Differentiation: CJC 2024 Promo Q10

(a)

$$\begin{aligned} \text{Volume of cone}&=\frac{1}{3}\pi r^2 h=20\pi\\ \frac{r^2 h}{3} &= 20\\ h &= \frac{60}{r^2}. \end{aligned}$$

(b)

Method 1⃝: Considering $A^2$ $$\begin{aligned} A &= \pi rl\\ &=\pi r\sqrt{r^2+h^2}\\ &=\pi r\sqrt{r^2+\left(\frac{60}{r^2}\right)^2}\\ &=\pi r\sqrt{r^2+\frac{3600}{r^4}}\\ A^2&=\pi^2r^2\left(r^2+\frac{3600}{r^4}\right)\\ &=\pi^2\left(r^4+\frac{3600}{r^2}\right)\\ &=\pi^2(r^4+3600r^{-2}). \end{aligned}$$ Differentiating implicitly with respect to $r$: $$\begin{aligned} 2A\frac{\,\mathrm{d} A}{\,\mathrm{d} r} &= \pi^2\left(4r^3 - 7200r^{-3}\right)\\ A\frac{\,\mathrm{d} A}{\,\mathrm{d} r} &= \pi^2\left(2r^3 - \frac{3600}{r^3}\right). \quad\text{((shown))} \end{aligned}$$

Method 2⃝: “Otherwise” Method $$\begin{aligned} A&=\pi rl\\ &=\pi r\sqrt{r^2+h^2}\\ &=\pi r\sqrt{r^2+\left(\frac{60}{r^2}\right)^2}\\ &=\pi r\sqrt{r^2+\frac{3600}{r^4}}\\ &=\pi r(r^2+3600r^{-4})^{1/2}. \end{aligned}$$ Differentiating with respect to $r$: $$\begin{aligned} \frac{\,\mathrm{d} A}{\,\mathrm{d} r} &=\pi\left[r\cdot\frac12(r^2+3600r^{-4})^{-1/2}(2r-4\cdot3600r^{-5})+(r^2+3600r^{-4})^{1/2}\right]\\ &=\pi(r^2+3600r^{-4})^{-1/2}\left[r^2-7200r^{-4}+r^2+3600r^{-4}\right]\\ &=\pi(r^2+3600r^{-4})^{-1/2}\left[2r^2-3600r^{-4}\right]\\ &=\frac{\pi}{\sqrt{r^2+3600r^{-4}}}\left(2r^2-\frac{3600}{r^4}\right),\\ A\frac{\,\mathrm{d} A}{\,\mathrm{d} r} &=\pi r\sqrt{r^2+3600r^{-4}}\cdot\frac{\pi}{\sqrt{r^2+3600r^{-4}}}\left(2r^2-\frac{3600}{r^4}\right)\\ &=\pi^2r\left(2r^2-\frac{3600}{r^4}\right)\\ &=\pi^2\left(2r^3-\frac{3600}{r^3}\right).\quad\text{((shown))} \end{aligned}$$

(c)

At stationary point, $\dfrac{\,\mathrm{d} A}{\,\mathrm{d} r}=0$ $$\begin{aligned} \pi^2\left(2r^3 - \frac{3600}{r^3}\right) &= 0\\ 2r^3-\frac{3600}{r^3}&=0\\ 2r^3&=\frac{3600}{r^3}\\ r^6&=1800\\ r&=(1800)^{1/6}. \end{aligned}$$ To prove minimum:

Method 1⃝: Using 1st derivative test

$r=(1800)^{1/6}$ gives the minimum $A$.

Method 2⃝: Using 2nd derivative test $$\begin{aligned} \text{From the result }A\frac{\,\mathrm{d} A}{\,\mathrm{d} r}&=\pi^2\left(2r^3-\frac{3600}{r^3}\right),\ \text{using implicit differentiation to obtain:}\\ A\frac{\,\mathrm{d}^2A}{\,\mathrm{d} r^2}+\left(\frac{\,\mathrm{d} A}{\,\mathrm{d} r}\right)^2&=\pi^2\left(6r^2+\frac{10800}{r^4}\right)\\ \text{At }\frac{\,\mathrm{d} A}{\,\mathrm{d} r}=0\text{ and }r=(1800)^{1/6},\quad\frac{\,\mathrm{d}^2A}{\,\mathrm{d} r^2}&=\frac{\pi^2}{A}\left(6r^2+\frac{10800}{r^4}\right)>0\quad\text{(minimum)} \end{aligned}$$ $r=(1800)^{1/6}$ gives the minimum $A$.

Method 1⃝: $$\begin{aligned} \frac rh&=\frac{r}{\left(\frac{60}{r^2}\right)}\\ &=\frac{r^3}{60}\\ &=\frac{\left(1800^{1/6}\right)^3}{60}\\ &=\frac{\sqrt{1800}}{60}\\ &=\frac{30\sqrt2}{60}\\ &=\frac{\sqrt2}{2},\quad\text{where }k=\frac12. \end{aligned}$$

Method 2⃝: $$\begin{aligned} h&=\frac{60}{r^2}\\ &=\frac{60}{\left(1800^{1/6}\right)^2}\\ &=\frac{60}{1800^{1/3}},\\[2pt] \frac rh&=\frac{1800^{1/6}}{\dfrac{60}{1800^{1/3}}}\\ &=\frac{\sqrt{1800}}{60}\\ &=\frac{30\sqrt2}{60}\\ &=\frac{\sqrt2}{2},\quad\text{where }k=\frac12. \end{aligned}$$

(d)

Let $V$ be the volume of water when the radius of the water surface is $x$ cm and the depth is $y$ cm.

By similar triangles,

$$\begin{aligned} \frac{x}{y} &= \frac{r}{h} = \frac{\sqrt{2}}{2}\\ x&=\frac{\sqrt2}{2}y. \end{aligned}$$

$$\begin{aligned} V &= \frac{1}{3}\pi x^2 y = \frac{1}{3}\pi\left(\frac{\sqrt{2}}{2}y\right)^{\!2} y = \frac{1}{6}\pi y^3. \end{aligned}$$ $$\begin{aligned} \frac{\,\mathrm{d} V}{\,\mathrm{d} y} &= \frac{1}{2}\pi y^2 \end{aligned}$$ $$\begin{aligned} \frac{\,\mathrm{d} V}{\,\mathrm{d} t} &= \frac{\,\mathrm{d} V}{\,\mathrm{d} y}\times\frac{\,\mathrm{d} y}{\,\mathrm{d} t} = \frac{1}{2}\pi y^2\frac{\,\mathrm{d} y}{\,\mathrm{d} t}. \end{aligned}$$ When $y = 2$, $$\begin{aligned} -3 &= \frac{1}{2}\pi(2)^2\frac{\,\mathrm{d} y}{\,\mathrm{d} t}\\ \frac{\,\mathrm{d} y}{\,\mathrm{d} t} &= -\frac{3}{2\pi}. \end{aligned}$$ Rate of decrease of depth of water is $\dfrac{3}{2\pi}$ cm/s or $0.477$ cm/s (3 s.f.)