Q: If $x$ and $y$ are connected parametrically by the equations $x=a \sec \theta, y=b \tan \theta$, without eliminating the parameter, find $\frac{d y}{d x}$

Given, $x=a \sec \theta, y=b \tan \theta$ Then, $$ \begin{aligned} & \frac{d x}{d \theta}=a \cdot \frac{d}{d \theta}(\sec \theta)=a \sec \theta \tan \theta \\ & \frac{d y}{d \theta}=b \cdot \frac{d}{d \theta}(\tan \theta)=b \sec ^{2} \theta \end{aligned} $$ Therefore, $$ \begin{aligned} \frac{d y}{d x} & =\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)} \\ & =\frac{b \sec ^{2} \theta}{a \sec \theta \tan \theta} \\ & =\frac{b}{a} \sec \theta \cot \theta \\ & =\frac{b \cos \theta}{a \cos \theta \sin \theta} \\ & =\frac{b}{a} \times \frac{1}{\sin \theta} \\ & =\frac{b}{a} \operatorname{cosec} \theta \end{aligned} $$ Question 10: If $x$ and $y$ are connected parametrically by the equations $x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)$, without eliminating the parameter, find $\frac{d y}{d x}$ \section*{Solution:} Given, $x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)$ Then, $$ \begin{aligned} \frac{d x}{d \theta} & =a\left[\frac{d}{d \theta} \cos \theta+\frac{d}{d \theta}(\theta \sin \theta)\right] \\ & =a\left[-\sin \theta+\theta \frac{d}{d \theta}(\sin \theta)+\sin \theta \frac{d}{d \theta}(\theta)\right] \\ & =a[-\sin \theta+\theta \cos \theta+\sin \theta] \\ & =a \theta \cos \theta \\ \frac{d y}{d \theta} & =a\left[\frac{d}{d \theta}(\sin \theta)-\frac{d}{d \theta}(\theta \cos \theta)\right]=a\left[\cos \theta-\left\{\theta \frac{d}{d \theta}(\cos \theta)+\cos \theta \cdot \frac{d}{d \theta}(\theta)\right\}\right] \\ & =a[\cos \theta+\theta \sin \theta-\cos \theta] \\ & =a \theta \sin \theta \end{aligned} $$ Therefore, $$ \begin{aligned} \frac{d y}{d x} & =\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)} \\ & =\frac{a \theta \sin \theta}{a \theta \cos \theta} \\ & =\tan \theta \end{aligned} $$

Q: If $x$ and $y$ are connected parametrically by the equations $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$, without eliminating the parameter, find $\frac{d y}{d x}$

Given, $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$ Then, $$ \begin{aligned} \frac{d x}{d t} & =a \cdot\left[\frac{d}{d t}(\cos t)+\frac{d}{d t}\left(\log \tan \frac{t}{2}\right)\right] \\ & =a\left[-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \frac{d}{d t}\left(\tan \frac{t}{2}\right)\right] \\ & =a\left[-\sin t+\cot \frac{t}{2} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{d}{d t}\left(\frac{t}{2}\right)\right] \\ & =a\left[-\sin t+\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} \times \frac{1}{\cos ^{2} \frac{t}{2}} \times \frac{1}{2}\right] \\ & =a\left[-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right] \\ & =a\left(-\sin t+\frac{1}{\sin t}\right) \\ & =a\left(\frac{-\sin t^{2} t+1}{\sin t}\right) \\ & =a\left(\frac{\cos ^{2} t}{\sin t}\right) \\ \frac{d y}{d t} & =a \frac{d}{d t}(\sin t)=a \cos t \end{aligned} $$ Therefore, $\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a \cos t}{\left(a \frac{\cos ^{2} t}{\sin t}\right)}=\frac{\sin t}{\cos t}=\tan t$

Q: If $x$ and $y$ are connected parametrically by the equations $x=a \cos \theta, y=b \cos \theta$, without eliminating the parameter, find $\frac{d y}{d x}$

Given, $x=a \cos \theta, y=b \cos \theta$ Then, $\frac{d x}{d \theta}=\frac{d}{d \theta}(a \cos \theta)=a(-\sin \theta)=-a \sin \theta$ $\frac{d y}{d \theta}=\frac{d}{d \theta}(b \cos \theta)=b(-\sin \theta)=-b \sin \theta$ $\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{-b \sin \theta}{-a \sin \theta}=\frac{b}{a}$

Q: If $x$ and $y$ are connected parametrically by the equations $x=\cos \theta-\cos 2 \theta, y=\sin \theta-\sin 2 \theta$, without eliminating the parameter, find $\frac{d y}{d x}$

Given, $x=\cos \theta-\cos 2 \theta, y=\sin \theta-\sin 2 \theta$ Then, $$ \begin{aligned} & \frac{d x}{d \theta}=\frac{d}{d \theta}(\cos \theta-\cos 2 \theta)=\frac{d}{d \theta}(\cos \theta)-\frac{d}{d \theta}(\cos 2 \theta) \\ & =-\sin \theta-(-2 \sin 2 \theta)=2 \sin 2 \theta-\sin \theta \\ & \frac{d y}{d \theta}=\frac{d}{d \theta}(\sin \theta-\sin 2 \theta)=\frac{d}{d \theta}(\sin \theta)-\frac{d}{d \theta}(\sin 2 \theta) \\ & =\cos \theta-2 \cos 2 \theta \\ & \therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta} \end{aligned} $$

Q: If $x$ and $y$ are connected parametrically by the equations $x=a(\theta-\sin \theta), y=a(1+\cos \theta)$, without eliminating the parameter, find $\frac{d y}{d x}$

Given, $x=a(\theta-\sin \theta), y=a(1+\cos \theta)$ Then, $\frac{d x}{d \theta}=a\left[\frac{d}{d \theta}(\theta)-\frac{d}{d \theta}(\sin \theta)\right]=a(1-\cos \theta)$ $\frac{d y}{d \theta}=a\left[\frac{d}{d \theta}(1)+\frac{d}{d \theta}(\cos \theta)\right]=a[0+(-\sin \theta)]=-a \sin \theta$ $\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{-a \sin \theta}{a(1-\cos \theta)}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}=\frac{-\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}=-\cot \frac{\theta}{2}$

Question 1

If $x$ and $y$ are connected parametrically by the equations $x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$, without eliminating the parameter, find $\frac{d y}{d x}$

Accepted Answer

Given, $x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$

Then,

$\frac{d x}{d t}=\frac{d}{d t}\left[\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}ight]$

$=\frac{\sqrt{\cos 2 t} \cdot \frac{d}{d t}\left(\sin ^{3} tight)-\sin ^{3} t \cdot \frac{d}{d t} \sqrt{\cos 2 t}}{\cos 2 t}$

$=\frac{\sqrt{\cos 2 t} \cdot 3 \sin ^{2} t \cdot \frac{d}{d t}(\sin t)-\sin ^{3} t 	imes \frac{1}{2 \sqrt{\cos 2 t}} \cdot \frac{d}{d t}(\cos 2 t)}{\cos 2 t}$

$=\frac{3 \sqrt{\cos 2 t} \cdot \sin ^{2} t \cdot \cos t-\frac{\sin ^{3} t}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t}$

$=\frac{3 \cos 2 t \cdot \sin ^{2} t \cos t+\sin ^{3} t \cdot \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}}$

$\frac{d y}{d t}=\frac{d}{d t}\left[\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}ight]$

$=\frac{\sqrt{\cos 2 t} \cdot \frac{d}{d t}\left(\cos ^{3} tight)-\cos ^{3} t \cdot \frac{d}{d t}(\sqrt{\cos 2 t})}{\cos 2 t}$

$=\frac{\sqrt{\cos 2 t} \cdot 3 \cos ^{2} t \cdot \frac{d}{d t}(\cos t)-\cos ^{3} t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot \frac{d}{d t}(\cos 2 t)}{\cos 2 t}$

$=\frac{3 \sqrt{\cos 2 t} \cdot \cos ^{2} t(-\sin t)-\cos ^{3} t \cdot \frac{1}{\sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t}$

$=\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t \cdot \sin 2 t}{\cos 2 t \cdot \sqrt{\cos 2 t}}$

$	herefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}ight)}{\left(\frac{d x}{d t}ight)}=\frac{\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t \sin 2 t}{\cos 2 t \cdot \sqrt{\cos 2 t}}}{\frac{3 \cos 2 t \cdot \sin ^{2} t \cdot \cos t+\sin ^{3} t \sin 2 t}{\cos 2 t \cdot \sqrt{\cos 2 t}}}$

$=\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t \sin 2 t}{3 \cos 2 t \cdot \sin ^{2} t \cdot \cos t+\sin ^{3} t \sin 2 t}$

$=\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t(2 \sin t \cos t)}{3 \cos 2 t \cdot \sin ^{2} t \cdot \cos t+\sin ^{3} t(2 \sin t \cos t)}$

$=\frac{\sin t \cos t\left[-3 \cos 2 t \cdot \cos t+2 \cos ^{3} tight]}{\sin t \cos t\left[3 \cos 2 t \sin t+2 \sin ^{3} tight]}$

$=\frac{\left[-3\left(2 \cos ^{2} t-1ight) \cos t+2 \cos ^{3} tight]}{\left[3\left(1-2 \sin ^{2} tight) \sin t+2 \sin ^{3} tight]} \quad\left[\begin{array}{l}\cos 2 t=\left(2 \cos ^{2} t-1ight) \ \cos 2 t=\left(1-2 \sin ^{2} tight)\end{array}ight]$

$=\frac{-4 \cos ^{3} t+3 \cos t}{3 \sin t-4 \sin ^{3} t} \quad\left[\begin{array}{l}\cos 3 t=4 \cos ^{3} t-3 \cos t \ \sin 3 t=3 \sin t-4 \sin ^{2} t\end{array}ight]$

$=\frac{-\cos 3 t}{\sin 3 t}=-\cot 3 t$

Question 2

If $x$ and $y$ are connected parametrically by the equations $x=a \sec 	heta, y=b 	an 	heta$, without eliminating the parameter, find $\frac{d y}{d x}$

Accepted Answer

Given, $x=a \sec 	heta, y=b 	an 	heta$

Then,

$$

\begin{aligned}

& \frac{d x}{d 	heta}=a \cdot \frac{d}{d 	heta}(\sec 	heta)=a \sec 	heta 	an 	heta \

& \frac{d y}{d 	heta}=b \cdot \frac{d}{d 	heta}(	an 	heta)=b \sec ^{2} 	heta

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{\left(\frac{d y}{d 	heta}ight)}{\left(\frac{d x}{d 	heta}ight)} \

& =\frac{b \sec ^{2} 	heta}{a \sec 	heta 	an 	heta} \

& =\frac{b}{a} \sec 	heta \cot 	heta \

& =\frac{b \cos 	heta}{a \cos 	heta \sin 	heta} \

& =\frac{b}{a} 	imes \frac{1}{\sin 	heta} \

& =\frac{b}{a} \operatorname{cosec} 	heta

\end{aligned}

$$

Question 10:

If $x$ and $y$ are connected parametrically by the equations $x=a(\cos 	heta+	heta \sin 	heta), y=a(\sin 	heta-	heta \cos 	heta)$, without eliminating the parameter, find $\frac{d y}{d x}$

\section*{Solution:}

Given, $x=a(\cos 	heta+	heta \sin 	heta), y=a(\sin 	heta-	heta \cos 	heta)$

Then,

$$

\begin{aligned}

\frac{d x}{d 	heta} & =a\left[\frac{d}{d 	heta} \cos 	heta+\frac{d}{d 	heta}(	heta \sin 	heta)ight] \

& =a\left[-\sin 	heta+	heta \frac{d}{d 	heta}(\sin 	heta)+\sin 	heta \frac{d}{d 	heta}(	heta)ight] \

& =a[-\sin 	heta+	heta \cos 	heta+\sin 	heta] \

& =a 	heta \cos 	heta \

\frac{d y}{d 	heta} & =a\left[\frac{d}{d 	heta}(\sin 	heta)-\frac{d}{d 	heta}(	heta \cos 	heta)ight]=a\left[\cos 	heta-\left\{	heta \frac{d}{d 	heta}(\cos 	heta)+\cos 	heta \cdot \frac{d}{d 	heta}(	heta)ight\}ight] \

& =a[\cos 	heta+	heta \sin 	heta-\cos 	heta] \

& =a 	heta \sin 	heta

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{\left(\frac{d y}{d 	heta}ight)}{\left(\frac{d x}{d 	heta}ight)} \

& =\frac{a 	heta \sin 	heta}{a 	heta \cos 	heta} \

& =	an 	heta

\end{aligned}

$$

Question 3

If $x$ and $y$ are connected parametrically by the equations $x=a\left(\cos t+\log 	an \frac{t}{2}ight), y=a \sin t$, without eliminating the parameter, find $\frac{d y}{d x}$

Accepted Answer

Given, $x=a\left(\cos t+\log 	an \frac{t}{2}ight), y=a \sin t$

Then,

$$

\begin{aligned}

\frac{d x}{d t} & =a \cdot\left[\frac{d}{d t}(\cos t)+\frac{d}{d t}\left(\log 	an \frac{t}{2}ight)ight] \

& =a\left[-\sin t+\frac{1}{	an \frac{t}{2}} \cdot \frac{d}{d t}\left(	an \frac{t}{2}ight)ight] \

& =a\left[-\sin t+\cot \frac{t}{2} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{d}{d t}\left(\frac{t}{2}ight)ight] \

& =a\left[-\sin t+\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} 	imes \frac{1}{\cos ^{2} \frac{t}{2}} 	imes \frac{1}{2}ight] \

& =a\left[-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}ight] \

& =a\left(-\sin t+\frac{1}{\sin t}ight) \

& =a\left(\frac{-\sin t^{2} t+1}{\sin t}ight) \

& =a\left(\frac{\cos ^{2} t}{\sin t}ight) \

\frac{d y}{d t} & =a \frac{d}{d t}(\sin t)=a \cos t

\end{aligned}

$$

Therefore,

$\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}ight)}{\left(\frac{d x}{d t}ight)}=\frac{a \cos t}{\left(a \frac{\cos ^{2} t}{\sin t}ight)}=\frac{\sin t}{\cos t}=	an t$

Question 4

If $x$ and $y$ are connected parametrically by the equations $x=2 a t^{2}, y=a t^{4}$, without eliminating the parameter, find $\frac{d y}{d x}$

Accepted Answer

Given, $x=2 a t^{2}, y=a t^{4}$

Then,

$\frac{d x}{d t}=\frac{d}{d t}\left(2 a t^{2}ight)=2 a \cdot \frac{d}{d t}\left(t^{2}ight)=2 a \cdot 2 t=4 a t$

$\frac{d y}{d t}=\frac{d}{d t}\left(a t^{4}ight)=a \cdot \frac{d}{d t}\left(t^{4}ight)=a \cdot 4 \cdot t^{3}=4 a t^{3}$

$	herefore \frac{d y}{d t}=\frac{\left(\frac{d y}{d t}ight)}{\left(\frac{d x}{d t}ight)}=\frac{4 a t^{3}}{4 a t}=t^{2}$

Question 5

If $x$ and $y$ are connected parametrically by the equations $x=a \cos 	heta, y=b \cos 	heta$, without eliminating the parameter, find $\frac{d y}{d x}$

Accepted Answer

Given, $x=a \cos 	heta, y=b \cos 	heta$

Then,

$\frac{d x}{d 	heta}=\frac{d}{d 	heta}(a \cos 	heta)=a(-\sin 	heta)=-a \sin 	heta$

$\frac{d y}{d 	heta}=\frac{d}{d 	heta}(b \cos 	heta)=b(-\sin 	heta)=-b \sin 	heta$

$	herefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d 	heta}ight)}{\left(\frac{d x}{d 	heta}ight)}=\frac{-b \sin 	heta}{-a \sin 	heta}=\frac{b}{a}$

Question 6

If $x$ and $y$ are connected parametrically by the equations $x=\sin t, y=\cos 2 t$, without eliminating the parameter, find $\frac{d y}{d x}$

Accepted Answer

Given, $x=\sin t, y=\cos 2 t$

Then, $\frac{d x}{d t}=\frac{d}{d t}(\sin t)=\cos t$

$\frac{d y}{d t}=\frac{d}{d t}(\cos 2 t)=-\sin 2 t \cdot \frac{d}{d t}(2 t)=-2 \sin 2 t$

$	herefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}ight)}{\left(\frac{d x}{d t}ight)}=\frac{-2 \sin 2 t}{\cos t}=\frac{-2.2 \sin t \cos t}{\cos t}=-4 \sin t$

Question 7

If $x$ and $y$ are connected parametrically by the equations $x=4 t, y=\frac{4}{t}$, without eliminating the parameter, find $\frac{d y}{d x}$

Accepted Answer

Given, $x=4 t, y=\frac{4}{t}$

$\frac{d x}{d t}=\frac{d}{d t}(4 t)=4$

$\frac{d y}{d t}=\frac{d}{d t}\left(\frac{4}{t}ight)=4 \cdot \frac{d}{d t}\left(\frac{1}{t}ight)=4 \cdot\left(\frac{-1}{t^{2}}ight)=\frac{-4}{t^{2}}$

$	herefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}ight)}{\left(\frac{d x}{d t}ight)}=\frac{\left(\frac{-4}{t^{2}}ight)}{4}=\frac{-1}{t^{2}}$

Question 8

If $x$ and $y$ are connected parametrically by the equations $x=\cos 	heta-\cos 2 	heta, y=\sin 	heta-\sin 2 	heta$, without eliminating the parameter, find $\frac{d y}{d x}$

Accepted Answer

Given, $x=\cos 	heta-\cos 2 	heta, y=\sin 	heta-\sin 2 	heta$

Then,

$$

\begin{aligned}

& \frac{d x}{d 	heta}=\frac{d}{d 	heta}(\cos 	heta-\cos 2 	heta)=\frac{d}{d 	heta}(\cos 	heta)-\frac{d}{d 	heta}(\cos 2 	heta) \

& =-\sin 	heta-(-2 \sin 2 	heta)=2 \sin 2 	heta-\sin 	heta \

& \frac{d y}{d 	heta}=\frac{d}{d 	heta}(\sin 	heta-\sin 2 	heta)=\frac{d}{d 	heta}(\sin 	heta)-\frac{d}{d 	heta}(\sin 2 	heta) \

& =\cos 	heta-2 \cos 2 	heta \

& 	herefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d 	heta}ight)}{\left(\frac{d x}{d 	heta}ight)}=\frac{\cos 	heta-2 \cos 2 	heta}{2 \sin 2 	heta-\sin 	heta}

\end{aligned}

$$

Question 9

If $x$ and $y$ are connected parametrically by the equations $x=a(	heta-\sin 	heta), y=a(1+\cos 	heta)$, without eliminating the parameter, find $\frac{d y}{d x}$

Accepted Answer

Given, $x=a(	heta-\sin 	heta), y=a(1+\cos 	heta)$

Then, $\frac{d x}{d 	heta}=a\left[\frac{d}{d 	heta}(	heta)-\frac{d}{d 	heta}(\sin 	heta)ight]=a(1-\cos 	heta)$

$\frac{d y}{d 	heta}=a\left[\frac{d}{d 	heta}(1)+\frac{d}{d 	heta}(\cos 	heta)ight]=a[0+(-\sin 	heta)]=-a \sin 	heta$

$	herefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d 	heta}ight)}{\left(\frac{d x}{d 	heta}ight)}=\frac{-a \sin 	heta}{a(1-\cos 	heta)}=\frac{-2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2 \sin ^{2} \frac{	heta}{2}}=\frac{-\cos \frac{	heta}{2}}{\sin \frac{	heta}{2}}=-\cot \frac{	heta}{2}$

Question 10

If $x=\sqrt{a^{\sin ^{-1} t}}, y=\sqrt{a^{\cos ^{-1} t}}$, show that $\frac{d y}{d x}=-\frac{y}{x}$

Accepted Answer

Given, $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$

Hence,

$x=\sqrt{a^{\sin ^{-1} t}}=\left(a^{\sin ^{-1} t}ight)^{\frac{1}{2}}=a^{\frac{1}{2} \sin ^{-1} t}$ and $y=\sqrt{a^{\cos ^{-1} t}}=\left(a^{\cos ^{-1} t}ight)^{\frac{1}{2}}=a^{\frac{1}{2} \cos ^{-1} t}$

Consider $x=a^{\frac{1}{2} \sin ^{-1} t}$

Taking log on both sides, we get

$\log x=\frac{1}{2} \sin ^{-1} t \log a$

Therefore,

$\Rightarrow \frac{1}{x} \cdot \frac{d x}{d t}=\frac{1}{2} \log a \cdot \frac{d}{d t}\left(\sin ^{-1} tight)$

$\Rightarrow \frac{d x}{d t}=\frac{x}{2} \log a \cdot \frac{1}{\sqrt{1-t^{2}}}$

$\Rightarrow \frac{d x}{d t}=\frac{x \log a}{2 \sqrt{1-t^{2}}}$

Now, $y=a^{\frac{1}{2} \cos ^{-1} t}$

Taking log on both sides, we get

$\log x=\frac{1}{2} \cos ^{-1} t \log a$

Therefore,

$$

\begin{aligned}

& \Rightarrow \frac{1}{y} \cdot \frac{d y}{d t}=\frac{1}{2} \log a \cdot \frac{d}{d t}\left(\cos ^{-1} tight) \

& \Rightarrow \frac{d y}{d t}=\frac{y}{2} \log a \cdot \frac{-1}{\sqrt{1-t^{2}}} \

& \Rightarrow \frac{d y}{d t}=\frac{-y \log a}{2 \sqrt{1-t^{2}}}

\end{aligned}

$$

Hence,

$$

\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}ight)}{\left(\frac{d x}{d t}ight)}=\frac{\left(\frac{-y \log a}{2 \sqrt{1-t^{2}}}ight)}{\left(\frac{x \log a}{2 \sqrt{1-t^{2}}}ight)}=-\frac{y}{x}

$$

\section*{EXERCISE 5.7}