MATH2010 Mock A — solutions and marking

Award marks for justified mathematical steps. Equivalent valid methods receive the same credit. Within a part, keep credit for correct later reasoning when an earlier arithmetic slip has been carried through consistently.

Question paper · Revision route

Q1. A plane, a projection and a volume — 16 marks

The two given directions are vectors in three-dimensional space. A point on the plane is obtained by starting at $P$ and adding multiples of those directions. To recognize the plane with one equation, we need a vector perpendicular to both directions. That same normal then supplies the perpendicular line needed for the distance problem.

(a) [6] A cross product gives

$$u\times v=(2\cdot1-0\cdot(-1),\ 0\cdot2-1\cdot1,\ 1\cdot(-1)-2\cdot2).$$ $$n=u\times v=(2,-1,-5),\qquad \|n\|^2=30.$$

As checks, $n\cdot u=2-2+0=0$ and $n\cdot v=4+1-5=0$. It is nonzero, so the two directions are independent. For a point $X=(x,y,z)$, the displacement from $P$ to $X$ lies along the plane precisely when it is perpendicular to this normal. The plane equation is $n\cdot((x,y,z)-P)=0$, hence

$$2(x-1)-(y+1)-5(z-2)=0, \qquad 2x-y-5z+7=0.$$

Marking: cross product 3; point-normal equation 2; consistent simplified equation 1. The opposite normal gives the same plane.

(b) [6] Write the plane's scalar left-hand side as $N(x,y,z)=2x-y-5z+7$. The normal $n$ is a vector, whereas $N$ assigns a real number to a point. Its value $N(O)=7$ measures a signed normal displacement with the normal's scale still included. Dividing by the normal's length removes that arbitrary scale. The distance from $O$ is

$$\frac{|N(O)|}{\|n\|}=\frac7{\sqrt{30}}.$$

To find the foot rather than just its distance, use the perpendicular line from $O$: $X=\lambda n=(2\lambda,-\lambda,-5\lambda)$. Substituting these three coordinates in the plane equation gives

$$2(2\lambda)-(-\lambda)-5(-5\lambda)+7 =(4+1+25)\lambda+7=30\lambda+7=0.$$

Thus $\lambda=-7/30$, and the foot is

$$B=-\frac7{30}(2,-1,-5) =\left(-\frac7{15},\frac7{30},\frac76\right).$$

Substitution checks that $N(B)=(-28-7-175+210)/30=0$, and $OB\parallel n$ by construction. The parameter $\lambda$ is not a distance: moving one parameter unit moves by the non-unit vector $n$. The length is $|\lambda|\|n\|=7/\sqrt{30}$, agreeing with the earlier result. The negative parameter says that the foot lies in the direction opposite to the chosen normal; a distance itself is nonnegative. Marking: distance 2; normal-line construction 2; correct foot 2.

(c) [4] The scalar triple product is

$$(u\times v)\cdot w=(2,-1,-5)\cdot(0,1,1)=-6.$$

Here the parallelogram base has area $\|u\times v\|=\sqrt{30}$. The perpendicular height contributed by $w$ is the magnitude of its projection onto a unit normal, namely $6/\sqrt{30}$. Base area times height is therefore $6$. This explains why the volume is the absolute scalar triple product. Its negative sign records orientation; swapping two edges reverses that sign without changing the solid's volume. Marking: triple product 2; absolute volume and orientation explanation 2.

If this question was difficult, revisit vectors and projections and planes and volumes.

Q2. A set and its edges — 16 marks

First translate the inequalities into geometry: the squared-radius bounds give radii from $1$ to $2$, and $y>0$ keeps the upper half-plane. The radial edges are included, but the horizontal edges are excluded. All interior, closure and boundary tests use balls in the ambient plane, centred at the point being tested.

(a) [8] The strict inequalities describe the interior, and adding the missing horizontal edges gives the closure:

$$\operatorname{Int}S=\{1<x^2+y^2<4,\ y>0\}, \qquad \overline S=\{1\le x^2+y^2\le4,\ y\ge0\}.$$

The boundary is

$$\partial S= \{x^2+y^2\in\{1,4\},\ y\ge0\} \ \cup\, \{(x,0):1\le |x|\le2\}.$$

For the interior, the strict radial and vertical gaps give room to move a little in every direction and remain in $S$. On either circle, moving across the circle leaves the permitted radial interval, so those included points are not interior points.

For the closure, replacing the strict vertical inequality by a non-strict one adds exactly the missing horizontal edges. Every added point can be approached with the same radius and an upper-half-plane angle tending to $0$ or $\pi$. A point outside the stated closure violates at least one closed inequality and has a neighbourhood disjoint from $S$.

The boundary is the closure minus the interior: the two upper semicircular arcs and the two axis segments. The axis segment between $-1$ and $1$ is not another boundary edge; its interior points lie inside the excluded hole and have small balls that miss $S$. For a concrete test, the ball of radius $1/2$ centred at the origin cannot meet $S$, whose points have radius at least $1$. The origin is therefore an exterior point, not a boundary point. Points beyond radius $2$ are also separated from $S$. This accounts for every piece rather than treating the entire axis as boundary. Marking: interior 2; closure 2; arcs and both horizontal segments 2; neighbourhood/approach justification 2.

(b) [6] $S$ is not open: $(0,1)\in S$ lies on the inner edge, and every ball there contains points of radius less than $1$. It is not closed: the points $(1,1/k)$, for positive integers $k$, belong to $S$ and tend to $(1,0)\notin S$. It is bounded, since $\|(x,y)\|\le2$; for example, the open ball of radius $3$ centred at the origin contains the whole set.

For path-connectedness, write any two points in their unique upper-half-plane polar form $(r_0,\theta_0),(r_1,\theta_1)$, with $1\le r_i\le2$ and $0<\theta_i<\pi$. For $0\le s\le1$ set

$$r(s)=(1-s)r_0+sr_1,\qquad \theta(s)=(1-s)\theta_0+s\theta_1, \qquad \gamma(s)=(r(s)\cos\theta(s),r(s)\sin\theta(s)).$$

The radius is a weighted average of two numbers in $[1,2]$, so it remains in that interval. The angle is a weighted average of two numbers strictly between $0$ and $\pi$, so it remains strictly between those bounds. Thus the second coordinate stays positive. Sine, cosine and the interpolated coordinates are continuous, and the path has the required endpoints. This verifies every part of the path-connectedness definition.

A straight segment is a different construction and need not work: a segment between opposite sides of the half-annulus can pass through the hole. The chosen polar path changes radius and angle while enforcing the constraints throughout. Marking: open/closed decisions with witnesses 2; boundedness 1; valid path and containment 3.

(c) [2] The point is

$$\left(\frac32\cos\frac{2\pi}3,\frac32\sin\frac{2\pi}3\right) =\left(-\frac34,\frac{3\sqrt3}4\right).$$

The cosine is $-1/2$ and the sine is $\sqrt3/2$, explaining both products. The squared radius is $9/16+27/16=9/4$, which lies between $1$ and $4$, and the second coordinate is positive. Thus the point belongs to $S$. Its negative first coordinate also agrees with the specified upper-left angle. Marking: coordinates 1; membership check 1.

Revisit point sets and boundary tests and coordinate conversion.

Q3. Which approaches prove a limit? — 18 marks

The first goal is to control every nearby point, not to collect examples of successful paths. For the second function, a path with a different answer is enough to disprove the limit. Keeping these two goals separate tells us which method to seek.

(a) [6] Expanding the nonnegative square $(x^2-y^2)^2\ge0$ gives $x^4-2x^2y^2+y^4\ge0$. Adding $4x^2y^2$ to both sides gives $4x^2y^2\le(x^2+y^2)^2$. Away from the origin the denominator is positive, so dividing gives, with $r=\sqrt{x^2+y^2}$,

$$0\le F(x,y)\le\frac{x^2+y^2}{4}=\frac{r^2}{4}\longrightarrow0.$$

For $\varepsilon>0$, choose $\delta=2\sqrt\varepsilon$. Then $0<r<\delta$ implies $r^2/4<\delta^2/4=\varepsilon$, and hence $|F(x,y)-F(0,0)|<\varepsilon$. For example, an output tolerance of $0.01$ is guaranteed by an input distance less than $0.2$. The same radius works for every approach because no angular or slope factor remains. The joint limit is zero; the assigned origin value is also zero, which is the additional fact needed for continuity. Marking: valid uniform bound 3; limit/continuity 1; explicit $\delta$ and verification 2.

(b) [4] On $y=mx$, if $m\ne1$,

$$G(x,mx)=\frac{(m-1)x^3}{x^4+(m-1)^2x^2} =\frac{(m-1)x}{x^2+(m-1)^2}\quad(x\ne0).$$ $$G(x,mx)=\frac{(m-1)x}{x^2+(m-1)^2}\longrightarrow0.$$

The slope is fixed during this limit, so the denominator tends to the positive number $(m-1)^2$. For $m=1$, use the original quotient: its numerator is zero and its denominator is $x^4>0$ away from the origin. On the vertical line $x=0$, the numerator is also zero and the denominator is $y^2>0$. Thus every line gives limit zero. This still leaves paths whose slope changes as they approach the origin. Marking: finite slopes other than 1, 2; slope 1 and vertical line, 2.

(c) [4] To find a useful curve, compare the denominator terms $x^4$ and $(y-x)^2$. Making $y-x=x^2$ balances them and makes the numerator the same order as well. This gives the curve $y=x+x^2$. For $x\ne0$,

$$G(x,x+x^2)=\frac{x^4}{x^4+x^4}=\frac12.$$

This differs from the zero obtained on the line $y=x$, so the joint limit does not exist. The slope of the straight line from the origin to a point on this curve is $y/x=1+x$. It varies with the point, so this curve is not one of the fixed-slope lines checked above. The value assigned at the origin cannot repair this missing joint limit. Marking: valid curve and value 2; comparison and conclusion 2.

(d) [4] With $x\ne0$ fixed, continuity of the quotient at $y=0$ gives

$$\lim_{y\to0}G(x,y)=G(x,0)=-\frac{x}{1+x^2}.$$

In that first step the denominator tends to $x^4+x^2>0$, so substituting $y=0$ is justified. The outer limit as $x\to0$ is then zero. In the other order, hold $y\ne0$ fixed. As $x\to0$, the numerator tends to zero and the denominator tends to $y^2>0$, giving $\lim_{x\to0}G(x,y)=0$; the outer limit is again zero.

Each iterated procedure finishes one coordinate limit while the other coordinate is fixed. It does not test simultaneous curved motion. Their agreement therefore does not prove a joint limit; part (c) is an explicit counterexample to that inference. A genuine joint-limit proof would need a condition such as the uniform bound used in part (a). Marking: each iterated limit 1; correct logical interpretation 2.

Revisit uniform estimates and paths and continuous extension.

Q4. Differentiating in the correct variables — 16 marks

The input has two real coordinates, and the output is one real number. Each partial derivative is a scalar rate in one coordinate direction. The gradient collects the two rates into a vector; a mixed partial is another scalar derivative of a rate function.

(a) [6] When differentiating with respect to $x$, regard $y$ as constant. The exponent $xy$ has derivative $y$, while $\log(1+y)$ is a constant factor. When differentiating with respect to $y$, the exponent has derivative $x$ and $\log(1+y)$ has derivative $1/(1+y)$. Thus

$$H_x=ye^{xy}+2x\log(1+y),\qquad H_y=xe^{xy}+\frac{x^2}{1+y},\qquad \nabla H(1,0)=(0,2).$$

For the point evaluation, $xy=0$, $e^0=1$, $\log1=0$ and $1+y=1$. Consequently $H_x(1,0)=0\cdot1+2\cdot1\cdot0=0$ and $H_y(1,0)=1\cdot1+1^2/1=2$. Another way to see the second rate is to fix only $x=1$, leaving the changing variable visible: $H(1,y)=e^y+\log(1+y)$. Its derivative at zero is $e^0+1/(1+0)=2$. Substituting both coordinates at the start would give only the single value $H(1,0)=1$ and discard the nearby values needed to find a rate. Marking: each first derivative 2; gradient at the correct point 2.

(b) [6] Here $H_{xy}$ means differentiate $H_x$ with respect to $y$. The product $ye^{xy}$ contributes two terms:

$$\partial_y(ye^{xy})=e^{xy}+xye^{xy},\qquad \partial_y(2x\log(1+y))=\frac{2x}{1+y}.$$

For $H_{yx}$, differentiate $H_y$ with respect to $x$:

$$\partial_x(xe^{xy})=e^{xy}+xye^{xy},\qquad \partial_x\!\left(\frac{x^2}{1+y}\right)=\frac{2x}{1+y}.$$

Both calculations therefore give

$$H_{xy}=e^{xy}(1+xy)+\frac{2x}{1+y},\qquad H_{yx}=e^{xy}(1+xy)+\frac{2x}{1+y}.$$

At $(1,0)$, the common expression is $1(1+0)+2/1=3$. Keeping both product-rule terms is essential: differentiating only the exponential would miss the standalone exponential term. Marking: each mixed derivative 2; evaluations 2.

(c) [4] One sufficient theorem is: if $H$ is $C^2$ on an open neighbourhood of a point, then the two mixed second partials agree there. This is a guarantee under regularity hypotheses, rather than a rule that permits every mixed derivative to be swapped without checking.

Here $D$ is open, $1+y>0$, and the exponential, polynomial and logarithmic expressions and all their derivatives are continuous on $D$. Thus $H$ is $C^\infty$ on $D$, in particular $C^2$. For the evaluated point, for example, a ball of radius $1/2$ around $(1,0)$ lies wholly in $D$. The line $y=-1$ is outside the function's domain, so no conclusion at that line is asserted. Marking: valid sufficient hypotheses 2; verification of open domain and continuity 2. Equality of two computed numbers at a point alone would not verify the theorem's hypotheses.

Revisit partial and mixed derivatives.

Q5. Linearization with a controlled error — 18 marks

The purpose of linearization is to replace the square-root calculation near the base point by a constant plus a linear predicted change. The increment has two coordinates, while the predicted change and the error are each single real numbers. Here the base point is the origin, so $\rho=\sqrt{x^2+y^2}$ is the increment length. The symbol $t=x+2y$ is a scalar substitution used to simplify the algebra; it is not a time or a distance. To justify the approximation, we must compare the error with the length of the whole increment.

(a) [4] The scalar derivative of the square root gives

$$f_x(x,y)=\frac1{2\sqrt{9+x+2y}},\qquad f_y(x,y)=\frac2{2\sqrt{9+x+2y}}.$$

The factor $2$ in the second numerator comes from differentiating $x+2y$ with respect to $y$. At the origin $f=3$, $f_x=1/6$ and $f_y=1/3$. Therefore

$$L(x,y)=3+\frac{x+2y}{6}=3+\frac t6.$$

The constant $3$ is the base output; $x/6+y/3$ predicts its change. The affine function $L$ includes both. Confusing that function with just the derivative's linear part would lose the base value. Marking: value and partials 2; affine approximation 2.

(b) [6] Rationalize before subtracting the linear term:

$$\sqrt{9+t}-3=\frac{t}{\sqrt{9+t}+3},$$

because multiplying by the conjugate replaces the numerator by $(9+t)-9=t$. Now use a common denominator to subtract $t/6$, and rationalize $3-\sqrt{9+t}$ once more:

$$\begin{aligned} E&=f-L\\ &=\frac{t}{\sqrt{9+t}+3}-\frac t6\\ &=\frac{6t-t(\sqrt{9+t}+3)}{6(\sqrt{9+t}+3)}\\ &=\frac{t(3-\sqrt{9+t})}{6(\sqrt{9+t}+3)}\\ &=-\frac{t^2}{6(\sqrt{9+t}+3)^2}. \end{aligned}$$

The denominator is positive on the stated domain, and no division by $t$ is needed, so the identity also holds at $t=0$. The negative square in the numerator shows that $f\le L$: the approximation overestimates the true value. We have obtained both the size and the direction of the error from exact algebra. Marking: first rationalization 2; common denominator 2; second rationalization and final expression 2.

(c) [4] Choose $\rho\le1$. Cauchy–Schwarz applied to $(1,2)$ and $(x,y)$ gives $|t|\le\sqrt5\rho\le\sqrt5<9$, so $9+t>0$ and the whole neighbourhood lies in the domain. The denominator in the identity is at least $6\cdot3^2=54$. Also

$$t^2=(x+2y)^2\le(1^2+2^2)(x^2+y^2)=5\rho^2.$$

Hence, for $0<\rho\le1$,

$$|E|\le\frac{t^2}{54}\le\frac{5\rho^2}{54}, \qquad \frac{|E|}{\rho}\le\frac{5\rho}{54}\longrightarrow0.$$

This is the differentiability condition. The bound says that the error is at most a constant times the square of the increment length; dividing by that length leaves a quantity tending to zero. Merely showing that the error tends to zero would establish a weaker fact. Dividing by $x$ alone would also be unsuitable, because increments with $x=0$ must be tested. The constant $54$ need not be the sharpest possible lower bound for the denominator to prove this limit. Marking: valid neighbourhood and bound 2; normalized remainder limit 2.

(d) [4] Substitute the actual increment first: $t=0.06+2(-0.015)=0.03$. The predicted change is $0.03/6=0.005$, so the approximation is $L=3.005$. Since $t>0$, $\sqrt{9+t}+3>6$. Thus

$$0<L-f<\frac{0.03^2}{216}=\frac1{240000}.$$

The bound follows from $0.03^2=0.0009$ and $0.0009/216=1/240000$. In particular $3.005-1/240000<f(0.06,-0.015)<3.005$. Thus the actual value is below the approximation by less than about $0.00000417$. This interval interprets the error estimate; it is not an assertion that the affine approximation equals the square root. Marking: approximation 2; valid absolute-error bound 2.

Revisit differentiability and linearization.

Q6. Direction, speed and differentiability — 16 marks

Here the input is a position in three-dimensional space and the output is a scalar. The gradient's three entries are local rates in the three coordinate directions. A unit direction specifies orientation without speed; the given non-unit vector must be normalized before asking for change per unit distance.

(a) [4] The polynomial is differentiable, and

$$\nabla g=(2x+2y,2x+6y,2z-2),\qquad \nabla g(A)=(0,-4,2).$$

For example, the mixed term $2xy$ contributes $2y$ to the first partial and $2x$ to the second. Substitution gives the three components $2-2=0$, $2-6=-4$ and $4-2=2$. Because $\|v\|=\sqrt{2^2+(-1)^2+2^2}=3$, the unit direction is $v/3$. Hence

$$D_{v/3}g(A)=(0,-4,2)\cdot(2/3,-1/3,2/3)=\frac83.$$

The dot product consists of $0+4/3+4/3=8/3$. It is positive, so the scalar output initially increases in this unit direction. The unnormalized dot product $8$ is a rate for parameter velocity $v$: that motion travels three distance units per parameter unit. Dividing by three removes the extra speed and gives the requested rate per unit distance. Marking: gradient 2; normalization and directional rate 2.

(b) [6] Restricting to unit directions compares changes over the same spatial distance. If speed could be multiplied by an arbitrary positive constant, a positive directional rate could be made arbitrarily large, so there would be no finite fastest rate under that different question.

The gradient length is $\sqrt{0^2+(-4)^2+2^2}=\sqrt{20}=2\sqrt5$. For $\|u\|=1$, Cauchy–Schwarz gives $|\nabla g(A)\cdot u|\le\|\nabla g(A)\|=2\sqrt5$. Equality requires a direction parallel to the gradient. Normalize it for increase, and reverse its sign for decrease:

$$u_+=\left(0,-\frac2{\sqrt5},\frac1{\sqrt5}\right),\qquad u_-=-u_+.$$

Each vector has squared length $4/5+1/5=1$. Their dot products with the gradient give $2\sqrt5$ and $-2\sqrt5$, respectively. Thus the steepest increasing direction moves toward decreasing $y$ and increasing $z$ at this point. The zero first gradient component means no first-order contribution from the $x$ direction here; it does not mean that the polynomial is independent of $x$. Marking: bound and equality condition 2; each correct direction/rate pair 2.

(c) [6] We cannot use the gradient dot-product formula before establishing differentiability. Start with the directional-derivative definition instead. For a unit vector $u=(a,b)$ and $s\ne0$, direct substitution gives $q(sa,sb)=s a^2b/(a^2+3b^2)$, so

$$\frac{q(sa,sb)-q(0,0)}s=\frac{a^2b}{a^2+3b^2}.$$

The denominator is positive because $u$ is nonzero. The expression does not depend on $s$ or its sign, so the two-sided limit exists in every unit direction. In the two coordinate directions, either $a=0$ or $b=0$, giving zero; hence $\nabla q(0,0)=(0,0)$. In contrast, for $u=(1,1)/\sqrt2$, the numerator is $1/(2\sqrt2)$ and the denominator is $2$, so

$$D_uq(0,0)=\frac1{4\sqrt2}\ne0.$$

If $q$ were differentiable, the coordinate partials would force its linear predicted change to be zero. Every unit directional derivative would then equal $\nabla q(0,0)\cdot u=0$, contradicting the diagonal result. This one incompatible direction is already enough to disprove differentiability. Each line having its own slope is not enough: those slopes must come from one compatible linear map, together with a small remainder for all increments. Even a successful list of directional checks would not replace that uniform remainder condition.

The same failure is visible directly. On $(x,y)=(s,s)$, the output is $q(s,s)=s/4$ and the increment length is $\sqrt2|s|$. The absolute error of the only possible zero linear approximation, divided by that length, is therefore $1/(4\sqrt2)$ for every nonzero $s$. It does not tend to zero. In fact $|q(x,y)|\le |y|\le\sqrt{x^2+y^2}$, so the function is continuous at the origin; continuity plus all directional derivatives still has not produced differentiability. Marking: all-direction formula and existence 2; coordinate partials 1; nonzero diagonal derivative and valid contradiction 3.

Revisit the small-remainder condition and the unit directional derivative and fastest-direction parts of directional derivatives.

Choose the next revision step

Record each part's earned marks and one concrete error type: arithmetic, definition, hypothesis, uniformity, or direction normalization. Rework the corresponding note example with the answer covered. Then solve that part again on a blank page and justify the step that previously failed. A high total score should still be followed by revisiting any unproved claim or missing boundary case.