Differentation (Approximate change/ Rate of change)

CIE 2020 /0606/Winter /Paper 11 

4. It is given that $y=\dfrac{\tan 3x}{\sin x}$.

(a) Find the exact value of $\dfrac{dy}{dx}$ when $x=\dfrac{\pi}{3}.$ [4]

(b) Hence find the approximate change in $y$ as $x$ increases from $\dfrac{\pi}{3}$ to $\dfrac{\pi}{3}+h,$ where $h$ is small. [1]

(c) Given that $x$ is increasing at the rate of 3 units per second, find the corresponding rate of change in $y$ when $x =\dfrac{\pi}{3}$, giving your answer in its simplest surd form.[2]


*********math solution*************
(a) $\begin{array}[t]{rcll}\dfrac{\mathrm{d} y}{\mathrm{~d} x} &=&\dfrac{\sin x\dfrac{d}{dx}(\tan 3x)-\tan 3x\dfrac{d}{dx}(\sin x)}{\sin^2x}\\ &=&\dfrac{\sin x \times 3 \mbox{sec }^{2} 3 x-\tan 3 x \cos x}{\sin ^{2} x} \\ \text { When } x=\dfrac{\pi}{3}\\\dfrac{\mathrm{d} y}{\mathrm{~d} x}&=&\dfrac{\sin \left(\dfrac{\pi}{3}\right) \times 3 \mbox{sec }^{2} 3 \left(\dfrac{\pi}{3}\right)-\tan 3 \left(\dfrac{\pi}{3}\right) \cos \left(\dfrac{\pi}{3}\right)}{\sin ^{2} \left(\dfrac{\pi}{3}\right)}\\&=&2 \sqrt{3} \end{array}$

(b) $\Delta y=\dfrac{dy}{dx}\times h=2 \sqrt{3} h $

(c) Since $\dfrac{dx}{dt}=3,$
$\begin{array}{rcl}\dfrac{dy}{dt}&=& \dfrac{\mathrm{d} y}{\mathrm{~d} x} \times \dfrac{\mathrm{d} x}{\mathrm{~d} t} \\&=& 2 \sqrt{3} \times 3=6\sqrt 3\end{array} $

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