$\begin{array}{l}y=4\sqrt{x^3}=4{\left(x^3\right)}^{\frac{1}{2}}=4x^{\frac{3}{2}}\\ \frac{dy}{dx}=\frac{3}{2}\left(4x^{\frac{3}{2}-1}\right)\\ =6x^{\frac{1}{2}}=6\sqrt{x}\end{array}$

$\begin{array}{l}\text{Apabila }x=-2,\\ \frac{d^{2}y}{d{x}^2}=12\left(-2\right)+6=-18<0\text{ (negatif)}\\ \text{Oleh itu, titik pusingan }\left(\text{-2, 27}\right)\text{ adalah titik maksimum}.\end{array}$

Soalan 9 (6 markah):
Rajah 4 menunjukkan pandangan hadapan sebahagian daripada laluan ‘roller coaster’ di sebuah taman replika.

Bahagian lengkung laluan ‘roller coaster’ itu diwakili oleh persamaan $y=\frac{1}{64}{x}^3-\frac{3}{16}{x}^2$ , dengan titik A sebagai asalan.
Cari jarak tegak terpendek, dalam m, dari laluan itu ke aras tanah.


Penyelesaian:
$\begin{array}{l}y=\frac{1}{64}{x}^3-\frac{3}{16}{x}^2\mathrm{...............}\left(1\right)\\ \frac{dy}{dx}=3\left(\frac{1}{64}\right)x^2-2\left(\frac{3}{16}\right)x^1\\ =\frac{3}{64}{x}^2-\frac{3}{8}x\\ \\ \text{Pada titik pusingan, }\frac{dy}{dx}=0\\ \frac{3}{64}{x}^2-\frac{3}{8}x=0\\ x\left(\frac{3}{64}x-\frac{3}{8}\right)=0\\ x=0\text{ atau}\\ \frac{3}{64}x-\frac{3}{8}=0\\ \frac{3}{64}x=\frac{3}{8}\\ x=\frac{3}{8}\times \frac{64}{3}\\ x=8\\ \\ \text{Gantikan nilai-nilai }x\text{ ke dalam (1):}\\ \text{Apabila }x=0,\\ y=\frac{1}{64}{\left(0\right)}^3-\frac{3}{16}{\left(0\right)}^2\\ y=0\\ \\ \text{Apabila }x=8,\\ y=\frac{1}{64}{\left(8\right)}^3-\frac{3}{16}{\left(8\right)}^2\\ y=-4\\ \\ \text{Oleh itu, titik-titik pusingan : }\left(0,\text{ 0}\right)\text{ dan }\left(8,-4\right)\end{array}$

$\begin{array}{l}\frac{dy}{dx}=\frac{3}{64}{x}^2-\frac{3}{8}x\\ \frac{d^{2}y}{d{x}^2}=2\left(\frac{3}{64}\right)x-\frac{3}{8}\\ =\frac{3}{32}x-\frac{3}{8}\\ \\ \text{Apabila }x=0,\\ \frac{d^{2}y}{d{x}^2}=\frac{3}{32}\left(0\right)-\frac{3}{8}\\ =-\frac{3}{8}\left(<0\right)\\ \left(0,0\right)\text{ ialah titik maksimum}\text{.}\\ \\ \text{Apabila }x=8,\\ \frac{d^{2}y}{d{x}^2}=\frac{3}{32}\left(8\right)-\frac{3}{8}\\ =\frac{3}{8}\left(>0\right)\\ \left(8,-4\right)\text{ ialah titik minimum}\text{.}\\ \\ \text{Jarak tegak terpendek dari laluan ke aras tanah}\\ \text{adalah pada titik minimum}\text{.}\\ \text{Jarak tegak terpendek}\\ =5-4\\ =1\text{ m}\end{array}$

Soalan 7:
Diberi persamaan suatu lengkung ialah y = 2x (1 – x)⁴ dan lengkung itu melalui T (2, 4).
Cari
(a) kecerunan lengkung pada titik T.
(b) persamaan garis normal kepada lengkung pada titik T.

Penyelesaian:
(a)
$\begin{array}{l}y=2x{\left(1-x\right)}^4\\ \frac{dy}{dx}=2x\times 4{\left(1-x\right)}^3\left(-1\right)+{\left(1-x\right)}^4\times 2\\ =-8x{\left(1-x\right)}^3+2{\left(1-x\right)}^4\\ \\ \text{Pada }T\left(2,4\right),x=2.\\ \frac{dy}{dx}=-16\left(-1\right)+2\left(1\right)\\ =16+2\\ =18\end{array}$

(b)
$\begin{array}{l}\text{Persamaan normal kepada lengkung:}\\ y-y_1=-\frac{1}{\frac{dy}{dx}}\left(x-x_1\right)\\ y-4=-\frac{1}{18}\left(x-2\right)\\ 18y-72=-x+2\\ x+18y=74\end{array}$

Soalan 8 (7 markah):
Diberi bahawa persamaan suatu lengkung ialah $y=\frac{5}{x^2}.$  
(a) Cari nilai $\frac{dy}{dx}$  apabila x = 3.
(b) Seterusnya, anggarkan nilai bagi $\frac{5}{{\left(2.98\right)}^2}.$  

Penyelesaian:
(a)
$\begin{array}{l}y=\frac{5}{x^2}=5x^{-2}\\ \frac{dy}{dx}=-10x^{-3}\\ =-\frac{10}{x^3}\\ \text{Apabila }x=3\\ \frac{dy}{dx}=-\frac{10}{3^3}=-\frac{10}{27}\end{array}$

(b)
$\begin{array}{l}\delta x=2.98-3=-0.02\\ \delta y=\frac{dy}{dx}.\delta x\\ =-\frac{10}{27}\times \left(-0.02\right)\\ =0.007407\\ \\ \text{Nilai bagi }\frac{5}{{\left(2.98\right)}^2}\\ =y+\delta y\\ =\frac{5}{x^2}+\left(0.007407\right)\\ =\frac{5}{3^2}+\left(0.007407\right)\\ =0.56296\end{array}$

Soalan 6:
Diberi persamaan suatu lengkung ialah:
y = x² (x – 3) + 1
(a) Cari kecerunan lengkung itu apabila x = –1.
(b) Cari koordinat-koordinat titik pusingan.

Penyelesaian:
(a)
$\begin{array}{l}y=x^2\left(x-3\right)+1\\ y=x^3-3x^2+1\\ \frac{dy}{dx}=3x^2-6x\\ \text{Apabila }x=-1\\ \frac{dy}{dx}=3{\left(-1\right)}^2-6\left(-1\right)\\ =9\\ \text{Keceruan lengkung ialah 9}\text{.}\end{array}$

(b)
Pada titik pusingan, $\frac{dy}{dx}=0$
3x² – 6x = 0
x² – 2x = 0
x (x – 2) = 0
x = 0, 2

y = x² (x – 3) + 1
Apabila x = 0, y = 1
Apabila x = 2,
y = 2² (2 – 3) + 1
y = 4 (–1) + 1 = –3
Maka, koordinat bagi titik-titik pusingan ialah (0, 1) dan (2, –3).

$\begin{array}{l}\frac{dA}{dr}=2\pi r\\ \\ \text{Perubahan kecil dalam luas kepada jejari,}\\ \frac{\delta A}{\delta r}\approx \frac{dA}{dr}\\ \delta A=\frac{dA}{dr}\times \delta r\end{array}$
δA = 0.08π cm²

Soalan 22 (3 markah):
Cari nilai bagi
$\begin{array}{l}\left(\text{a}\right)\underset{x\to 1}{\text{had}}\left(7-x^2\right),\\ \left(\text{b}\right)f\text{'}\text{'}\left(2\right)\text{ jika }f\text{'}\left(x\right)=2x^3-4x+3.\end{array}$

Penyelesaian:
(a)
$\begin{array}{l}\underset{x\to 1}{had}\left(7-x^2\right)\\ =7-{\left(1\right)}^2\\ =6\end{array}$

(b)
$\begin{array}{l}f\text{'}\left(x\right)=2x^3-4x+3\\ f\text{'}\text{'}\left(x\right)=6x^2-4\\ f\text{'}\text{'}\left(2\right)=6{\left(2\right)}^2-4\\ =24-4\\ =20\end{array}$

Soalan 23 (4 markah):
Diberi bahawa L = 4t – t² dan x = 3 + 6t.
(a) Ungkapkan $\frac{dL}{dx}$  dalam sebutan t.
(b) Cari perubahan kecil bagi x, apabila L berubah daripada 3 kepada 3.4 pada ketika t = 1.

Penyelesaian:
(a)
$\begin{array}{l}\text{Diberi }L=4t-t^2\text{ dan }x=3+6t\\ L=4t-t^2\\ \frac{dL}{dt}=4-2t\\ \\ x=3+6t\\ \frac{dx}{dt}=6\\ \\ \frac{dL}{dx}=\frac{dL}{dt}\times \frac{dt}{dx}\\ \frac{dL}{dx}=\left(4-2t\right)\times \frac{1}{6}\\ =\frac{4-2t}{6}\\ =\frac{2-t}{3}\end{array}$


(b)
$\begin{array}{l}\delta L=3.4-3=0.4\\ \frac{\delta L}{\delta x}\approx \frac{dL}{dx}\\ \delta x=\delta L÷\frac{\delta L}{\delta x}\\ \delta x=\delta L\times \frac{\delta x}{\delta L}\\ =0.4\times \frac{3}{2-t}\\ =\frac{2}{5}\times \frac{3}{2-t}\\ =\frac{6}{5\left(2-t\right)}\\ \\ \text{Apabila }t=1,\\ \delta x=\frac{6}{5\left(2-1\right)}=\frac{6}{5}\end{array}$