Solution

The given equations are $x=2 a t^{2}$ and $y=a t^{4}$

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

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

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

Solution

The given equations are $x=a \cos \theta$ and $y=b \cos \theta$

Then, $\frac{d x}{d \theta}=\frac{d}{d \theta}(a \cos \theta)=a(-\sin \theta)=-a \sin \theta$

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

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d \theta})}{(\frac{d x}{d \theta})}=\frac{-b \sin \theta}{-a \sin \theta}=\frac{b}{a}$

Solution

The given equations are $x=\sin t$ and $y=\cos 2 t$

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

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

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{-2 \sin 2 t}{\cos t}=\frac{-2 \cdot 2 \sin t \cos t}{\cos t}=-4 \sin t$

Solution

The given equations are $x=4 t$ and $y=\frac{4}{t}$

$ \begin{aligned} & \frac{d x}{d t}=\frac{d}{d t}(4 t)=4 \\ & \frac{d y}{d t}=\frac{d}{d t}(\frac{4}{t})=4 \cdot \frac{d}{d t}(\frac{1}{t})=4 \cdot(\frac{-1}{t^{2}})=\frac{-4}{t^{2}} \\ & \therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{(\frac{-4}{t^{2}})}{4}=\frac{-1}{t^{2}} \end{aligned} $

Solution

The given equations are $x=\cos \theta-\cos 2 \theta$ and $y=\sin \theta-\sin 2 \theta$

Then, $\frac{d x}{d \theta}=\frac{d}{d \theta}(\cos \theta-\cos 2 \theta)=\frac{d}{d \theta}(\cos \theta)-\frac{d}{d \theta}(\cos 2 \theta)$

$ =-\sin \theta-(-2 \sin 2 \theta)=2 \sin 2 \theta-\sin \theta $

$\frac{d y}{d \theta}=\frac{d}{d \theta}(\sin \theta-\sin 2 \theta)=\frac{d}{d \theta}(\sin \theta)-\frac{d}{d \theta}(\sin 2 \theta)$

$=\cos \theta-2 \cos 2 \theta$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d \theta})}{(\frac{d x}{d \theta})}=\frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta}$

Solution

The given equations are $x=a(\theta-\sin \theta)$ and $y=a(1+\cos \theta)$

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

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

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d \theta})}{(\frac{d x}{d \theta})}=\frac{-a \sin \theta}{a(1-\cos \theta)}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}=\frac{-\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}=-\cot \frac{\theta}{2}$

Solution

The given equations are $x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}$ and $y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$

$ \begin{aligned} & \text{ Then, } \frac{d x}{d t}=\frac{d}{d t}[\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}] \\ & =\frac{\sqrt{\cos 2 t} \cdot \frac{d}{d t}(\sin ^{3} t)-\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 \times \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 \cos t-\frac{\sin ^{3} t}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t} \\ & =\frac{3 \cos 2 t \sin ^{2} t \cos t+\sin ^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \\ & \frac{d y}{d t}=\frac{d}{d t}[\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}] \\ & =\frac{\sqrt{\cos 2 t} \cdot \frac{d}{d t}(\cos ^{3} t)-\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}{2 \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 \sin 2 t}{\cos 2 t \cdot \sqrt{\cos 2 t}} \end{aligned} $

$ \begin{matrix} \therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})} & =\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t \sin 2 t}{3 \cos 2 t \sin ^{2} t \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 \sin ^{2} t \cos t+\sin ^{3} t(2 \sin t \cos t)} & \\ & =\frac{\sin t \cos t[-3 \cos 2 t \cdot \cos t+2 \cos ^{3} t]}{\sin t \cos t[3 \cos 2 t \sin t+2 \sin ^{3} t]} & { \begin{bmatrix} \cos 2 t=(2 \cos ^{2} t-1), \\ \cos 2 t=(1-2 \sin ^{2} t) \end{bmatrix} } \\ & =\frac{[-3(2 \cos { }^{2} t-1) \cos t+2 \cos ^{3} t]}{[3(1-2 \sin ^{2} t) \sin t+2 \sin ^{3} t]} & { \begin{bmatrix} \cos 3 t=4 \cos ^{3} t-3 \cos ^{3} t \\ \sin 3 t=3 \sin ^{3} t-4 \sin ^{3} t \end{bmatrix} } \\ & =\frac{-4 \cos t+3 \cos t}{3 \sin t-4 \sin ^{3} t} & \end{matrix} $

Solution

The given equations are $x=a(\cos t+\log \tan \frac{t}{2})$ and $y=a \sin t$

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

Solution

The given equations are $x=a \sec \theta$ and $y=b \tan \theta$

Then, $\frac{d x}{d \theta}=a \cdot \frac{d}{d \theta}(\sec \theta)=a \sec \theta \tan \theta$

$\frac{d y}{d \theta}=b \cdot \frac{d}{d \theta}(\tan \theta)=b \sec ^{2} \theta$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d \theta})}{(\frac{d x}{d \theta})}=\frac{b \sec ^{2} \theta}{a \sec \theta \tan \theta}=\frac{b}{a} \sec \theta \cot \theta=\frac{b \cos \theta}{a \cos \theta \sin \theta}=\frac{b}{a} \times \frac{1}{\sin \theta}=\frac{b}{a} cosec \theta$

Solution

The given equations are $x=a(\cos \theta+\theta \sin \theta)$ and $y=a(\sin \theta-\theta \cos \theta)$

Then, $\frac{d x}{d \theta}=a[\frac{d}{d \theta} \cos \theta+\frac{d}{d \theta}(\theta \sin \theta)]=a[-\sin \theta+\theta \frac{d}{d \theta}(\sin \theta)+\sin \theta \frac{d}{d \theta}(\theta)]$ $=a[-\sin \theta+\theta \cos \theta+\sin \theta]=a \theta \cos \theta$

$ \begin{aligned} \frac{d y}{d \theta}=a[\frac{d}{d \theta}(\sin \theta)-\frac{d}{d \theta}(\theta \cos \theta)] & =a[\cos \theta-{\theta \frac{d}{d \theta}(\cos \theta)+\cos \theta \cdot \frac{d}{d \theta}(\theta)}] \\ & =a[\cos \theta+\theta \sin \theta-\cos \theta] \\ & =a \theta \sin \theta \end{aligned} $

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d \theta})}{(\frac{d x}{d \theta})}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta$

Solution

The given equations are $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$

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

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

$\Rightarrow x=a^{\frac{1}{2} \sin ^{-1} t}$ and $y=a^{\frac{1}{2} \cos ^{-1} t}$

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

Taking logarithm on both the sides, we obtain

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

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

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

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

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

Taking logarithm on both the sides, we obtain

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

$\therefore \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{2} \log a \cdot \frac{d}{d t}(\cos ^{-1} t)$

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

$\Rightarrow \frac{d y}{d t}=\frac{-y \log a}{2 \sqrt{1-t^{2}}}$ $\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{(\frac{-y \log a}{2 \sqrt{1-t^{2}}})}{(\frac{x \log a}{2 \sqrt{1-t^{2}}})}=-\frac{y}{x}$.

Hence, proved.