Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 1

Write $\mathbf{r}=\mu \mathbf{n}+\mathbf{v}$. Since $\mathbf{n}$ is a unit vector, the equation becomes $\mathbf{v}\cdot \mathbf{n}=0$. Thus $\mathbf{v}$ ranges over the two-dimensional orthogonal complement of $\mathbf{n}$: the locus is an affine hyperplane through $\mu \mathbf{n}$, with unit normal $\mathbf{n}$. Conversely, every such vector satisfies the equation. Its perpendicular distance from the origin is $\boxed{|\mu |}$; $\mu$ is the signed distance in the chosen normal direction.

The normal vector must be parallel to $\mathbf{b}$, and the plane must contain $\mathbf{a}$. Therefore one choice isThe nonzero scalar triple product ensures $\mathbf{b}\ne 0$. Replacing both $\mathbf{n}$ and $\mu$ by their negatives describes the same affine hyperplane.

Two directions in the plane are $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$. Their cross product $\mathbf{N}=(\mathbf{b}-\mathbf{a})\times (\mathbf{c}-\mathbf{a})$ is a normal vector. Moreover,so $\mathbf{N}$ cannot vanish. The required unit normal and offset areTaking dot products with $\mathbf{a},\mathbf{b},\mathbf{c}$ gives the same offset for all three points.

This plane contains the direction $\mathbf{b}-\mathbf{a}$ and is parallel to $\mathbf{c}$. Put $\mathbf{N}=(\mathbf{b}-\mathbf{a})\times \mathbf{c}$. Its scalar triple product with $\mathbf{a}$ is again $D\ne 0$, proving $\mathbf{N}\ne 0$. HenceThe dot product with $\mathbf{c}$ vanishes, while the dot products with $\mathbf{a}$ and $\mathbf{b}$ equal $\mu$ after normalization. These verify both incidence and parallelism.

The periodically extended $x^2$ is even, so every sine coefficient is zero. With the convention $f=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$, the Fourier coefficients areFor the second integral, two applications of integration by parts give ${\int }_0^{\pi }{x}^2\cos nxdx=2\pi (-1{)}^n/n^2$. ConsequentlyThe periodic extension is continuous even at the joined endpoints, where both one-sided values equal ${\pi }^2$. The pointwise convergence of a piecewise smooth Fourier series therefore gives the displayed values throughout the interval; the coefficient bound also gives uniform convergence by the Weierstrass M-test. Setting $x=\pi$ and then $x=0$ yields

At an interior local extremum of a differentiable function, $f^{\prime }(x)=0$. The second-derivative test classifies a stationary point: a positive second derivative gives a strict local minimum, and a negative second derivative gives a strict local maximum. A zero second derivative is inconclusive: $x^4$, $-x^4$, and $x^3$ at zero illustrate a minimum, a maximum, and neither.

HereThus $f^{\prime \prime }(1/3)=-8$ and $f^{\prime \prime }(3)=8$. Substitution into $f$ gives

Use implicit differentiation on $F(x,y)=x{y}^3-y{x}^3-6=0$. Along the curve,At $(1,2)$, $F_x=2$ and $F_y=11\ne 0$, so the slope is $-2/11$. The tangent line is

For $F(x,y)=e^{xy}+y\ln x-x^2$, the real domain has $x>0$. Its partial derivatives are $F_x=y{e}^{xy}+y/x-2x$ and $F_y=x{e}^{xy}+\ln x$. At $(1,0)$ they equal $-2$ and $1$, respectively. Thus implicit differentiation gives $y^{\prime }=2$, and the tangent line is

For two independent smooth constraints $g=h=0$, the tangent direction is orthogonal to both gradients. At a constrained stationary point, the gradient of the objective is also orthogonal to that direction, hence lies in their span: $\nabla f=\lambda \nabla g+\mu \nabla h$. This is the two-constraint Lagrange multiplier condition. Alternatively, distance extrema on a central plane section of an ellipsoid follow from a two-dimensional eigenvalue problem.

Minimizing distance is equivalent to minimizing $f=x^2+y^2+z^2$. Put $g=x-y+z$ and $h=x^2+(y^2+z^2)/4-1$. The Lagrange multiplier equations areAdding the latter two gives $(2-\mu /2)(y+z)=0$. If $\mu =4$, those equations give $\lambda =x=0$; the constraints then give $y=z=\pm \sqrt{2}$ and $f=4$. Otherwise $z=-y$ and $x=2y$. The ellipsoidal constraint gives $y=\pm \sqrt{2}/3$, and henceThe intersection is compact. The two constraint gradients are independent there: if $\nabla h=t\nabla g$, then $(x,y,z)=(t/2,-2t,2t)$, and $g=0$ forces $t=0$, contradicting $h=0$. Thus every global extremum is among these candidates. Taking square roots gives

Let $I={\int }_0^{\infty }{e}^{-x^2}dx$. Positivity allows the squared integral to be written as a first-quadrant double integral. In polar coordinates the Jacobian determinant is $r$, soUsing integration by parts, with the boundary term $x{e}^{-x^2}$ vanishing at both endpoints,The substitution $u=x^2$ also gives the two antiderivativesThese values also follow from the Gaussian moment recurrence and can be checked by differentiation.

The upper boundary follows $y=x$ until its intersection with $y=1/x$ at $(1,1)$, then follows the hyperbola to $(2,1/2)$. The lower boundary is $y=0$. Therefore the two vertical-slice descriptions are $0\le x\le 1$, $0\le y\le x$, and $1\le x\le 2$, $0\le y\le 1/x$.

Integrating with respect to $y$ first reduces the double integral toUsing the antiderivatives from part (a),

Direct matrix multiplication givesConsequently $\boxed{A^2-6A+5I=0}$, the Cayley-Hamilton theorem for its characteristic polynomial $t^2-6t+5$. Rearranging gives $A(6I-A)=5I$, so

For $\mathbf{u}=(1,1{)}^T$ and $\mathbf{v}=(1,-1{)}^T$, the outer products areTheir inner product is zero and both squared norms are two. Thus $P_u=\mathbf{u}{\mathbf{u}}^T/2$ and $P_v=\mathbf{v}{\mathbf{v}}^T/2$ are orthogonal spectral projectors, with $P_u+P_v=I$. Since $A\mathbf{u}=5\mathbf{u}$ and $A\mathbf{v}=\mathbf{v}$, the eigenvalues are $\lambda =5$, $\mu =1$, andMultiplying these decompositions uses $P_u{P}_v=0$ and $P_u^2=P_u$, $P_v^2=P_v$, giving the identity matrix.

On an interval avoiding $x=\pm 1$, separation of variables gives, for $y\ne -1$,Absorb the sign into the integration constant and include the equilibrium solution $y=-1$ by allowing $C=0$. The complete real family isEach solution is considered on a connected interval in $\mathbb{R}\setminus \{-1,1\}$, where the original equation is defined.

Divide by $\cos x$ on an interval where it is nonzero. For the resulting first-order linear differential equation, use the integrating factor $M=\sec x+\tan x$, since $M^{\prime }=\sec xM$. The right-hand side after division is also $M^{\prime }$, soThe identity $(\sec x+\tan x)(\sec x-\tan x)=1$ shows $M$ never vanishes on the interval, givingAn integrating factor may have either fixed sign; if obtained by integrating $\sec x$, the real logarithm is $\ln |\sec x+\tan x|$.

The Bernoulli differential equation becomes linear after $z=1/y$: multiplying by $-y^{-2}$ gives $z^{\prime }-2z=-x^2$. A quadratic particular solution is $z_p=x^2/2+x/2+1/4$, while the homogeneous solution is $C{e}^{2x}$. The initial condition gives $z(0)=1$ and $C=3/4$. HenceThe denominator is positive for all real $x$, since $2x^2+2x+1=2(x+1/2{)}^2+1/2$. Thus this initial value problem has the displayed solution on the entire real line. The separate solution $y=0$ does not satisfy its initial condition.

Let $N=\mathbf{a}{\mathbf{b}}^T$. The orthogonality condition gives $N^2=\mathbf{a}({\mathbf{b}}^T\mathbf{a}){\mathbf{b}}^T=0$: this is the nilpotent part of a transvection when both vectors are nonzero. In two dimensions,Expanding its determinant, the fourth-degree terms cancel:These formulas also cover either vector being zero.

Since $M^T=I-\mathbf{a}{\mathbf{b}}^T=I-N$, nilpotence already gives $(I+N)(I-N)=I$. Explicitly, in index notation,Thus $\boxed{L{M}^T=I}$, or equivalently $M^T=L^{-1}$. Here ${\delta }_{ij}$ is the Kronecker delta.

For the specified coordinate vectors,Choose the orthogonal matrix representing a quarter-turn,Then $S^{T}S=I$ and direct matrix multiplication gives $S^{T}LS=\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)=M$. Thus these two shear mappings are orthogonally similar.

In Cartesian components, $\mathbf{F}=(z-y,x,-x)$, soIt cannot be a conservative vector field. For the second field, put $\Phi (\mathbf{r})=-(1+|\mathbf{r}{|}^2{)}^{-1/2}$. Its gradient is $\nabla \Phi =\mathbf{r}/(1+|\mathbf{r}{|}^2{)}^{3/2}=\mathbf{G}$. This smooth global scalar function proves $\mathbf{G}$ is conservative, with $\boxed{\nabla \times \mathbf{G}=0}$. The following path integrals exhibit the different behavior directly.

Parametrize the segment by $\mathbf{r}(t)=(1-t,t,0)$, $0\le t\le 1$. Its tangent is $(-1,1,0)$ and $\mathbf{F}=(-t,1-t,t-1)$, so $\mathbf{F}\cdot {\mathbf{r}}^{\prime }=1$. Hence its line integral isFor the other line integral, $|\mathbf{r}{|}^2=(1-t{)}^2+t^2$ and $\mathbf{r}\cdot {\mathbf{r}}^{\prime }=2t-1$. ThusThis also equals the change in $\Phi$, by the fundamental theorem for line integrals.

On each of the two radial legs, $\mathbf{F}=\mathbf{a}\times \mathbf{r}$ is perpendicular to the tangent. Each contribution is zero, giving $\boxed{\int \mathbf{F}\cdot d\mathbf{r}=0}$. This differs from the straight-path answer, confirming path dependence.

For $\mathbf{G}$, use $\mathbf{r}=(s,0,0)$ with $s$ decreasing from one to zero, followed by $\mathbf{r}=(0,s,0)$ with $s$ increasing from zero to one. The two contributions areThey cancel: $\boxed{\int \mathbf{G}\cdot d\mathbf{r}=0}$. This agreement is a consequence of the conservative vector field property already proved, rather than evidence sufficient by itself to establish it.

Put $\tau =t+b$ and $q=x+a$. For $\theta =f(t)e^{-q^2/(4\tau )}$, partial differentiation givesThe heat equation therefore requires $f^{\prime }=-f/(2\tau )$, giving $f=C/\sqrt{|t+b|}$ on an interval not containing $-b$. A decaying spatial Gaussian function uses $t+b>0$, so $f=C/\sqrt{t+b}$ there.

The initial profile is a difference of two translated Gaussian functions. Set $b=1/4$, $a=-2$ and $a=2$, with respective constants $C=1/2$ and $C=-1/2$. The superposition principle givesThis is an odd pair of Gaussian profiles under diffusion, obtained from Gaussian initial data for the heat equation. It is odd in $x$, vanishes at zero, is positive for $x>0$ and negative for $x<0$. The two lobes broaden and decrease in height, and $\theta (x,t)\to 0$ at every fixed $x$ as $t\to \infty$. Their extrema move; they do not remain exactly at $\pm 2$.

Complex conjugation reverses the argument, reciprocation also inverts the modulus, and multiplication by $-1$ adds $\pi$ to the argument. With all arguments chosen in $[0,2\pi )$ and $0<\varphi <\pi /2$,Thus $z$ is in the first quadrant, $z^{\ast }$ and $z^{-1}$ lie on the same fourth-quadrant ray at radii two and one-half, and $-z$ is diametrically opposite $z$.

Euler's formula and its version with $-\theta$ giveSquaring $e^{i\theta }=\cos \theta +i\sin \theta$ and comparing real and imaginary parts with $e^{2i\theta }$ yields the double-angle formulas

Factor $e^{i\varphi /2}$ from the numerator and denominator:For $0<\varphi <\pi$, the tangent ranges over all positive real values. Thus $\boxed{\mathrm{Re}w=0,\mathrm{Im}w>0}$. Using the principal complex logarithm,The real part ranges over the entire real line, while the imaginary part is fixed: the locus is the horizontal line $\mathrm{Im}\log w=\pi /2$. Any other fixed logarithm branch gives the parallel line $\mathrm{Im}\log w=\pi /2+2\pi k$ for one integer $k$; allowing every branch gives the union of those lines.

For $k>0$, the two ordinary differential equations have solutionsTheir products $h(x)g(y)$ and $g(x)h(y)$ solve Laplace's equation, since the two second derivatives contribute opposite $k^2$ terms. For $k=0$, both factors are linear functions. This is separation of variables with positive, negative and zero separation constants.

The zero value at $x=0$ removes the even factor in $x$, and the zero value at $y=0$ removes the even factor in $y$. Up to multiplicative constants, the nonzero separated modes areThe zero-separation case also supplies $\boxed{xy}$. Each is a harmonic function and vanishes on both axes.

For the first family, vanishing at $y=\pi$ requires $\sin (k\pi )=0$, so $k=n$ is a positive integer. The second family would require $\sinh (k\pi )=0$, impossible for real $k>0$; the mode $xy$ also fails unless its coefficient is zero. Conveniently normalized surviving modes areAt $x=\pi$, these become the Fourier sine series basis functions $\sin (ny)$.

Expand the constant right-edge value in a Fourier sine series. Its coefficients areThe superposition principle combines the normalized modes, givingThis is a harmonic function in a rectangle with one constant boundary value, with the coordinate roles interchanged. The series and its derivatives converge uniformly on compact subsets of $0<x<\pi$, so it satisfies Laplace's equation there. It vanishes on the other three edges and approaches one at every interior point of the right edge. At the two right corners the one-sided prescribed values disagree; no continuous extension can satisfy both. The stated open-edge condition avoids that incompatibility, and isolated corner assignments do not change the bounded interior Dirichlet problem.