MATH2010 Complementary Mock B — worked solutions

Complete worked solutions and marking for the original, non-official 100-mark practice paper. Suggested timing and scope are practice choices, not examination rules.

Revision route · Question paper

Q1. Worked solution

Question

Why this method: The curve records a motion, not only a geometric image. Differentiate to obtain velocity, take its norm to obtain speed, and integrate the speed to count travelled length. Endpoint subtraction instead measures straight-line displacement.

(a) Componentwise differentiation gives

$h'(t)=(-2\sin t,2\cos t,1),\qquad \|h'(t)\|=\sqrt{4\sin^2t+4\cos^2t+1}=\sqrt5.$

At $t=\pi/2$, the point is $(0,2,\pi/2)$ and the nonzero velocity is $(-2,0,1)$. Thus the tangent line is

$\ell(\lambda)=(0,2,\pi/2)+\lambda(-2,0,1),\qquad\lambda\in\mathbb R.$

A tangent line needs both its point and its direction; writing only $(-2,0,1)$ is not a complete line.

Marking [8]: derivative 3; speed 2; point 1; complete line 2.

Paired paper

(b) Integrate speed:

$L=\int_0^\pi\sqrt5\,dt=\pi\sqrt5.$

The endpoints are $(2,0,0)$ and $(-2,0,\pi)$, so

$D=\|(-4,0,\pi)\|=\sqrt{16+\pi^2}.$

Both numbers are positive. Compare their squares:

$L^2-D^2=5\pi^2-(16+\pi^2)=4(\pi^2-4)>0,$

because $\pi>2$. Hence $L>D$. Comparing squares proves this strictly, rather than relying on how long the curve looks in a sketch. Marking [8]: length setup/result 3; displacement/distance 3; strict comparison 2.

Paired paper

(c) Distance from the initial point along the curve is

$s(t)=\int_0^t\sqrt5\,d\tau=\sqrt5\,t.$

Therefore $t=s/\sqrt5$ and

$H(s)=\left(2\cos\frac{s}{\sqrt5},2\sin\frac{s}{\sqrt5},\frac{s}{\sqrt5}\right),\qquad0\le s\le\pi\sqrt5.$

Direct differentiation gives

$H'(s)=\left(-\frac2{\sqrt5}\sin\frac{s}{\sqrt5},\frac2{\sqrt5}\cos\frac{s}{\sqrt5},\frac1{\sqrt5}\right),\qquad\|H'(s)\|^2=\frac45+\frac15=1.$

Each unit increase of the new parameter travels one unit of length along the curve. Merely renaming the parameter $s$ does not make the speed equal to one.

Marking [8]: $s(t)$ and inverse 3; formula and domain 3; direct unit-speed check 2.

Paired paper

(d) Chain rule supplies the minus sign:

$q'(u)=-h'(\pi-u)=(2\sin(\pi-u),-2\cos(\pi-u),-1).$

Thus $\|q'(u)\|=\sqrt5$ and $L_q=\int_0^\pi\sqrt5\,du=\pi\sqrt5$. Velocity reverses relative to the original velocity at the corresponding point, but its norm is nonnegative. In general the speed factor is $|dt/du|$, not $dt/du$. The minus sign changes the direction of travel, not the sign of the length. Marking [6]: velocity 2; speed and length 2; absolute-value explanation 2.

Paired paper

Common errors: integrating $h'$ and calling the result length; forgetting the tangent point; keeping the old parameter interval after switching to arc length; using a negative “speed” after reversal.

Q2. Worked solution

Question

Why this method: A theorem with sufficient hypotheses gives a safe way to conclude something, but failure of a hypothesis does not prove the conclusion false. Here we can directly compute both lengths and inspect the straight-line image.

(a) Since $u\mapsto u^3$ maps $[-1,1]$ bijectively onto $[-1,1]$, both images are the segment $\{(x,0):-1\le x\le1\}$.

$a'(u)=(3u^2,0),\quad\|a'(u)\|=3u^2;\qquad b'(t)=(1,0),\quad\|b'(t)\|=1.$

At zero, $a'(0)=(0,0)$ while $b'(0)=(1,0)$. The same image need not describe the same motion or speed. Marking [6]: common image with reason 2; derivatives/speeds 3; zero comparison 1.

Paired paper

(b) Both curves are continuously differentiable, so their lengths may be computed directly:

$L_a=\int_{-1}^{1}3u^2\,du=[u^3]_{-1}^{1}=2,\qquad L_b=\int_{-1}^{1}1\,dt=2.$

The image is a straight segment, so its tangent line at the interior point $(0,0)$ is the $x$-axis. Every secant line through $(0,0)$ and $a(u)$ for $u\ne0$ is already the $x$-axis. But $(0,0)+\lambda a'(0)$ is only the single point $(0,0)$, not a line. The regular tangent formula requires a nonzero derivative; at this parameter the formula is inconclusive, not a proof that the image has no tangent. Marking [6]: two integrals 3; tangent and secant reason 2; formula limitation 1.

Paired paper

(c) The substitution is continuously differentiable and strictly increasing, but

$\phi'(u)=3u^2,\qquad\phi'(0)=0.$

It fails the stated nonzero-derivative hypothesis. Its inverse is $\psi(t)=\sqrt[3]{t}$. At zero its difference quotient is

$\frac{\psi(t)-\psi(0)}{t-0}=\frac{\sqrt[3]{t}}{t}=\frac1{|t|^{2/3}},\qquad t\ne0,$

which tends to $+\infty$, not a finite derivative. Hence the inverse is not differentiable at zero in the usual real-valued sense. The stated regular-reparametrization theorem therefore cannot be invoked as written. Nevertheless, the direct integrals in (b) prove equality for this example. “This sufficient theorem does not apply” and “lengths are unequal” are different claims. Marking [8]: failed hypothesis 2; inverse/quotient/no finite derivative 3; logically correct reconciliation 3.

Paired paper

Common errors: treating increasing as equivalent to everywhere positive derivative; calling infinity an ordinary derivative; concluding “no tangent” from zero velocity; concluding the result false simply because one sufficient theorem cannot be applied.

Q3. Worked solution

Question

Why this method: Cylindrical radius measures distance to the $z$-axis, whereas spherical radius measures distance to the origin. Work with signs and the chosen polar-angle convention before squaring.

(a) The planar radius is

$r=\sqrt{(-1)^2+(\sqrt3)^2}=2.$

The horizontal projection lies in quadrant II, so $\theta=2\pi/3$. Therefore cylindrical coordinates are

$(r,\theta,z)=(2,2\pi/3,-2).$

The spherical radius is $\rho=\sqrt{1+3+4}=2\sqrt2$. Since $\cos\phi=z/\rho=-1/\sqrt2$ and $0\le\phi\le\pi$, we obtain $\phi=3\pi/4$. Thus

$(\rho,\theta,\phi)=(2\sqrt2,2\pi/3,3\pi/4).$

Check:

$\rho\sin\phi=2,\quad x=2\cos(2\pi/3)=-1,\quad y=2\sin(2\pi/3)=\sqrt3,\quad z=2\sqrt2\cos(3\pi/4)=-2.$

Using only $\arctan(y/x)$ can select the wrong quadrant. Measuring $\phi$ from the $xy$-plane instead of the positive vertical axis would also use the wrong convention.

Marking [10]: cylindrical radius/azimuth/height 3; spherical radius/polar angle 3; sign/quadrant reasoning 2; substitution 2.

Paired paper

(b) The radial bounds give $1\le\rho\le3$, so $\rho>0$. Also

$z=\rho\cos\phi,\qquad\sqrt{x^2+y^2}=\rho\sin\phi,$

where $\sin\phi\ge0$ on $[0,\pi]$. Hence the cone condition becomes $\cos\phi\ge\sin\phi$. In particular $\cos\phi\ge0$, so $\phi\in[0,\pi/2]$. On this interval the inequality is equivalent to $\phi\le\pi/4$. Therefore

$1\le\rho\le3,\qquad0\le\phi\le\pi/4,\qquad0\le\theta<2\pi.$

This is the portion of the spherical shell in the upper cone, including all its stated boundaries. At $\phi=0$ the azimuth can be any representative; different azimuths there describe the same axis point. Repeated coordinate descriptions do not add geometric points to the set. Marking [10]: radial restriction 2; substitution 2; signed angular derivation 4; complete ranges/boundary or axis explanation 2.

Paired paper

(c) In cylindrical coordinates the same set is

$r\ge0,\qquad0\le\theta<2\pi,\qquad1\le r^2+z^2\le9,\qquad z\ge r.$

The radial condition is squared legitimately because $\sqrt{r^2+z^2}\ge0$. By contrast, replacing $z\ge r$ with $z^2\ge r^2$ would also permit $z\le-r$. For example $(r,\theta,z)=(0,0,-2)$ satisfies the squared restrictions but corresponds to $(0,0,-2)\notin E$. Squaring has discarded the upper-cone sign condition. Marking [6]: complete cylindrical description 4; valid unwanted-point counterexample 2.

Paired paper

(d) $x=r\cos(\pi/2)=0$ and $y=r\sin(\pi/2)=r\ge0$, while $z$ is free. The set is the closed half-plane $\{(0,y,z):y\ge0\}$, including the $z$-axis. It excludes points with $y<0$, so it is not the whole plane. Marking [4]: coordinate substitution 2; half-plane/boundary and exclusion 2.

Paired paper

Common errors: swapping $r$ and $\rho$; wrong quadrant; losing the upper-cone restriction after squaring; deleting axis points because azimuth is not unique; changing a fixed azimuth half-plane into a full plane.

Q4. Worked solution

Question

Why this method: Equality conditions are part of a theorem, not optional decoration. Squaring hides the sign of an inner product; triangle equality requires the positive sign.

(a) For $u\ne0$, complete the square:

$\|v-tu\|^2=\|u\|^2\left(t-\frac{\langle u,v\rangle}{\|u\|^2}\right)^2+\|v\|^2-\frac{\langle u,v\rangle^2}{\|u\|^2}.$

At $t_*=\langle u,v\rangle/\|u\|^2$, the minimum is a squared norm and hence nonnegative. Therefore

$\langle u,v\rangle^2\le\|u\|^2\|v\|^2,\qquad|\langle u,v\rangle|\le\|u\|\|v\|.$

Equality holds exactly when the minimizing residual is zero: $v=t_*u$. The scalar may be positive, negative or zero. If $u=0$ (or $v=0$), both sides of Cauchy–Schwarz are zero, so equality holds directly. Equivalently, equality holds exactly when the pair is linearly dependent, including zero-vector cases. Marking [8]: square expansion/minimizer 3; nonnegative minimum and inequality 2; residual equality 2; zero cases 1.

Paired paper

(b) Here $\|u\|=3$, $\|v\|=6$, $\langle u,v\rangle=-2\|u\|^2=-18$, and $u+v=-u$, so $\|u+v\|=3$.

Cauchy–Schwarz: $|-18|=18=3\cdot6$, equality. Triangle inequality: $3<3+6=9$, strict. Oppositely directed vectors can attain equality in the absolute-value Cauchy–Schwarz inequality without making the length of their sum equal to the sum of their lengths. Marking [6]: four quantities 4; two explicit comparisons 2.

Paired paper

(c) Correct statement: if either vector is zero, triangle equality holds; if both are nonzero, equality holds exactly when one is a positive scalar multiple of the other.

Indeed,

$\|u+v\|^2=\|u\|^2+2\langle u,v\rangle+\|v\|^2.$

Comparing with $(\|u\|+\|v\|)^2$ shows equality is equivalent to $\langle u,v\rangle=\|u\|\|v\|$. For nonzero vectors this requires Cauchy–Schwarz equality and a positive inner product. Write $v=cu$; then $c\|u\|^2=|c|\|u\|^2$, and since $c\ne0$, this means $c>0$. Conversely, if $v=cu$ with $c>0$, then $\|u+v\|=(1+c)\|u\|=\|u\|+\|v\|$. Zero vectors satisfy equality by direct substitution. Marking [6]: precise corrected condition 2; necessity 2; sufficiency and zero cases 2.

Paired paper

Common errors: dropping the absolute value in Cauchy–Schwarz; saying all dependence gives triangle equality; dividing by the norm of a zero vector; proving necessity without proving sufficiency.

---

How to use your result

• Q1 mistakes: revisit velocity versus speed, arc-length integration and the chain rule for changes of parameter.
• Q2 mistakes: distinguish regularity of a parametrization from geometry of its image, and sufficient conditions from necessary conditions.
• Q3 mistakes: write the full coordinate convention first, then keep signs and range restrictions.
• Q4 mistakes: redo both equality cases with positive, negative and zero scalar multiples.

Award method marks even after a minor arithmetic slip if the later reasoning is coherent; do not award the final exact-answer point for an incorrect value. Do not penalize the same arithmetic slip repeatedly. Equivalent parametrizations and correct alternative proofs earn full credit.

Scope and relationship to Mock A

Mock A's six questions cover plane/projection/volume; annular-set topology with one polar conversion; limits; partial derivatives; differentiability; and gradient/directional derivatives. Mock B does not replace it. Q1–Q2 directly add arc length and reparametrization; Q3 adds cylindrical/spherical conversion and signed cone restrictions. Q4 intentionally overlaps the Cauchy–Schwarz idea used in A-Q6, but tests its proof and exact equality conditions rather than repeating the gradient-maximization calculation. Neither paper alone nor the pair is asserted to cover every course topic.

Topic basis: existing course notes 1.1 Vectors, inner products and distance and 1.3 Coordinates, curves and surfaces, within the known Weeks1–3 material. These are learning references, not official examination-scope certification. No original school question or lecture image is reproduced.