Differentiation: ASRJC 2024 Promo Q11

Anderson Serangoon Junior College2024 PromoPromo Q11●●● Challenging14 marks

What this question tests

Differentiation (Applications).

Question

A sphere has radius $r$ cm, surface area $S$ cm $^2$ and volume $V$ cm $^3$ .
1. Show that $\dfrac{\,\mathrm{d} V}{\,\mathrm{d} S} = \dfrac{r}{2}$ . [2]
2. Given that the surface area of the sphere is increasing at a constant rate of $3$ cm $^2$ /s, find the radius when the rate of increase of the volume is $9$ cm $^3$ /s. [2]
$\left[\text{Volume of sphere, } V = \dfrac{4}{3}\pi r^3; \text{ Surface area of sphere, } S = 4\pi r^2.\right]$
CarsExtreme is hosting a competition where competitors must drive from Town $A$ to Town $C$ in the shortest time possible.
$Anderson Serangoon Junior College 2024 Promo Q11: question diagram$
Town $A$ is located $250$ km to the west of Town $B$ , while Town $B$ is positioned $80$ km away from Town $C$ at a bearing of $330^\circ$ . A straight road connects Town $A$ and Town $B$ , as shown in the picture above. The three towns are in a desert and competitors must traverse this desert to reach their destinations. To ensure safety, a speed limit of $110$ km/h is enforced within the desert, except on the straight road where the speed limit is $130$ km/h. It is assumed that competitors will always drive at the maximum permitted speed and will take the shortest possible route when crossing the desert.
Competitor $P$ and $Q$ have different strategies for the competition. Competitor $P$ plans to drive along the straight road from Town $A$ to a point $x$ km before reaching Town $B$ , and then cut diagonally across the desert to Town $C$ . On the other hand, Competitor $Q$ intends to drive directly from Town $A$ to Town $C$ , traveling entirely through the desert.
1. Show that the time taken, $T$ hours, for Competitor $P$ is $T = \frac{1}{1430}\left(2750 - 11x + 13\sqrt{x^2 + 80x + 6400}\right).$ [3]
2. Use differentiation to find the minimum time that Competitor $P$ will take, giving your answer correct to the nearest minute. (You need not show that your answer gives a minimum.) [4]
3. However, prior to the start of the competition, weather changes in the desert prompts the organisers to adjust the speed limit to $M$ km/h for desert driving. Competitor $P$ then decides to drive to Town $B$ before heading to Town $C$ , while Competitor $Q$ sticks to his original plan. Find the range of values of $M$ if Competitor $P$ arrives at Town $C$ before $Q$ . [3]

Show full worked solution▾

(a)

(i) $\begin{aligned} V=\frac43\pi r^3&\Rightarrow\frac{\,\mathrm{d} V}{\,\mathrm{d} r}=4\pi r^2\\ S=4\pi r^2&\Rightarrow\frac{\,\mathrm{d} S}{\,\mathrm{d} r}=8\pi r\\ \frac{\,\mathrm{d} V}{\,\mathrm{d} S} &=\frac{\,\mathrm{d} V}{\,\mathrm{d} r}\div\frac{\,\mathrm{d} S}{\,\mathrm{d} r} =\frac{4\pi r^2}{8\pi r}=\frac r2 \end{aligned}$

(ii) $\begin{aligned} \frac{\,\mathrm{d} V}{\,\mathrm{d} t} &=\frac{\,\mathrm{d} V}{\,\mathrm{d} S}\frac{\,\mathrm{d} S}{\,\mathrm{d} t} =\frac r2(3)=\frac{3r}{2} \end{aligned}$ When $\dfrac{\,\mathrm{d} V}{\,\mathrm{d} t}=9$ , $9=\frac{3r}{2}\Rightarrow r=6$

$Anderson Serangoon Junior College 2024 Promo Q11: solution diagram$

(b)

(i) Time taken on straight road $=\dfrac{250-x}{130}$

Distance from straight road to Town $C$ , $d$ $\text{(Method 1): }=\sqrt{x^2+80^2-2x(80)\cos\frac{2\pi}{3}}\quad(\text{cosine rule})$ $\text{(Method 2): }=\sqrt{(x+40)^2+(40\sqrt3)^2}\quad(\text{Pythagoras' Thm})$ Distance from straight road to Town $C=\sqrt{x^2+80x+6400}$

$\text{Time taken on desert}=\frac{\sqrt{x^2+80x+6400}}{110}$ $\text{Total time taken by competitor P,}\quad T=\frac{250-x}{130}+\frac{\sqrt{x^2+80x+6400}}{110}$ $\begin{aligned} T&=11\left(\frac{250-x}{1430}\right) +13\left(\frac{\sqrt{x^2+80x+6400}}{1430}\right)\\ T&=\frac1{1430}\left(2750-11x+13\sqrt{x^2+80x+6400}\right)\qquad(\text{shown}) \end{aligned}$

(ii) $\begin{aligned} \frac{\,\mathrm{d} T}{\,\mathrm{d} x} &=\frac1{1430}\left(-11+13\left(\frac12\right)(x^2+80x+6400)^{-1/2}(2x+80)\right)\\ \frac{\,\mathrm{d} T}{\,\mathrm{d} x} &=\frac1{1430}\left(-11+13(x+40)(x^2+80x+6400)^{-1/2}\right) \end{aligned}$ To find minimum time, $\dfrac{\,\mathrm{d} T}{\,\mathrm{d} x}=0$ $\frac1{1430}\left(-11+13(x+40)(x^2+80x+6400)^{-1/2}\right)=0$ $\frac{13(x+40)}{\sqrt{x^2+80x+6400}}=11$

Method 1: Using GC

$Anderson Serangoon Junior College 2024 Promo Q11: solution diagram$

$x=70$

Method 2: Algebra $\begin{aligned} 13(x+40)&=11\sqrt{x^2+80x+6400}\\ 169(x+40)^2&=121(x^2+80x+6400)\\ 169(x^2+80x+1600)-121(x^2+80x+6400)&=0\\ 169x^2+13520x+270400-121x^2-9680x-774400&=0\\ 48x^2+3840x-504000&=0\\ x&=70\quad\text{or}\quad x=-150\quad(\text{rej as }x\geq0) \end{aligned}$ Sub $x=70$ into $T$ $\begin{aligned} T&=\frac1{1430}\left(2750-11(70)+13\sqrt{(70)^2+80(70)+6400}\right)\\ &=\frac{3670}{1430}\\ &=2.566\text{ hr}\\ &=154\text{ min or }2\text{ hr }34\text{ min (nearest minute)} \end{aligned}$

(iii) $\text{Time taken by Competitor }Q =\frac{\sqrt{250^2+80^2-2(250)(80)\cos\dfrac{2\pi}{3}}}{M} =\frac{\sqrt{88900}}M$ $\text{Time taken by Competitor }P=\frac{250}{130}+\frac{80}M$ Since Competitor $P$ is faster than Competitor $Q$ $\begin{aligned} \frac{\sqrt{88900}}M&>\frac{250}{130}+\frac{80}M\\ \frac{\sqrt{88900}-80}M&>\frac{25}{13}\\ M&<113.44 \end{aligned}$ Also, $M>0$

Hence $M$ lies between $0$ and $113$ .

Answers

(a)(ii) $r = 6$ cm
(b)(ii)minimum time $= 154$ min ( $2$ hr $34$ min) (b)(iii) $0 < M < 113$

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