2.7.3 Pembezaan, SPM Praktis (Kertas 1 Soalan 21 – 30)

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

$\begin{array}{l}y=15x+\frac{24}{x^3}\\ y=15x+24x^{-3}\\ \frac{dy}{dx}=15-72x^{-4}\\ \frac{dy}{dx}=15-\frac{72}{x^4}\\ \\ \text{Apabila }x=2\\ \frac{dy}{dx}=15-\frac{72}{2^4}=10.5\end{array}$
δy = 10.5k

$\begin{array}{l}\delta t=\frac{dt}{dx}\times \delta x\\ \delta t=\frac{1}{5}\times \left(5.01-5\right)\\ \delta t=0.002\end{array}$

 

$\frac{dL}{dr}=2\pi$
$\begin{array}{l}2\pi j=88\\ j=\frac{88}{2\pi }=\frac{44}{\pi }\\ \\ \text{Oleh itu, jejari bulatan selepas }5s\\ =\frac{44}{\pi }+5\left(0.0477\right)\\ =14.24\text{ cm}\end{array}$

Soalan 25:
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).

Soalan 26:
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 27 (SPM 2018 - 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 28 (SPM 2018 - 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)
$\begin{array}{l}\delta L=3.4-3=0.4\\ \frac{\delta L}{\delta x}\approx \frac{dL}{dx}\\ \delta x=\delta L÷\frac{\delta L}{\delta x}\\ \delta x=\delta L\times \frac{\delta x}{\delta L}\\ =0.4\times \frac{3}{2-t}\\ =\frac{2}{5}\times \frac{3}{2-t}\\ =\frac{6}{5\left(2-t\right)}\\ \\ \text{Apabila }t=1,\\ \delta x=\frac{6}{5\left(2-1\right)}=\frac{6}{5}\end{array}$