Exercise 11.5
Question 7
$\frac{d y}{d x}$ ज्ञात करे जब[Find $\frac{d y}{d x}$ when]
Sol :
$x=\sqrt{1+t^{2}}$
Differentiating w.r.t t
$\frac{d x}{d t}=\frac{1}{2 \sqrt{1+t^2}} \times 2t$
$\frac{d x}{d t}=\frac{t}{\sqrt{1+t^{2}}}$..(i)
अब, $y=\sqrt{1-t^{2}}$
Differentiating w.r.t t
$\frac{d y}{d t}=\frac{1}{2 \sqrt{1-t^{2}}} \times(-2 t)$
$\frac{d y}{d t}=\frac{-t}{\sqrt{1-t^{2}}}$..(ii)
समीकरण (ii) मे (i) से भाग देने पर ,
$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{-t}{\sqrt{1-t^{2}}}}{\frac{t}{\sqrt{1+t^{2}}}}$
$\frac{d y}{d x}=-\frac{\sqrt{1+t^{2}}}{\sqrt{1-t^{2}}}$
(ii) $x=t+\frac{1}{t}, y=t-\frac{1}{t}$
Sol :
$x=t+\frac{1}{t}$
Differentiating w.r.t t
$\frac{d x}{d t}=1-\frac{1}{t^{2}}$
$\frac{d x}{d t}=\frac{t^{2}-1}{t^{2}}$..(i)
अब ,$y=t-\frac{1}{t}$
Differentiating w.r.t t
$\frac{dy}{d t}=1+\frac{1}{t^{2}}$
$\frac{d y}{d t}=\frac{t^{2}+1}{t^{2}}$..(ii)
समीकरण (ii) मे (i) से भाग देने पर ,
$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{t^{2}+1}{t^{2}}}{\frac{t^{2}-1}{t^{2}}}$
$\frac{d y}{d x}-\frac{t^{2}+1}{t^{2}-1}$
Question 8
$\frac{d y}{d x}$ ज्ञात करे जब[Find $\frac{d y}{d x}$ when]
(i) x=et+sint,t=logt
Sol :
Differentiating w.r.t t
$\frac{d x}{d t}=e^{t}+\cos t$..(i)
$\frac{d y}{d t}=\frac{1}{t}$..(ii)
समीकरण (ii) मे (i) से भाग देने पर ,
$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{1}{t}}{e^{t}+\cos t}$
$\frac{d_{y}}{d x}=\frac{1}{t\left(e^{t}+\cos t\right)}$
(ii) $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$
Sol :
Differentiating w.r.t t
$\frac{d x}{d t}=a\left(-\sin t+\frac{1}{\tan \frac{t}{2}} \times \sec ^{2} \frac{t}{2} \times \frac{1}{2}\right)$
$\frac{d x}{d t}=a\left(-\sin t+\frac{1}{\frac{\sin t / 2}{\cos t / 2}} \times \frac{1}{\cos ^{2} \frac{t}{2}} \times \frac{1}{2}\right)$
$\frac{d x}{d t}=a\left(-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right)$
$\frac{d x}{d t}=a\left(-\sin t+\frac{1}{\sin t}\right)$
$\frac{d x}{d t}=a\left(\frac{-\sin ^{2} t+1}{\sin t}\right)$
$\frac{d x}{d t}=a \frac{\cos ^{2} t}{\sin t}$..(i)
अब,y=asint
Differentiating w.r.t t
$\frac{d y}{d t}=a \cos t$..(ii)
समीकरण (ii) मे (i) से भाग देने पर ,
$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a \cos t}{\frac{a \cos^{2} t}{sin t}}$
$\frac{d y}{d x}=\tan t$
Question 9
$\frac{d y}{d x}$ ज्ञात करे जब[Find $\frac{d y}{d x}$ when]
$x=\frac{2 t}{1+t^{2}}, y=\frac{1-t^{2}}{1+t^{2}}$
Differentiating w.r.t t
$\frac{d x}{d t}=\frac{2 \cdot\left(1+t^{2}\right)-2 t(2 t)}{\left(1+t^{2}\right)^{2}}$
$\frac{d x}{d t}=\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}}$
$\frac{d x}{d t}=\frac{2-2 t^{2}}{\left(1+t^{2}\right)^{2}}$
$\frac{d x}{d t}=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}$..(i)
Differentiating w.r.t t
$\frac{d y}{d t}=\frac{-2 t\left(1+t^{2}\right)-\left(1-t^{2}\right) \cdot 2 t}{\left(1+t^{2}\right)^{2}}$
$=\frac{-2 t-2 t^{3}-2 t+2 t^{3}}{\left(1+t^{2}\right)^{2}}$
$\frac{d y}{d t}=-\frac{4 t}{\left(1+t^{2}\right)^{2}}$..(ii)
समीकरण (ii) मे (i) से भाग देने पर ,
$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{-4 t}{\left(1+t^{2}\right)^{2}}}{\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}}$
$\frac{d y}{d x}=\frac{-2 t}{1-t^{2}}$
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