KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.5 (Q10-Q12)

Exercise 11.5

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Question 10

यदि(If) $x=\frac{1-t^{2}}{1+t^{2}}, y=\frac{2 t}{1+t^{2}}$ , सिद्ध करे कि (prove that) $\frac{d y}{d x}+\frac{x}{y}=0$
Sol :
$x=\frac{1-t^{2}}{1+t^{2}}$

Differentiating w.r.t t

$\frac{d x}{d t}=\frac{-2 t\left(1+t^{2}\right)-\left(1-t^{2}\right) \cdot 2t}{\left(1+t^{2}\right)^{2}}$

$=\frac{-2 t-2 t^{3}-2 t+2t^{3}}{\left(1+t^{2}\right)^{2}}$

$\frac{d x}{d t}=\frac{-4 t}{\left(1+t^{2}\right)^{2}}$..(i)

अब , $y=\frac{2 t}{1+t^{2}}$

Differentiating w.r.t t

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

$=\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}}$

$\frac{d y}{d t}=\frac{2-2 t^{2}}{\left(1+t^{2}\right)^{2}}$

$\frac{d y}{d t}=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d{y}}{d t}}{\frac{dx}{d t}}=\dfrac{\frac{2(1-t^2)}{(1+t^2)^2}}{\frac{-4t}{(1+t^2)^2}}$

$\frac{d y}{d x}=-\frac{1-t^{2}}{2 t}$

$\frac{d y}{d x}=-\frac{x}{y}$

$\frac{dy}{d x}+\frac{x}{y}=0$

Question 11

यदि(If) y=acos²θ,x=bsin²θ , सिद्ध करे कि (prove that) $\frac{d y}{d x}+\frac{a}{b}=0$
Sol :
y=acos²θ

Differentiating w.r.t θ

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

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

अब , x=bsin²θ

Differentiating w.r.t θ

$\frac{d x}{d \theta}=2b \sin \theta \cdot \cos \theta$

$\frac{d x}{d \theta}=b\sin2 \theta$

समीकरण (i) मे (ii) से भाग देने पर ,

$\frac{\frac{d y}{d\theta}}{\frac{d x}{d\theta}}=\frac{-a \sin 2\theta}{b \sin 2 \theta}$

$\frac{dy}{dx}+\frac{a}{b}=0$

Question 12

$t=\frac{\pi}{3}$ पर $\frac{d y}{d x}$ निकले जब (Find $\frac{d y}{d x}$ at $t=\frac{\pi}{3}$ when) x=a(cost+tsint) and y=a(sint-tcost)
Sol :
x=a(cost+tsint)

Differentiating w.r.t t

$\frac{d x}{d t}=a(-\sin t+1.\sin t+t \cos t)$

$\frac{d x}{d t}=a t \cos t$..(i)

अब,

y=a(sint-tcost)

Differentiating w.r.t t

$\frac{d y}{d t}=a[\cos t-1 \cdot \cos t-t(-\sin t)]$

$\frac{d y}{d t}=a t \sin t$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{dy}{d t}}{\frac{d x}{dt}}=\frac{at\sin t}{a t \cos t}$

$\frac{d y}{d x}=\tan t$

At $t=\frac{\pi}{3}$

$\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{3}}=\tan \frac{\pi}{3}=\sqrt{3}$