterexample.

True. Distribute the matrix product while retaining its order:Only addition is commuted; no commutation of $A$ and $B$ is assumed.

The permutation matrix cycles the three coordinate vectors, so three applications give $C^3=I$. Consequently $C^{2}C=C{C}^2=I$, proving $C^{-1}=C^2$. Since $16=5\cdot 3+1$,

Expansion of the determinant gives $\det M=-\mu -\mu =-2\mu$. The matrix inverse exists precisely when

For $\mu \ne 0$, the adjugate matrix formula, or elimination on the three unit right-hand sides, givesMultiplication by $M$ gives the identity matrix, checking all signs and coefficients.

Subtracting the first and third equations gives $\mu (x-z)=0$. Since $\mu \ne 0$, $x=z$, and the second equation gives $x=z=1$. The first then gives $y=1-\mu$, so the unique solution of the system of linear equations is

At $\mu =0$, the first and third equations both reduce to $y=1$, while the second gives $x+z=2$. All solutions form the straight lineThe limit of the nonzero-$\mu$ solution selects only the point $(1,1,1)$ on this whole line; setting a parameter to its singular value can enlarge the solution set.

Use the real Fourier series convention $f(x)\sim a_0/2+{\sum }_{n\ge 1}[a_n\cos (n\pi x)+b_n\sin (n\pi x)]$ for period two. The Fourier coefficients areThe last two integrals follow by integration by parts. Thus the Fourier series of a half-wave rectified ramp isOn $(-1,1)$, $f(x)=|x|+x$. Its even part is $f_e(x)=|x|$ and its odd part is $f_o(x)=x$, so the cosine and sine terms separate:At a jump, the Fourier series converges to the average of the one-sided limits. Thus the first series for $f$ takes the value $1$ at odd integers, while the Fourier series of a periodic sawtooth for $f_o$ takes the value $0$ there. The even function has a continuous periodic extension.

At $x=0$, the formula for $f_e$ gives $0=1/2-(4/{\pi }^2){\sum }_{r\ge 0}(2r+1{)}^{-2}$. At $x=1/2$, the formula for $f_o$ has only odd-index terms and gives $1/2=(2/\pi ){\sum }_{r\ge 0}(-1{)}^r/(2r+1)$. Hence

Set $g(x,y)=y-(x-2{)}^2$. The gradient of $g$ never vanishes, so the Lagrange multiplier equations $\nabla f=\lambda \nabla g$ apply at a constrained stationary point. They are $4x=-2\lambda (x-2)$, $2y=\lambda$, and $y=(x-2{)}^2$. Eliminating $\lambda ,y$ givesWith $t=x-2$, this becomes $t^3+t+2=(t+1)(t^2-t+2)=0$. The quadratic factor is $(t-1/2{)}^2+7/4>0$, so the only real possibility is $t=-1$. ThereforeThe level curve through this point is the ellipse $2x^2+y^2=3$, meeting the coordinate axes at $(\pm \sqrt{3/2},0)$ and $(0,\pm \sqrt{3})$. Its tangent line at $(1,1)$ has slope $-2x/y=-2$, equal to the derivative $2(x-2)$ of the constraint parabola there. The two curves are tangent.

The parabola touches this ellipse from outside; every other point on the constraint lies on a larger contour. An exact check of this geometric argument iswith equality only at $x=1$. Thus the point is the unique constrained global minimum.

Substitute $u(y,t)=Y(y)T(t)$ into the diffusion equation and divide on a region where both factors are nonzero. This gives $Y^{\prime \prime }/Y=T^{\prime }/(\nu T)$. The two sides depend on different variables, so they equal a separation constant $a$:These identities then extend through the zeros of the separated factors. The zero Dirichlet boundary conditions require $Y(0)=Y(1)=0$. For a nontrivial solution, $a$ must be negative: $a=0$ gives only a vanishing linear function and $a>0$ gives only a vanishing hyperbolic-sine solution. Writing $a=-k^2$ gives $Y=A\sin (ky)$, and $k=n\pi$, $n=1,2,\dots$. Consequently the separated modes of free viscous decay between parallel plates are $\sin (n\pi y)e^{-\nu n^2{\pi }^{2}t}$.

The first initial condition selects the lowest mode:The superposition principle for a homogeneous linear differential equation states that any linear combination of solutions is again a solution, with the same combination of homogeneous boundary conditions and initial data. It therefore givesThe higher spatial mode decays nine times as fast, so the velocity profile approaches a multiple of the first sine mode before vanishing altogether.