हल

दिए गए समीकरण $x=2 a t^{2}$ और $y=a t^{4}$ हैं

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

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

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

हल

दिए गए समीकरण $x=a \cos \theta$ और $y=b \cos \theta$ हैं

तब, $\dfrac{d x}{d \theta}=\dfrac{d}{d \theta}(a \cos \theta)=a(-\sin \theta)=-a \sin \theta$

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

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

हल

दिए गए समीकरण $x=\sin t$ और $y=\cos 2 t$ हैं

तब, $\dfrac{d x}{d t}=\dfrac{d}{d t}(\sin t)=\cos t$

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

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

हल

दिए गए समीकरण $x=4 t$ और $y=\dfrac{4}{t}$ हैं

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

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

Solution

दिए गए समीकरण $x=\cos \theta-\cos 2 \theta$ और $y=\sin \theta-\sin 2 \theta$ हैं

तब, $\dfrac{d x}{d \theta}=\dfrac{d}{d \theta}(\cos \theta-\cos 2 \theta)=\dfrac{d}{d \theta}(\cos \theta)-\dfrac{d}{d \theta}(\cos 2 \theta)$

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

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

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

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

Solution

दिए गए समीकरण $x=a(\theta-\sin \theta)$ और $y=a(1+\cos \theta)$ हैं

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

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

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

Solution

दिए गए समीकरण $x=\dfrac{\sin ^{3} t}{\sqrt{\cos 2 t}}$ और $y=\dfrac{\cos ^{3} t}{\sqrt{\cos 2 t}}$ हैं

$ \begin{aligned} & \text{ तब, } \dfrac{d x}{d t}=\dfrac{d}{d t}[\dfrac{\sin ^{3} t}{\sqrt{\cos 2 t}}] \\ & =\dfrac{\sqrt{\cos 2 t} \cdot \dfrac{d}{d t}(\sin ^{3} t)-\sin ^{3} t \cdot \dfrac{d}{d t} \sqrt{\cos 2 t}}{\cos 2 t} \\

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

हल

दिए गए समीकरण $x=a(\cos t+\log \tan \dfrac{t}{2})$ और $y=a \sin t$ हैं

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

हल

दिए गए समीकरण $x=a \sec \theta$ और $y=b \tan \theta$ हैं

तब, $\dfrac{d x}{d \theta}=a \cdot \dfrac{d}{d \theta}(\sec \theta)=a \sec \theta \tan \theta$

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

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

हल

दिए गए समीकरण $x=a(\cos \theta+\theta \sin \theta)$ और $y=a(\sin \theta-\theta \cos \theta)$ हैं

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

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

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

हल

दिए गए समीकरण $x=\sqrt{a^{\sin ^{-1} t}}$ और $y=\sqrt{a^{\cos ^{-1} t}}$ हैं

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

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

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

मान लीजिए $x=a^{\dfrac{1}{2} \sin ^{-1} t}$

दोनों ओर लघुगणक लेने पर, हम प्राप्त करते हैं

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

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

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

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

अब, मान लीजिए $y=a^{\dfrac{1}{2} \cos ^{-1} t}$

दोनों ओर लघुगणक लेने पर, हम प्राप्त करते हैं

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

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

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

$\Rightarrow \dfrac{d y}{d t}=\dfrac{-y \log a}{2 \sqrt{1-t^{2}}}$

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

इसलिए, सिद्ध कर दिया गया है।