**Soalan 7:**Diberi persamaan suatu lengkung ialah

*y*= 2

*x*(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)**

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

**(b)**

$\begin{array}{l}\text{Persamaannormalkepadalengkung:}\\ y-{y}_{1}=-\frac{1}{\frac{dy}{dx}}\left(x-{x}_{1}\right)\\ y-4=-\frac{1}{18}\left(x-2\right)\\ 18y-72=-x+2\\ x+18y=74\end{array}$

**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)**

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

**(b)**

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