KIX1001 2020/2021 Semester 1 — Paper 1

Original Question Paper (PDF)

Paper details

Academic session 2020/2021, Semester 1 · 2 questions · 6 printed pages · 1.5 hours. Questions are reproduced in English from the bilingual original. Solutions are worked in full; numbers are given to a sensible precision.

Set selection — read this first

For Questions 1 and 2, students are divided into 3 groups, and each group should solve only its own set of questions:

Set	Student ID ending with
Set 1	0, 3, 6 and 8
Set 2	4, 1 and 9
Set 3	7, 5 and 2

The printed paper contains only one version of Questions 1 and 2: pages 3–6 carry a single set of parts, 1(a)–(c) and 2(a)–(b), and no set-specific variants are printed anywhere in the paper. The three sets therefore differ only in which student ID numbers are told to attempt them — not in the questions themselves. The solutions below are the answers for all three sets.

Reference only — these are not official answers

These solutions are revision aids, not examiner answers. They were produced with the help of a large language model and then checked, but an LLM can still be wrong: expect the occasional arithmetic slip, a different method than your lecturer's, or a misreading of the printed question. The question wording is reproduced by hand from a scan of the original paper, so it too may differ in small ways from what was printed.

Use this page as a worked suggestion to check, not as ground truth. Always compare against the original paper (linked above) and your own lecture notes, and if something here disagrees with your lecturer, trust your lecturer. Marks shown are the paper's allocations, not a marking scheme.

Question 1

Question 1(a) (4 marks)

Question

A hyperboloid of one sheet is constructed by connecting two concentric vertically offset rings with tilted wires in the construction of power plant's cooling tower. In order to analyze the force acting on the cooling tower, determine the equation of tangent plane and normal plane at the point $(2,4,6)$. The support function of the hyperboloid of one sheet structure is

$$\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}-\frac{z^{2}}{2^{2}}=1$$

Solution

Technique. Write the surface as a level surface $F(x,y,z)=0$. The gradient $\nabla F$ is normal to the surface, so the tangent plane at a point $P$ is $\nabla F(P)\cdot\left[(x,y,z)-P\right]=0$, and the normal line is $(x,y,z)=P+t\,\nabla F(P)$.

Take $F(x,y,z)=\frac{x^{2}}{25}+\frac{y^{2}}{9}-\frac{z^{2}}{4}-1=0$.

Step 1 — gradient.

$$\nabla F=\left(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z}\right)=\left(\frac{2x}{25},\frac{2y}{9},-\frac{z}{2}\right).$$

At $P(2,4,6)$:

$$\nabla F(2,4,6)=\left(\frac{4}{25},\frac{8}{9},-3\right).$$

Step 2 — tangent plane.

$$\frac{4}{25}(x-2)+\frac{8}{9}(y-4)-3(z-6)=0 .$$

Multiply through by $225$ (the LCM of $25$ and $9$):

$$36(x-2)+200(y-4)-675(z-6)=0$$$$36x-72+200y-800-675z+4050=0$$$$36x+200y-675z+3178=0 .$$

Check: $36(2)+200(4)-675(6)+3178=72+800-4050+3178=0$. ✔

Step 3 — normal line. The normal line passes through $P$ with direction $\nabla F(P)=\left(\tfrac{4}{25},\tfrac{8}{9},-3\right)$, i.e. with direction proportional to $(36,200,-675)$:

$$\frac{x-2}{4/25}=\frac{y-4}{8/9}=\frac{z-6}{-3}\qquad\Longleftrightarrow\qquad \frac{x-2}{36}=\frac{y-4}{200}=\frac{z-6}{-675},$$

or in parametric form

$$(x,y,z)=(2,4,6)+t\left(\tfrac{4}{25},\tfrac{8}{9},-3\right),\qquad t\in\mathbb{R}.$$

Printing check on the given point

Substituting $(2,4,6)$ into the printed support function gives

$$\frac{4}{25}+\frac{16}{9}-\frac{36}{4}=-\frac{1589}{225}\approx-7.06\neq 1,$$

so the printed point does not lie on the printed hyperboloid. The standard tangent-plane formula has nevertheless been applied at the printed point: it gives the tangent plane of the level surface $F=-1814/225$ that does pass through $(2,4,6)$, using the normal direction $\nabla F(2,4,6)$. Had the point been on the surface, only the constant term would change.

About the "normal plane"

The paper asks for the "normal plane". For a surface the object normally intended is the normal line — the line through the point in the direction of $\nabla F$ — and that is what is given above. (The only plane through the point whose normal vector is $\nabla F$ is the tangent plane itself, so a separate "normal plane" would carry no extra information.)

:::

Answer: Tangent plane $36x+200y-675z+3178=0$, equivalently $\frac{4}{25}(x-2)+\frac{8}{9}(y-4)-3(z-6)=0$. Normal line $\frac{x-2}{36}=\frac{y-4}{200}=\frac{z-6}{-675}$ (direction $(36,200,-675)$).

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Question 1(b) (4 marks)

Question

As a structure engineer, you are required to design a stable structure in the form of cuboid and surrounded by pyramid ends. In order to analyze the stability of the structure, determine the percentage change in volume of the structure. Given that the width, $a$ of 4 m is increased by 0.01 m, while the length, $l$ of 10 m is increased by 0.02 m and the height, $h$ of 6 m is increased by 0.03 m. Please Refer Figure 1.

Solution

Technique. Use the total differential of a function of three variables, $\mathrm{d}V=\frac{\partial V}{\partial a}\,\mathrm{d}a+\frac{\partial V}{\partial l}\,\mathrm{d}l+\frac{\partial V}{\partial h}\,\mathrm{d}h$, and divide by $V$ to get the percentage change.

Step 1 — volume of the structure. From Figure 1 the structure is a cuboid of cross-section $a\times a$ and length $l$, capped at both ends by square pyramids of base $a$ and height $h$:

$$V=a^{2}l+2\left(\tfrac{1}{3}a^{2}h\right)=a^{2}l+\tfrac{2}{3}a^{2}h .$$

Step 2 — partial derivatives.

$$\frac{\partial V}{\partial a}=2al+\frac{4}{3}ah,\qquad \frac{\partial V}{\partial l}=a^{2},\qquad \frac{\partial V}{\partial h}=\frac{2}{3}a^{2}.$$

At $a=4$, $l=10$, $h=6$:

$$\frac{\partial V}{\partial a}=2(4)(10)+\frac{4}{3}(4)(6)=80+32=112,$$$$\frac{\partial V}{\partial l}=4^{2}=16,\qquad \frac{\partial V}{\partial h}=\frac{2}{3}(4)^{2}=\frac{32}{3}.$$

Step 3 — total differential. With $\mathrm{d}a=0.01$, $\mathrm{d}l=0.02$, $\mathrm{d}h=0.03$ (all in m):

$$\mathrm{d}V=112(0.01)+16(0.02)+\frac{32}{3}(0.03)=1.12+0.32+0.32=1.76\ \text{m}^{3}.$$

Step 4 — percentage change. The original volume is

$$V=a^{2}l+\tfrac{2}{3}a^{2}h=16(10)+\tfrac{2}{3}(16)(6)=160+64=224\ \text{m}^{3}.$$

Hence

$$\frac{\mathrm{d}V}{V}\times 100\%=\frac{1.76}{224}\times 100\%=\frac{11}{14}\%\approx 0.786\%.$$

Step 5 — exact check. Using the new dimensions $a=4.01$, $l=10.02$, $h=6.03$:

$$V'=(4.01)^{2}(10.02)+\tfrac{2}{3}(4.01)^{2}(6.03)=225.7646\ \text{m}^{3},$$

so the true increase is $225.7646-224=1.7646\ \text{m}^{3}$, i.e. $0.788\%$. The linear (total-differential) estimate $0.786\%$ is therefore accurate to three significant figures, as expected for such small increments.

Answer: $\mathrm{d}V=1.76\ \text{m}^{3}$ out of $V=224\ \text{m}^{3}$, a percentage increase of $\frac{11}{14}\%\approx 0.786\%$ (exact value from the new dimensions: $0.788\%$).

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Question 1(c) (7 marks)

Question

You are a manufacturer of metal cones. In order to minimize the material used in the manufacturing process, you need to fabricate the metal cones shown below. You are provided with 3 copper cones with total volume of 15 m³. Each cone consists of the flat base surface and lateral surface. Determine the possible dimensions of the lateral surface for each cone to estimate the amount of copper material needed to construct each cone's lateral surface. (Given the dimensions of cone are height of the cone, $h$, the slant length, $s$, and the radius of the base, $r$). Refer Figure 2.

Solution

Technique. This is a constrained optimisation: minimise the lateral surface area $A=\pi rs$ subject to a fixed volume, using Lagrange multipliers (one constraint).

Step 1 — set up. For one cone, $s=\sqrt{r^{2}+h^{2}}$, so the material needed for the lateral surface is $A(r,h)=\pi r\sqrt{r^{2}+h^{2}}$, and the volume constraint (each cone carries $15/3=5\ \text{m}^{3}$) is $g(r,h)=\tfrac{1}{3}\pi r^{2}h-5=0$.

Step 2 — Lagrange conditions. With $s=\sqrt{r^{2}+h^{2}}$,

$$\frac{\partial A}{\partial r}=\pi\sqrt{r^{2}+h^{2}}+\frac{\pi r^{2}}{\sqrt{r^{2}+h^{2}}}=\frac{\pi\left(2r^{2}+h^{2}\right)}{s},\qquad \frac{\partial A}{\partial h}=\frac{\pi rh}{s},$$$$\frac{\partial g}{\partial r}=\frac{2\pi rh}{3},\qquad \frac{\partial g}{\partial h}=\frac{\pi r^{2}}{3}.$$

Setting $\nabla A=\lambda\nabla g$:

$$\frac{\pi\left(2r^{2}+h^{2}\right)}{s}=\lambda\,\frac{2\pi rh}{3},\qquad \frac{\pi rh}{s}=\lambda\,\frac{\pi r^{2}}{3}.$$

From the second equation, $\lambda=\dfrac{3h}{rs}$. Substituting into the first:

$$\frac{2r^{2}+h^{2}}{s}=\frac{2rh}{3}\cdot\frac{3h}{rs}=\frac{2h^{2}}{s}\;\Longrightarrow\;2r^{2}+h^{2}=2h^{2}\;\Longrightarrow\;h^{2}=2r^{2},$$$$h=\sqrt{2}\,r .$$

Step 3 — apply the volume constraint.

$$\frac{1}{3}\pi r^{2}\left(\sqrt{2}\,r\right)=5\;\Longrightarrow\;r^{3}=\frac{15}{\sqrt{2}\,\pi}=\frac{15\sqrt{2}}{2\pi},$$$$r=\left(\frac{15\sqrt{2}}{2\pi}\right)^{1/3}\approx 1.500\ \text{m}.$$

Step 4 — the other two dimensions.

$$h=\sqrt{2}\,r=\left(\frac{30}{\pi}\right)^{1/3}\approx 2.122\ \text{m},$$$$s=\sqrt{r^{2}+h^{2}}=\sqrt{r^{2}+2r^{2}}=r\sqrt{3}\approx 2.598\ \text{m}.$$

Step 5 — copper required (lateral surface only).

$$A=\pi rs=\pi r^{2}\sqrt{3}=\pi\sqrt{3}\left(\frac{15\sqrt{2}}{2\pi}\right)^{2/3}\approx 12.25\ \text{m}^{2}\ \text{per cone},$$

so the three cones together need $3A\approx 36.74\ \text{m}^{2}$.

Step 6 — confirm this is a minimum. Eliminating $h$ with $h=\dfrac{15}{\pi r^{2}}$ gives

$$A(r)=\pi r\sqrt{r^{2}+\frac{225}{\pi^{2}r^{4}}},\qquad A(r)\to\infty\ \text{as}\ r\to 0^{+}\ \text{and as}\ r\to\infty .$$

The single stationary point $r=\left(\frac{15\sqrt{2}}{2\pi}\right)^{1/3}$ is therefore the global minimum of $A$. The optimum has the classic shape $h=\sqrt{2}\,r$ (apex angle $2\arctan(r/h)\approx 70.5^{\circ}$).

Assumption about the three cones

The paper gives only the total volume of the three cones, $15\ \text{m}^{3}$; the cones are taken to be identical, so each carries $5\ \text{m}^{3}$. (If the volume were free to be shared unequally, the total lateral area would be minimised by loading all $15\ \text{m}^{3}$ into a single cone, since $A\propto V^{2/3}$ is concave — not a physical design. The intended reading is three equal cones, so each has volume $5\ \text{m}^{3}$.)

:::

Answer: Each cone: radius $r=\left(\frac{15\sqrt{2}}{2\pi}\right)^{1/3}\approx 1.500\ \text{m}$, height $h=\sqrt{2}\,r=\left(\frac{30}{\pi}\right)^{1/3}\approx 2.122\ \text{m}$ and slant length $s=r\sqrt{3}\approx 2.598\ \text{m}$, giving a lateral surface area of $\pi r s\approx 12.25\ \text{m}^{2}$ per cone and $\approx 36.74\ \text{m}^{2}$ of copper for all three cones.

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Question 2

Question 2(a) (7 marks)

Question

In a distortion analysis, the eigenvalues

$$\begin{pmatrix}2&0&0\\0&-5&0\\0&0&7\end{pmatrix}$$

give the distortion induced by the transformation and the corresponding eigenvectors

$$\begin{pmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{pmatrix}$$

are the orientation of the distortion according to $AX=\lambda X$ where, $\lambda$ is the eigenvalues and $X$ is the corresponding eigenvectors. Find matrix $A$ before the distortion happened.

Solution

Technique. The columns of the eigenvector matrix are the eigenvectors, so with $P=\begin{pmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{pmatrix}$ (columns $X_1,X_2,X_3$) and $D=\operatorname{diag}(2,-5,7)$ we have $AP=PD$, hence $A=PDP^{-1}$.

Step 1 — determinant of $P$.

$$\det P=2(4-1)+1(-2+1)+1(1-2)=6-1-1=4\neq 0,$$

so $P$ is invertible.

Step 2 — inverse of $P$. The cofactor matrix is

$$C=\begin{pmatrix}3&1&-1\\1&3&1\\-1&1&3\end{pmatrix},\qquad P^{-1}=\frac{1}{\det P}\,C^{\mathsf T}=\frac{1}{4}\begin{pmatrix}3&1&-1\\1&3&1\\-1&1&3\end{pmatrix}.$$

Step 3 — $PD$. Multiplying column $j$ of $P$ by $\lambda_j$:

$$PD=\begin{pmatrix}4&5&7\\-2&-10&-7\\2&5&14\end{pmatrix}.$$

Step 4 — multiply out $A=PDP^{-1}=\frac{1}{4}(PD)C^{\mathsf T}$.

Row 1 of $PD$ times the columns of $C^{\mathsf T}$:

$$4(3)+5(1)+7(-1)=10,\quad 4(1)+5(3)+7(1)=26,\quad 4(-1)+5(1)+7(3)=22 .$$

Row 2:

$$-2(3)-10(1)-7(-1)=-9,\quad -2(1)-10(3)-7(1)=-39,\quad -2(-1)-10(1)-7(3)=-29 .$$

Row 3:

$$2(3)+5(1)+14(-1)=-3,\quad 2(1)+5(3)+14(1)=31,\quad 2(-1)+5(1)+14(3)=45 .$$

Therefore

$$A=\frac{1}{4}\begin{pmatrix}10&26&22\\-9&-39&-29\\-3&31&45\end{pmatrix}=\begin{pmatrix}2.5&6.5&5.5\\-2.25&-9.75&-7.25\\-0.75&7.75&11.25\end{pmatrix}.$$

Step 5 — verify. $A$ must reproduce $AX=\lambda X$ for each eigenvector, e.g.

$$A\begin{pmatrix}2\\-1\\1\end{pmatrix}=\frac{1}{4}\begin{pmatrix}20-26+22\\-18+39-29\\-6-31+45\end{pmatrix}=\frac{1}{4}\begin{pmatrix}16\\-8\\8\end{pmatrix}=\begin{pmatrix}4\\-2\\2\end{pmatrix}=2\begin{pmatrix}2\\-1\\1\end{pmatrix}\ \checkmark$$

Also $\operatorname{tr}A=2.5-9.75+11.25=4=2+(-5)+7$ and $\det A=2(-5)(7)=-70$, both matching the given eigenvalues. ✔

Answer: $A=\frac{1}{4}\begin{pmatrix}10&26&22\\-9&-39&-29\\-3&31&45\end{pmatrix}=\begin{pmatrix}2.5&6.5&5.5\\-2.25&-9.75&-7.25\\-0.75&7.75&11.25\end{pmatrix}$.

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Question 2(b) (8 marks)

Question

As a consultant engineer, you need to consult a client on the cost of repairing a gas pipeline. You have come out with proposal A, B and C. For proposal A, the installation cost is triple, the material cost is double and the painting cost is triple. The total cost of the repairment is RM 10 000. For proposal B, the installation cost is double, the material cost is 5 times more expensive and the painting cost is double. The total cost of the repairment is RM 12 000. For proposal C, the installation cost is cheaper by 1/3 in comparison to proposal A, the material cost is triple and the painting cost is double. The total cost of the repairment is RM 8 000. Find the installation cost, material cost and painting cost for each of the proposal respectively using Cramer's Rule.

Solution

Technique. Let $I$, $M$ and $P$ be the base installation, material and painting costs. Each proposal gives one linear equation (the multipliers of the base costs are the coefficients), so we have a $3\times3$ system to be solved by Cramer's rule $I=\frac{\det A_I}{\det A}$, $M=\frac{\det A_M}{\det A}$, $P=\frac{\det A_P}{\det A}$.

Step 1 — build the system exactly as printed.

Proposal	Installation	Material	Painting	Total (RM)
A	$3I$	$2M$	$3P$	10 000
B	$2I$	$5M$	$2P$	12 000
C	$2I$	$3M$	$2P$	8 000

$$A=\begin{pmatrix}3&2&3\\2&5&2\\2&3&2\end{pmatrix},\qquad \mathbf{b}=\begin{pmatrix}10000\\12000\\8000\end{pmatrix}.$$

Step 2 — the printed coefficients are singular. Expanding along the first row,

$$\det A=3(10-6)-2(4-4)+3(6-10)=12-0-12=0 .$$

Cramer's rule cannot be applied, and the system is in fact inconsistent: the first row satisfies $(3,2,3)=-\tfrac{5}{4}(2,5,2)+\tfrac{11}{4}(2,3,2)$, but the right-hand sides give $-\tfrac{5}{4}(12000)+\tfrac{11}{4}(8000)=-15000+22000=7000\neq 10000$. So no triple $(I,M,P)$ satisfies all three printed equations — see the note below for the reading adopted.

Step 3 — solve the corrected system by Cramer's rule. Reading proposal B's painting cost as triple (coefficient $3$), which keeps the installation column $3,2,2$ that the question fixes explicitly:

$$A=\begin{pmatrix}3&2&3\\2&5&3\\2&3&2\end{pmatrix}.$$

Determinant of $A$:

$$\det A=3(10-9)-2(4-6)+3(6-10)=3+4-12=-5 .$$

Replacing the $I$-column:

$$\det A_I=\det\begin{pmatrix}10000&2&3\\12000&5&3\\8000&3&2\end{pmatrix}=10000(10-9)-2(24000-24000)+3(36000-40000)=10000-0-12000=-2000,$$$$I=\frac{\det A_I}{\det A}=\frac{-2000}{-5}=400 .$$

Replacing the $M$-column:

$$\det A_M=\det\begin{pmatrix}3&10000&3\\2&12000&3\\2&8000&2\end{pmatrix}=3(24000-24000)-10000(4-6)+3(16000-24000)=0+20000-24000=-4000,$$$$M=\frac{\det A_M}{\det A}=\frac{-4000}{-5}=800 .$$

Replacing the $P$-column:

$$\det A_P=\det\begin{pmatrix}3&2&10000\\2&5&12000\\2&3&8000\end{pmatrix}=3(40000-36000)-2(16000-24000)+10000(6-10)=12000+16000-40000=-12000,$$$$P=\frac{\det A_P}{\det A}=\frac{-12000}{-5}=2400 .$$

Step 4 — cost of each proposal. Applying that proposal's multipliers to the base costs:

Proposal	Installation	Material	Painting	Total (RM)
A	$3(400)=1200$	$2(800)=1600$	$3(2400)=7200$	10 000 ✔
B	$2(400)=800$	$5(800)=4000$	$3(2400)=7200$	12 000 ✔
C	$2(400)=800$	$3(800)=2400$	$2(2400)=4800$	8 000 ✔

All three printed totals are reproduced exactly.

Printing error in proposal B

The printed data are inconsistent. Proposals A and C as printed are mutually consistent and force $M=800$ and $I+P=2800$; proposal B as printed would then need $2I+5M+2P=2(2800)+4000=9600$, not the printed RM 12 000, and its coefficient row $(2,5,2)$ is parallel to the other two rows, so $\det A=0$.

One coefficient of proposal B must therefore be $3$, not $2$. The solution above takes proposal B's painting cost as triple ($3$), because the installation column $3,2,2$ is fixed explicitly by the wording (A triple, B double, C "cheaper by $1/3$ in comparison to A", i.e. $3\times\tfrac{2}{3}=2$). Under the other equally plausible single-entry correction — proposal B's installation cost triple, giving rows $(3,2,3)$, $(3,5,2)$, $(2,3,2)$ and $\det A=5$ — the solution is instead $I=\text{RM }2400$, $M=\text{RM }800$, $P=\text{RM }400$, with proposal costs A: 7200, 1600, 1200; B: 7200, 4000, 800; C: 4800, 2400, 800. Note that in both readings $M=\text{RM }800$ and the printed A and C totals are satisfied exactly; only the split of the RM 2800 between $I$ and $P$ depends on which figure in proposal B is corrected.

:::

Answer: Installation $I=\text{RM }400$, material $M=\text{RM }800$, painting $P=\text{RM }2400$ per unit of the multipliers, i.e.

• Proposal A — installation RM 1 200, material RM 1 600, painting RM 7 200 (total RM 10 000);
• Proposal B — installation RM 800, material RM 4 000, painting RM 7 200 (total RM 12 000);
• Proposal C — installation RM 800, material RM 2 400, painting RM 4 800 (total RM 8 000).

TAMAT / END