KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.5 (Q4-Q6)

Exercise 11.5

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Question 4

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

(i) x=a(θ-sinθ), y=a(1-cosθ)
Sol :
Differentiating w.r.t θ

$\frac{d x}{d \theta}=a(1-\cos \theta)$

$\frac{d x}{d \theta}=a \cdot 2 \sin ^{2} \frac{\theta}{2}$

$\frac{d x}{d \theta}=2 a \sin ^{2} \frac{\theta}{2}$..(i)

अब , y=(1-cosθ)

Differentiating w.r.t θ

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

$\frac{d y}{d \theta}=2 a \sin \frac{\theta}{2} \cos \frac{\theta}{2}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{2 a \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 a \sin^2 \frac{\theta}{2}}$

$\frac{d y}{d x}=\cot \frac{\theta}{2}$


(ii) x=a(θ-sinθ),y=a(1+cosθ)
Sol :

Question 5

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

(i) x=sinθ,y=θ+cosθ
Sol :
Differentiating w.r.t θ

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

$\frac{d y}{d \theta}=1-\sin \theta$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d \theta}}{\frac{dx}{d\theta}}=\frac{1-\sin \theta}{\cos \theta}$

$\frac{d y}{d x}=\frac{1-\sin \theta}{\cos \theta}$


(ii) x=10(t-sint),y=12(1-cost),$-\frac{\pi}{2}<t \leq \frac{\pi}{2}$
Sol :
Differentiating w.r.t t

$\frac{d x}{d t}=10(1-\cos t)$

$\frac{d x}{d t}=10 \times 2 \sin ^{2} \frac{t}{2}$

$\frac{d x}{d t}=20 \sin^2 \frac{t}{2}$

अब , y=12(1-cost)

Differentiating w.r.t t

$\frac{d y}{d t}=12 \sin t=12 \times 2 \sin \frac{t}{2} \cos \frac{t}{2}$

$\frac{d y}{d t}=24 \sin \frac{t}{2} \cos \frac{t}{2}$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{24 \sin \frac{t}{2} \cos \frac{t}{2}}{20 \sin ^{2} \frac{t}{2}}$

$\frac{dy}{d t}=\frac{6}{5} \cos \frac{t}{2}$

Question 6

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

(i) x=3cosθ-cos³θ,y=3sinθ-sin³θ
Sol :
Differentiating w.r.t θ

$\frac{d x}{d \theta}=-3 \sin \theta-3 \cos ^{2} \theta(-\sin \theta)$

$\frac{d x}{d \theta}=-3 \sin \theta+3 \cos ^{2} \theta \sin \theta$

$\frac{d x}{d \theta}=-3 \sin \theta\left(1-\cos^{2} \theta\right)$

$\frac{d x}{d t}=-3 \sin^{3} \theta$..(i)

अब, y=3sinθ-sin³θ

Differentiating w.r.t θ

$\frac{d y}{d \theta}=3 \cos \theta-3 \sin ^{2} \theta \cdot \cos \theta$

=3cosθ(1-sin²θ)

$\frac{d y}{d \theta}=3 \cos ^{3} \theta$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 \cos ^{3} \theta}{-3 \sin^3 \theta}$

$\frac{d y}{d x}=-\cot ^{3} \theta$


(ii) x=cosθ-cos2θ,y=sinθ-sin³θ
Sol :
Differentiating w.r.t θ

$\frac{d x}{d \theta}=-\sin \theta+\sin 2 \theta \cdot 2$

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

अब , y=sinθ--sin³θ

Differentiating w.r.t θ

$\frac{d y}{d t}=\cos \theta-3 \sin ^{2} \theta \cdot \cos \theta$

$\frac{d y}{d \theta}=\cos \theta\left(1-3 \sin ^{2} \theta\right)$..(ii)

समीकरण (ii) मे (i) से भाग देने पर ,

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{\cos \theta\left(1-3 \sin ^{2} \theta\right)}{2\sin 2\theta-\sin \theta}$

$\frac{d y}{d x}=\frac{\cos \theta\left(1-3 \sin ^{2} \theta\right)}{2 \sin 2 \theta-\sin \theta}$