If x=a(1-cos θ), y=a(θ-sin θ) , then fracd^2 yd x^2=

Correct answer: (B) fraccosec θ2 a cos^2 ((θ)/2 ). (d x)/(d θ)=a sin θ, (d y)/(d θ)=a(1-cos θ) (d y)/(d x)=(a(1-cos θ))/(a sin θ)=frac2 sin^2 (θ)/22 sin (θ)/2 cos (θ)/2=tan (θ)/2 fracd^2 yd x^2=1/2 sec^2 (θ)/2 · (d θ)/(d x) =1/2 sec^2 (θ)/2 × 1/(a sin θ)=fraccosec θ2 cos^2 (θ)/2

Q57 / 150 · MHT CET 2020 (15 Oct Shift 1) · Mathematics

If $x=a(1-\cos \theta), \quad y=a(\theta-\sin \theta)$, then $\frac{d^{2} y}{d x^{2}}=$

A
$\frac{\cos ^{2}\left(\frac{\theta}{2}\right)}{2 \operatorname{acosec} \theta}$
Correct answer
B
$\frac{\operatorname{cosec} \theta}{2 a \cos ^{2}\left(\frac{\theta}{2}\right)}$
C
$\frac{\cos \left(\frac{\theta}{2}\right)}{2 \operatorname{asin} \theta}$
D
$\frac{\sin \left(\frac{\theta}{2}\right)}{2 \mathrm{a} \cos \theta}$

Solution

$\frac{d x}{d \theta}=a \sin \theta, \frac{d y}{d \theta}=a(1-\cos \theta)$ $\frac{d y}{d x}=\frac{a(1-\cos \theta)}{a \sin \theta}=\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\tan \frac{\theta}{2}$ $\frac{d^{2} y}{d x^{2}}=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \cdot \frac{d \theta}{d x}$ $=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \times \frac{1}{a \sin \theta}=\frac{\operatorname{cosec} \theta}{2 \cos ^{2} \frac{\theta}{2}}$

Hence, the correct answer is option: 'B'.

If $x=a(1-\cos \theta), \quad y=a(\theta-\sin \theta)$, then $\frac{d^{2} y}{d x^{2}}=$

Solution

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