Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1

The printed label 99E is treated as question 9.

In components, the scalar triple product isIt is the determinant with rows $\mathbf{a},\mathbf{b},\mathbf{c}$. A cyclic permutation of the rows has positive sign, so $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\mathbf{c}\cdot (\mathbf{a}\times \mathbf{b})=(\mathbf{a}\times \mathbf{b})\cdot \mathbf{c}$. Equivalently the Levi-Civita symbol expression ${ϵ}_{ijk}{a}_i{b}_j{c}_k$ is unchanged by that cyclic relabeling.

Take $P=(2,-1,2)$, $Q=(1,2,3)$ and $R=(4,1,0)$. The two plane directions are $Q-P=(-1,3,1)$ and $R-P=(2,2,-2)$. Their cross product is $(-8,0,-8)$, so a normal vector is $(1,0,1)$. The Cartesian equation of a plane is thereforeEach given point satisfies it, and the nonzero cross product confirms that the points determine a unique plane.

The volume of a parallelepiped is the absolute scalar triple product of its edge vectors. HereThe absolute value removes the orientation of the chosen edge ordering.

The comparison test for series says that $0\le v_n\le C{w}_n$ eventually and convergence of $\sum w_n$ imply convergence of $\sum v_n$. Conversely, an eventual positive lower comparison with a divergent positive series proves divergence. The ratio test uses $L={\lim }_{n\to \infty }{v}_{n+1}/v_n$: $L<1$ proves convergence, $L>1$ proves divergence, and $L=1$ is inconclusive. Removing finitely many initial zero terms does not affect these conclusions.

For $n\ge 1$, $\ln n\le n$ and $2n^3+1\ge 2n^3$. Hence $0\le \ln n/(2n^3+1)\le 1/(2n^2)$. The comparison p-series has exponent two and converges, so the series converges by the comparison test.

The summand $n/(n+4)$ tends to one rather than zero. The term test for divergence therefore proves the series diverges. A ratio limit equal to one would be inconclusive and is unnecessary here.

The elementary bound $|\sin t|\le |t|$ gives $0\le {\sin }^2(1/n)\le 1/n^2$. Comparison with a convergent p-series proves the series converges.

For $n\ge 1$, the ratio of positive terms isThe ratio test proves the series converges. Its zero term at $n=0$ has no effect.

For an oriented piecewise smooth surface $S$ with boundary $C=\partial S$ and a vector field $\mathbf{A}$ with continuous first derivatives near $S$, the Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal vector, $dS$ the positive area element, and $d\mathbf{r}$ the oriented tangent line element. The traversal of $C$ follows the right-hand rule relative to the surface orientation.

Differentiate the Cartesian components to calculate the curl:

Orient the three faces outward from the tetrahedron. The two coordinate-plane faces have normals parallel to $-{\mathbf{e}}_x$ or $-{\mathbf{e}}_y$, so their flux of the curl is zero. On the inclined face $z=1-x-y$, its upward vector area is $(1,1,1)dxdy$, with $x,y\ge 0$, $x+y\le 1$. Thus the direct surface integral isThis surface induces the counterclockwise boundary $O\to X\to Y\to O$ viewed from above, where $X=(1,0,0)$ and $Y=(0,1,0)$. At $z=0$, $\mathbf{A}=(2x-y,0,0)$. On $OX$, take $(x,y)=(t,0)$, giving integral ${\int }_0^{1}2tdt=1$. On $XY$, take $(x,y)=(1-t,t)$, giving ${\int }_0^1(-2+3t)dt=-1/2$. On $YO$, $dx=0$ and the integral vanishes. ThereforeBoth sides have been calculated directly, verifying the Stokes theorem with consistent orientations.

The partial fraction decomposition is $(x-1)/[x(x+1)]=-1/x+2/(x+1)$. Since both endpoints are positive, integration gives

Use ${\tan }^{2}x={\sec }^{2}x-1$ and the derivative of the tangent:The antiderivative applies on each interval avoiding the poles of $\tan x$.

For $1<x<e$, the natural logarithm satisfies $0<\ln x<1$. Squaring therefore decreases its value, while $\ln (x^2)=2\ln x$. Pointwise comparison gives $(\ln x{)}^2<\ln x<\ln (x^2)$, soAs a check, integration by parts gives $I_1=1$, $I_2=2$ and $I_3=e-2$, consistent with this order.

The cumulative distribution function must tend to one at infinity, so $\boxed{k=1}$. Differentiation gives the probability density function $f_X(x)=e^{-x}$ for $x>0$ and zero for $x<0$; there is no atom at zero. This is an exponential distribution of rate one. Its first two moments are ${\int }_0^{\infty }x{e}^{-x}dx=1$ and ${\int }_0^{\infty }{x}^2{e}^{-x}dx=2$, so

The continuous nonnegative part has total mass ${\lim }_{y\to \infty }(1-e^{-y})/2=1/2$. The remaining mass is at $-1$, giving $\boxed{m=1/2}$. This is a mixed probability distribution, specifically a mixture of a point mass and an exponential distribution; its atom must be included in normalization.

Include the negative atom as well as the continuous part in the expected value:

The second moment of the mixed probability distribution is $\mathbb{E}[Y^2]=\frac{1}{2}(-1{)}^2+\frac{1}{2}{\int }_0^{\infty }{y}^2{e}^{-y}dy=1/2+1=3/2$. Since the mean is zero,

The conditioning event selects the exponential component: $P(Y\ge 0)=1/2$ and $P(Y\ge 1)=e^{-1}/2$. Thus conditional probability givesThe atom at $-1$ is excluded by the conditioning event.

The determinant is $k-6$, so for $k\ne 6$ the inverse matrix isThe specified equations use $k=4$, with right-hand side $(2,3,1{)}^T$. Multiplying by $A^{-1}{|}_{k=4}$ givesIn particular the parameter in the inverse formula must be set to four, not left arbitrary for this numerical system.

For $k=1$, the required right-hand matrix has twice the first row of $A$, the same second row, and twice the third row. Left multiplication scales rows, soEquivalently, right multiply the given equation $BA=C$ by $A^{-1}$ to obtain the same answer.

At $k=6$, the second row is twice the first. The independent equations are $x+3y=0$ and $4x+z=0$. Therefore the null space isIt has dimension one, consistent with matrix rank two.

The trace is unchanged by transposition and is cyclic under a product. With $S^T=S$ and $A^T=-A$,Thus $\boxed{\mathrm{tr}(AS)=0}$. In components this is the vanishing contraction of a symmetric tensor with an antisymmetric tensor.

The Levi-Civita symbol vanishes when two indices coincide, so

The antisymmetric contribution is the negative of twice the cross-product matrix for $\mathbf{a}$:The cross-product matrix spectrum is $0,\pm i|\mathbf{a}|$, so $M$ has eigenvalues $1,1\pm 2i|\mathbf{a}|$. Multiplication givesDirect expansion of the displayed matrix yields the same real positive determinant.

Use index contraction:The last term vanishes because $a_k{a}_{\ell }$ is symmetric in the contracted indices while the Levi-Civita symbol is antisymmetric. Hence $\boxed{MN=N}$.

Taking determinants in $MN=N$ gives $(\det M-1)\det N=0$. Here $\det M-1=4|\mathbf{a}{|}^2>0$ because $\mathbf{a}\ne 0$, so $\boxed{\det N=0}$. This also agrees with $N=\mathbf{a}{\mathbf{b}}^T$ being an outer product of rank one.

On the straight path $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$, both fields restrict to $(3t^5,2t^5,t^5)$. Dotting with ${\mathbf{r}}^{\prime }=(1,1,1)$ gives

For $\mathbf{r}(t)=(t,t^2,t^2)$, the velocity is $(1,2t,2t)$. Because $y=z$ throughout this path, both fields again agree pointwise, now as $(3t^8,2t^7,t^7)$. ThusThese equal values alone do not establish path independence for either field on all of space.

Direct differentiation gives $\boxed{\Phi =x^3{y}^{2}z+C}$ with $\nabla \Phi =\mathbf{G}$, so $\mathbf{G}$ is a conservative vector field and its integral depends only on the endpoints. In contrast,is nonzero, for example at $(1,1,1)$. Thus $\mathbf{F}$ is not conservative. Both earlier paths lie in the surface $y=z$, where the two fields happen to coincide. This illustrates why coinciding path integrals need not imply a conservative field.

By Euler's formula, $\boxed{e^{2i}=\cos 2+i\sin 2}$, with the angle measured in radians.

Since the arguments of $-1$ are $(2k+1)\pi$, the complex logarithm has values $\boxed{\log (-1)=i(2k+1)\pi }$ for $k\in \mathbb{Z}$. The principal value is $\boxed{i\pi }$.

The modulus of $i$ is one and its arguments are $\pi /2+2k\pi$. Thus $\boxed{\log i=i(\pi /2+2k\pi )}$, with principal value $\boxed{i\pi /2}$.

For the positive real base ten, use its real natural logarithm. Then Euler's formula gives

Using a value of the complex logarithm, $i^i=\exp (i\log i)=\exp [-(\pi /2+2k\pi )]$. Thus every logarithmic-branch value is real and positive:The principal value is $\boxed{e^{-\pi /2}}$.

The hyperbolic cosine and hyperbolic sine are defined by $\boxed{\cosh x=(e^x+e^{-x})/2,\sinh x=(e^x-e^{-x})/2}$. Their quotient is $\tanh x=\sinh x/\cosh x$.

Expand the two squares using the exponential definitions:

The exponential definitions show that $\sinh$ is odd and $\cosh$ is even. Their quotient therefore satisfies $\boxed{\tanh (-x)=-\tanh x}$.

Differentiating the exponential definition gives $\boxed{(\cosh x{)}^{\prime }=(e^x-e^{-x})/2=\sinh x}$.

The fifth power of the odd function $\tanh x$ is odd and continuous. Its integral over the symmetric interval is therefore $\boxed{{\int }_{-1000}^{1000}{\tanh }^{5}xdx=0}$.

The derivative of $\ln \cosh x$ is $\tanh x$. Since $\cosh (\ln 2)=(2+1/2)/2=5/4$,

Take the real part of De Moivre's theorem. Put $c=\cos \theta$ and use ${\sin }^2\theta =1-c^2$:

The modulus equation says that $z$ has fixed distance eight from $-1-i$. Thus its locus is a circle of radius eight centered at $(-1,-1)$, with Cartesian equation $(x+1{)}^2+(y+1{)}^2=64$.

Writing $u=X+iY$ gives $X=2x$, $Y=y$. Substitute $x=X/2$, $y=Y$ into the original circle equation. This affine image of a circle is the ellipsecentered at $(-2,-1)$ with horizontal semiaxis sixteen and vertical semiaxis eight. The map covers the whole ellipse as $z$ covers the circle.