KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.5 (Q1-Q3)

Exercise 11.5










Question 1

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

(i) x=acosθ , y=asinθ
Sol :
Differentiate w.r.t θ

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

$\frac{d y}{d \theta}=a \cos \theta$..(ii)

समीकरण (i) मे (ii) से भाग देने पर ,

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

=-cotθ

(ii) x=acosθ , y=bcosθ
Sol :
Differentiate w.r.t θ

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

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

समीकरण (i) मे (ii) से भाग देने पर ,

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

$\frac{d y}{d x}=\frac{b}{a}$


Question 2

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

(i) x=at2,y=2at
Sol :
Differentiate w.r.t t

$\frac{d x}{d t}=2 a t$..(i)

$\frac{d y}{d t}=2 a$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2a}{2at}$

$\frac{d y}{d x}=\frac{1}{t}$



(ii) x=4t , $y=\frac{4}{t}$
Sol :
Differentiate w.r.t t

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

$\frac{d y}{d t}=-\frac{4}{t^{2}}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\dfrac{\frac{d y}{d t}}{\frac{dx}{dt}}=\dfrac{\frac{-4}{t^2}}{4}$

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


Question 3

$\frac{d y}{d x}$ ज्ञात करे जब
[Find $\frac{d y}{d x}$ when]

(i) x=sint,y=cos2t
Sol :
Differentiate w.r.t t

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


$\frac{d y}{d t}=-\sin 2 t \times 2$

$\frac{dy}{d t}=-2 .\sin 2t$

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-2 \cdot \cos t}{\cos t}$

$\frac{d y}{d x}=\frac{-2 \times 2\sin t \cos t}{\cos t}$

⇒-4sint

(ii) x=asecθ,y=btanθ
Sol :

Differentiating w.r.t θ

$\frac{d x}{d \theta}=a \sec \theta \tan \theta$..(i)

$\frac{d y}{d \theta}=b \sec ^{2} \theta$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{b \sec ^{2} \theta}{a \sec \theta \tan \theta}$

$\frac{d y}{d x}=\frac{b}{a}\times \frac{1}{\cos \theta}$

$=\frac{b}{a} \operatorname{cosec} \theta$


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