9.8.4 Pembezaan, SPM Praktis (Kertas 2)

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