Past exam of the mathematical methods I course of the University of Oxford / 2016 (A10285W1, CP3)

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The nonnegative modulus $r=|z|$ must be a zero of the sine function. The restriction $r<10$ permits $r=0,\pi ,2\pi ,3\pi$, since $3\pi <10<4\pi$. Thus the locus is the origin and three concentric circles, centred at the origin with radii $\pi$, $2\pi$ and $3\pi$. A circle of radius zero is just one point.

For any value of the complex logarithm, $\log z=\ln |z|+i\arg z$ up to adding an integer multiple of $2\pi i$. Its real part therefore satisfiesThe locus is the circle centred at the origin with radius $e^2$. Different logarithm values change only the imaginary part. If a particular analytic branch of the complex logarithm with a cut is imposed, restrict the circle to that branch's domain; the modulus equation itself is branch independent.

The complex sine identity isIt follows either from the exponential definition of sine or its addition formula, using $\sin (iy)=i\sinh y$ and $\cos (iy)=\cosh y$. ConsequentlyThe angles $2$ and $3$ in these functions are dimensionless; the real trigonometric argument is measured in radians.

The scalar product $\mathbf{r}\cdot \hat{\mathbf{n}}=d$ fixes the component of the position vector along the given unit normal vector. Thus the locus is a plane normal to $\hat{\mathbf{n}}$, through $d\hat{\mathbf{n}}$. Its signed distance from the origin along that normal is $d$, and its perpendicular distance from the origin is $|d|$. Any displacement lying within the plane is orthogonal to $\hat{\mathbf{n}}$.

Noncoplanarity makes the scalar triple product $\Delta$ nonzero, so the three displayed vectors form a reciprocal basis. Their scalar products obeywith all six cross-pairings zero. Therefore, assuming the nonzero $\alpha ,\beta ,\gamma$ needed to define the three given points,Every difference of two of these points is orthogonal to $\mathbf{X}$. The two independent differences span their plane, so $\mathbf{X}$ is a normal to the plane, whose equation is $\mathbf{X}\cdot \mathbf{r}=1$. Also $\mathbf{X}\ne 0$, because its scalar product with each specified point is $1$.

The equations are an eigenvalue problem for the matrixA nonzero solution requires $\det (\lambda I-A)=0$. Expanding givesRow reduction gives the complete eigenspaces:For $\lambda =1$ the solutions form the plane $x-2y+z=0$ through the origin; for $\lambda =-3$ they form a line through the origin. Exclude the zero vector when requiring a nontrivial solution. The repeated eigenvalue $1$ has a two-dimensional eigenspace; there are three independent eigenvectors in total. This matrix is not symmetric, so its different eigenspaces need not be mutually orthogonal.

Use direction $\mathbf{u}=(-3,-2,-1)$ and displacement $\mathbf{w}=(1,2,3)-(6,5,4)=(-5,-3,-1)$. The point-to-line distance is the height of the parallelogram spanned by these vectors:Equivalently the closest line parameter is $t=(\mathbf{w}\cdot \mathbf{u})/|\mathbf{u}{|}^2=11/7$, giving the foot $(9/7,13/7,17/7)$. Its separation from the point is orthogonal to $\mathbf{u}$ and has the same length.

Measure the car's vertical displacement $y$ from its static equilibrium, so gravity's constant contribution is already balanced. Both the spring and the viscous damper act on motion relative to the wheel. The base-excited oscillator therefore satisfiesFor dimensional clarity write the road profile as $B\sin (qx)$; the printed notation sets the implicit wave number $q=1$ in the chosen length units. Constant forward speed gives forcing angular frequency $\omega =qv$. With complex amplitudes $Y,Y_0$, the steady response isSet $q=1$ for the expression in the question's notation. Assuming $M,k>0$ and $\gamma >0$, the transient response decays, leaving this steady-state response.

For a useful sketch define ${\omega }_n=\sqrt{k/M}$, $r=\omega /{\omega }_n$ and damping ratio $\zeta =\gamma /(2\sqrt{kM})$. The displacement transmissibility isIt starts at $1$, rises to a finite maximum for finite positive damping, crosses $1$ at $r=\sqrt{2}$, then tends to zero. Differentiation with respect to $r^2$ gives the maximum at $r^2=2/[1+\sqrt{1+8{\zeta }^2}]$.

Very small damping permits large resonant oscillations near ${\omega }_n$, producing discomfort and potentially large suspension travel. At exactly zero damping the assumed decay of transients fails, and resonant forcing produces secular growth. Very large damping makes the car follow the bumps over a wide range of speeds, sacrificing vibration isolation: at each fixed forcing frequency, $Y/Y_0\to 1$ as $\gamma \to \infty$. For any fixed finite damping the response still eventually decreases at sufficiently high frequency. These two limits should not be interchanged.

Split the mixed quadratic terms evenly across a real symmetric matrix:Its characteristic polynomial isso $c=2$. Choose the orthonormal eigenvectorsThe orthogonal transformation $\mathbf{r}=P\mathbf{R}$ givesIndeed $P^{T}P=I$ and $P^{T}AP=\mathrm{diag}(-1,-5,4)$; preserving $Q_2$ is precisely the length-preserving property of an orthogonal matrix.

The requested plane has normal $(2,1,-2)$. An orthonormal basis within it isWrite a point of that plane as $\mathbf{r}=u{e}_1+w{e}_2$. Restricting the quadratic form gives $Q_1=-u^2-(13/9)w^2$, so its section with $Q_1=-5$ isIt is an ellipse centred at the origin, with semi-axes $\sqrt{5}$ along $e_1$ and $\sqrt{45/13}$ along $e_2$. Normalizing the in-plane basis is essential: coefficients in nonorthonormal coordinates would not directly give physical lengths.

The complete surface is the one-sheet hyperboloidIt is connected, extends indefinitely in both directions along its axis $u_3$, and has an elliptical waist at $Z=0$ with semi-axes $\sqrt{5}$ and $1$. At larger $|Z|$ its transverse sections widen; its asymptotic cone is $4Z^2=X^2+5Y^2$. The specified plane is oblique to that axis, rather than the waist plane.

For $x\ne 0$, the integrating factor for $y^{\prime }-y/x=3$ can be taken as $1/x$ on either half-line. Thus $(y/x{)}^{\prime }=3/x$, givingThe arbitrary constant may differ on the disconnected domains $x>0$ and $x<0$.

The initial point lies in $x>1$, where division by $\ln x$ is valid. The first-order linear differential equationhas integrating factor $(\ln x{)}^2$. HenceThe condition at $x=e$ sets $C=0$, soThe solution cannot be continued regularly through $x=1$: its numerator tends to $-1$ while the denominator tends to zero. The real logarithm also requires $x>0$.

The characteristic equation of the homogeneous ordinary differential equation is $r^2-2r+5=0$, with roots $1\pm 2i$. Therefore $y_h=e^x(C\cos 2x+D\sin 2x)$. For the forcing term, try the particular solution $y_p=e^{-x}(ax+b)$. Substitution gives $8a=1$, $8b-4a=0$, so $a=1/8$, $b=1/16$. The two initial conditions give $C=-1/16$ and $D=0$. ConsequentlyBoth the value and the first derivative vanish at zero; direct differentiation verifies the inhomogeneous equation.

For the Cauchy-Euler equation $x^2{y}^{\prime \prime }-x{y}^{\prime }+y=0$, the power trial $y=x^r$ on $x>0$ gives $(r-1{)}^2=0$. The repeated root supplies the logarithmic second solution. On either half-line,One may also obtain this by setting $y=xu$, which reduces the equation to $(x{u}^{\prime }{)}^{\prime }=0$. Constants on opposite sides of the singular point need not agree.

Expanding $R\sin (x+\varphi )$ and equating coefficients gives $R\sin \varphi =A$, $R\cos \varphi =B$. ThusThe two-argument arctangent fixes the quadrant; if $R=0$, the phase is arbitrary.

For the coupled ordinary differential equations, set $z=fg$. The first equation becomes $\dot{z}=-3z$, and the initial values give $z=2e^{-3t}$. The second then reduces to $f^{\prime \prime }+3f=12e^{-3t}$. Its initial value problem has solutionWhere $f\ne 0$, reconstruct $g(t)=2e^{-3t}/f(t)$. The scalar function $f$ is smooth for every real $t$, but the original coupled system cannot have a finite $g$ at its first zero, about $t=1.5146$. The requested sketch up to $t=3$ can display the analytic continuation of $f$, with that distinction marked. Initially $f=2$, $f^{\prime }=0$ and $f^{\prime \prime }=6$; it first rises, reaches a maximum, then decreases through zero and becomes negative within the plotted interval.

For the determinant of $M$, separate its upper-left block from the constant last diagonal entry. Using $fg=2e^{-3t}$ givesThus $\det M=1$ precisely when $2\cos (\sqrt{3}t-\pi /3)=2$. The smallest positive solution isIt occurs before the zero of $f$, so $g$ and $M$ are finite there. Put $f_{\ast }=2+e^{-\pi /\sqrt{3}}$. At this time the upper block isFor a rotation matrix, each column must have unit norm, giving $a^2=f_{\ast }^2$. Both signs are allowed by the stated assumption that $a$ is a constant:Both have determinant $1$ and preserve lengths. If a positive $a$ is intended, only the first row remains; positivity is not stated in the question.

The transpose of a matrix product reverses factor order:For the requested two-dimensional check, put $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and $B=\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)$. Both sides areMore generally, the entry formula $(AB{)}_{ji}={\sum }_k{A}_{jk}{B}_{ki}$ is exactly $(B^T{A}^T{)}_{ij}$.

The conjugate transpose also reverses factor order. If $U$ and $V$ are unitary matrices,Thus their product is unitary. This is closure under multiplication in the unitary group.

For a Hermitian matrix $H$, $H^{†}=H$. Reversing the product of $n$ identical factors has no effect, soHence every positive integer power is Hermitian. The identical-factor condition matters; a product of different Hermitian matrices need not be Hermitian.

If $A$ and $B$ are Hermitian matrices, then $(AB{)}^{†}=B^{†}{A}^{†}=BA$. ThereforeThe required condition is that the matrices commute.

At $t=0$ the power series gives $\mathbf{v}(0)={\mathbf{v}}_0$. Equate the coefficient of $t^n$ after differentiating the series and substituting into the linear system of ordinary differential equations:Mathematical induction yields ${\mathbf{v}}_n=(-iH{)}^n{\mathbf{v}}_0/n!$. Summing gives the matrix exponential solution $\mathbf{v}(t)=e^{-iHt}{\mathbf{v}}_0$.

For the specified diagonal Hermitian matrix, its powers and exponential act separately on the two components. ThusThis unitary matrix preserves ${\mathbf{v}}^{†}\mathbf{v}$: the two components acquire opposite phases of unit modulus. The coefficient recurrence applies to any constant finite square matrix $H$; unitarity here additionally uses its Hermitian property.