Sec 4 A Math: Differentiation: Applications, practice questions & worked solutions

$$\begin{aligned} \frac{dy}{dx} &= 2x - 4\\ 2x - 4 &= 1\\ x &= 2.5, y = 3.25\\ y - 3.25 &= 1(x - 2.5)\\ y &= x + 0.75 \end{aligned}$$

(a)

$$\begin{aligned} \frac{\mathrm{d}V}{\mathrm{d}h} &= \frac{2(5+2h)^2}{\pi}\\[4pt] \frac{\mathrm{d}h}{\mathrm{d}t} &= \frac{\mathrm{d}h}{\mathrm{d}V}\times\frac{\mathrm{d}V}{\mathrm{d}t}\\[4pt] &= \frac{\pi}{2(5+2h)^2}\times 42\\[4pt] &= \frac{21\pi}{(5+2h)^2} \end{aligned}$$

(b)

Method 1: As $h$ increases, $(5+2h)^2$ increases. Since $21\pi$ is a constant, $\dfrac{\mathrm{d}h}{\mathrm{d}t}=\dfrac{21\pi}{(5+2h)^2}$ will decrease.

Method 2: $$\begin{aligned} \frac{\mathrm{d}^2h}{\mathrm{d}t^2} &= -\frac{84\pi}{(5+2h)^3}\\ h>0 \;&\Rightarrow\; (5+2h)^3>0\\ &\Rightarrow\; -(5+2h)^3<0\\ &\Rightarrow\; \frac{\mathrm{d}^2h}{\mathrm{d}t^2}<0\\ &\Rightarrow\; \frac{\mathrm{d}h}{\mathrm{d}t}\text{ will decrease} \end{aligned}$$

(a)

$$\begin{aligned} V &= \frac{1}{3}\pi r^2\left(\frac{3}{4}r\right)\\ &= \frac{\pi}{4}r^3 \end{aligned}$$

(b)

$$\begin{aligned} \frac{dV}{dr} &= \frac{3\pi}{4}r^2\\ \frac{dV}{dt} &= \frac{dV}{dr} \times \frac{dr}{dt}\\ 16 &= \frac{3\pi}{4}(8)^2 \times \frac{dr}{dt}\\ \frac{dr}{dt} &= \frac{1}{3\pi} = 0.106\text{ cm/s} \end{aligned}$$

$$\begin{aligned} A &= \pi r^2\\ \frac{dA}{dr} &= 2\pi r\\ A=225\pi &= \pi r^2\\ r &= 15\\ \frac{dA}{dt} &= \frac{dA}{dr}\times\frac{dr}{dt}\\ \frac{dA}{dt} &= 2\pi(15)\times-0.2\\ \frac{dA}{dt} &= -6\pi\ \text{cm}^2/\text{s} \end{aligned}$$

$$\begin{aligned} y &= xe^{-3x}\\ \frac{\mathrm{d}y}{\mathrm{d}x} &= e^{-3x}(1)+x(-3e^{-3x})\\ &= e^{-3x}(1-3x)\\ e^{-3x}(1-3x) &= 0\\ x &= \frac{1}{3}\\ y &= \frac{1}{3e}\ \text{or}\ 0.123\\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= (1-3x)(-3e^{-3x})+e^{-3x}(-3) \end{aligned}$$ When $x=\dfrac{1}{3}$, $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}<0$.

$\left(\dfrac{1}{3},\,-\dfrac{1}{3e}\right)$ is a maximum point.

(a)

$$\begin{aligned} y &= 5(3x-1)^{-2}\\ \frac{\mathrm{d}y}{\mathrm{d}x} &= 5(-2)(3x-1)^{-3}(3)\\ &= \frac{-30}{(3x-1)^3} \end{aligned}$$ when $x=1$, $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{-30}{8}$. $$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}t} &= \frac{-30}{8}(-0.1)\\ &= \frac{3}{8}\ \text{units/s} \end{aligned}$$

(b)

$$\begin{aligned} \frac{3}{125} &= \frac{3}{(3x-1)^3}\\ (3x-1)^3 &= 125\\ 3x-1 &= 5\\ x &= 2 \end{aligned}$$

(a)

$$\begin{aligned} \text{External surface area, } \pi r^2+\frac{1}{2}(2\pi ry) &= 8500\\ \pi r^2+\pi ry &= 8500\\ y &= \frac{8500-\pi r^2}{\pi r} \end{aligned}$$ $$\begin{aligned} \text{Volume of the container, } V &= \frac{1}{2}\pi r^2 y\\ &= \frac{1}{2}\pi r^2\!\left(\frac{8500-\pi r^2}{\pi r}\right)\\ &= 4250r-\frac{1}{2}\pi r^3 \end{aligned}$$

(b)

$$\begin{aligned} V &= 4250r-\frac{1}{2}\pi r^3\\ \frac{\mathrm{d}V}{\mathrm{d}r} &= 4250-\frac{3}{2}\pi r^2 \end{aligned}$$ For stationary value of $V$, $\dfrac{\mathrm{d}V}{\mathrm{d}r}=0$: $$\begin{aligned} 4250-\frac{3}{2}\pi r^2 &= 0\\ \frac{3}{2}\pi r^2 &= 4250\\ r &= \sqrt{\frac{8500}{3\pi}}\ (r>0)\\ &= 30.031\ \text{cm (to 5 s.f.)} \end{aligned}$$ $$\begin{aligned} \frac{\mathrm{d}^2V}{\mathrm{d}r^2} &= -3\pi r \end{aligned}$$ When $r=30.031$, $\dfrac{\mathrm{d}^2V}{\mathrm{d}r^2}=-3\pi(30.031)<0$.

When $r=30.031$, $V$ is maximum. $$\begin{aligned} \text{Maximum } V &= 4250(30.031)-\frac{1}{2}\pi(30.031)^3\\ &= 85\,100\ \text{cm}^3\ (\text{to 3 s.f.}) \end{aligned}$$

(a)

When $x = h$, $y = \sqrt{3 - h}$ $$A = h\sqrt{3 - h}$$ $$\begin{aligned} \frac{dA}{dh} &= (h)\left(\tfrac{1}{2}\right)(3 - h)^{-\frac{1}{2}}(-1) + (3 - h)^{\frac{1}{2}}(1)\\ &= \frac{-h}{2\sqrt{3 - h}} + \sqrt{3 - h}\\ &= \frac{-h + 2(3 - h)}{2\sqrt{3 - h}}\\ \frac{dA}{dh} &= \frac{6 - 3h}{2\sqrt{3 - h}} \end{aligned}$$

(b)

$$\begin{aligned} \frac{dA}{dh} = \frac{6 - 3h}{2\sqrt{3 - h}} &= 0\\ 6 - 3h &= 0\\ h &= 2\\ A = 2\sqrt{3 - 2} &= 2 \text{ units}^2 \end{aligned}$$

Or

$$\frac{d^2 A}{dh^2} = \frac{2\sqrt{3 - h}(-3) - (6 - 3h)\left(\tfrac{1}{2}\right)(3 - h)^{-1/2}(-1)}{4(3 - h)} = -\frac{3}{2}$$ At $h = 2$, $\dfrac{d^2 A}{dh^2} < 0$

Therefore Stationary value $A = 2$ is a maximum value.

(a)

$$\begin{aligned} A &= 6x^2\\ \frac{dA}{dx} &= 12x\\ \frac{dA}{dt} &= \frac{dA}{dx} \times \frac{dx}{dt}\\ 0.2 &= (12x)\frac{dx}{dt}\\ \frac{dx}{dt} &= \frac{1}{60x}\\ V &= x^3\\ \frac{dV}{dx} &= 3x^2\\ \frac{dV}{dt} &= \frac{dV}{dx} \times \frac{dx}{dt}\\ \frac{dV}{dt} &= 3x^2\left(\frac{1}{60x}\right)\\ \frac{dV}{dt} &= \frac{x}{20} \end{aligned}$$ Subst. $x = 1$, $$\begin{aligned} \frac{dV}{dt} &= \frac{1}{20}\\ \frac{dV}{dt} &= 0.05\ \text{cm}^3/\text{s} \end{aligned}$$

(b)

$$\begin{aligned} y &= e^{1 - 8x}(\cos 2x)\\ \frac{dy}{dx} &= e^{1 - 8x}(-2\sin 2x) + (\cos 2x)(-8)e^{1 - 8x}\\ &= -2e^{1 - 8x}(\sin 2x + 4\cos 2x) \end{aligned}$$ When $y = 0$, $$\begin{aligned} e^{1 - 8x}(\cos 2x) &= 0\\ e^{1 - 8x} = 0 \ (\text{NA}) \ &\text{or} \ \cos 2x = 0\\ 2x &= \frac{\pi}{2}, \frac{3\pi}{2}\\ x &= \frac{\pi}{4}, \frac{3\pi}{4} \ (\text{NA})\\ \frac{dy}{dx} &= -2e^{1 - 8\left(\frac{\pi}{4}\right)}\left(\sin 2\left(\frac{\pi}{4}\right) + 2\cos 2\left(\frac{\pi}{4}\right)\right)\\ \frac{dy}{dx} &= -2e^{1 - 2\pi} \end{aligned}$$

(a)

$$\begin{aligned} 4x - 4y &= 3\\ y &= x - \frac{3}{4}\\ m \text{ of normal} &= 1\\ m \text{ of tangent} &= -1 \end{aligned}$$ At $x = \dfrac{1}{2}$ and $\dfrac{dy}{dx} = -1$, $$\begin{aligned} 6\!\left(\frac{1}{2}\right) - \frac{1}{k\!\left(\frac{1}{2}\right)^3} &= -1\\ 3 - \frac{8}{k} &= -1\\ k &= 2 \end{aligned}$$

(b)

At stat points, $$\begin{aligned} 6x - \frac{1}{2x^3} &= 0\\ 6x &= \frac{1}{2x^3}\\ x^4 &= \frac{1}{12}\\ x &= \pm\sqrt[4]{\frac{1}{12}}\\ x &= -0.537 \ \text{or} \ 0.537 \end{aligned}$$

(c)

$$\begin{aligned} \frac{d^2 y}{dx^2} &= 6 + \frac{3}{2x^4}\\ 6 &> 0\\ \frac{3}{2x^4} &> 0\\ \therefore \frac{d^2 y}{dx^2} = 6 + \frac{3}{2x^4} &> 0 \end{aligned}$$ Since $\dfrac{d^2 y}{dx^2} = 6 + \dfrac{3}{2x^4} > 0$, the gradient has no turning point.

(a)

$$\frac{dy}{dx} = x^4 + 9x^2 - 10$$ For stationary points, $$\begin{aligned} \frac{dy}{dx} &= 0\\ x^4 + 9x^2 - 10 &= 0\\ \left(x^2 + 10\right)\left(x^2 - 1\right) &= 0\\ x = 1 \ \text{or} \ x = -1 \ &\text{or} \ x^2 = -10\\ &\ (\text{rej.\ as } x^2 \neq -10) \end{aligned}$$ The $x$-coordinates are $x = 1$ and $x = -1$.

(b)

$$\begin{aligned} x = 1 \ &\text{or} \ x = -1\\ y = -6.8 \ &\text{or} \ 6.8 \end{aligned}$$ $$\begin{aligned} \text{Midpoint of } AB &= \left(\frac{1 - 1}{2},\ \frac{-6.8 + 6.8}{2}\right)\\ &= (0, 0) \end{aligned}$$ Let $m = x^4 + 9x^2 - 10$ $$\begin{aligned} \frac{dm}{dx} &= 4x^3 + 18x \end{aligned}$$ For max./min. gradient, $\dfrac{dm}{dx} = 0$. $$\begin{aligned} 2x\left(2x^2 + 9\right) &= 0\\ x = 0 \ \text{or} \ 2x^2 + 9 &= 0 \ (\text{rej.\ as } 2x^2 \neq -9)\\ \Rightarrow y &= 0 \end{aligned}$$ $$\frac{d^2 m}{dx^2} = 12x^2 + 18$$ Subt. $x = 0$, $\dfrac{d^2 m}{dx^2} = 18 > 0$

Therefore, $P = (0, 0)$ is a point on the curve where the gradient is a minimum.

Since the midpoint of $AB = (0, 0)$, $P$ and the midpoint of $AB$ are the same.

(a)

$$\begin{aligned} \mathrm{f}(x) &= x^2 e^{(x+2)}\\ \mathrm{f}'(x) &= 2x e^{(x+2)} + x^2 e^{(x+2)}\\ &= x e^{(x+2)}(2 + x)\\ &= x(x + 2)e^{(x+2)} \end{aligned}$$ For decreasing function, $$\begin{aligned} \mathrm{f}'(x) &< 0\\ x(x + 2)e^{(x+2)} &< 0 \end{aligned}$$ since $e^{(x+2)} > 0$ $$x(x + 2) < 0$$

$$-2 < x < 0$$

(b)

$$\begin{aligned} \mathrm{f}'(x) &= (x^2 + 2x)e^{(x+2)}\\ \mathrm{f}''(x) &= (2x + 2)e^{(x+2)} + (x^2 + 2x)e^{(x+2)}\\ \mathrm{f}''(x) &= (x^2 + 4x + 2)e^{(x+2)} \end{aligned}$$ For the least gradient, $$\begin{aligned} \mathrm{f}''(x) &= 0\\ (x^2 + 4x + 2)e^{(x+2)} &= 0\\ x^2 + 4x + 2 &= 0\\ x &= \frac{-4 \pm \sqrt{4^2 - 4(1)(2)}}{2(1)}\\ &= \frac{-4 \pm \sqrt{8}}{2}\\ &= \frac{-4 \pm 2\sqrt{2}}{2}\\ &= -2 \pm \sqrt{2} \end{aligned}$$ Since the least gradient lies in $-2 < x < 0$, $$x = -2 + \sqrt{2}$$ Least gradient $$\begin{aligned} \mathrm{f}'(x) &= x(x + 2)e^{(x+2)}\\ \mathrm{f}'\!\left(-2 + \sqrt{2}\right) &= \left(-2 + \sqrt{2}\right)\left(-2 + \sqrt{2} + 2\right)e^{\left(-2 + \sqrt{2} + 2\right)}\\ &= \left(-2 + \sqrt{2}\right)\sqrt{2}\,e^{\sqrt{2}}\\ &= \left(2 - 2\sqrt{2}\right)e^{\sqrt{2}} \end{aligned}$$