If a≠0, x=a(t+sin⁡t) and y=a(1−cos⁡t), then d2ydx2 at t=2π3 is

x=a(t+sin⁡t), y=a(1−cos⁡t)⇒dxdt=a(1+cos⁡t), dydt=asin⁡tdydx=dy/dtdx/dt=asin⁡ta(1+cos⁡t)=2sin⁡t/2cos⁡t/22cos2⁡t/2=tan⁡t2⇒d2ydx2=sec2⁡t2⋅12dtdx=12cos2⁡t2⋅1a(1+cos⁡t)=12acos2⁡t22cos2⁡t2=14acos4⁡t2⇒d2ydx2t=2π/3=14acos4⁡(π/3)=14a(1/2)4=4a.

If a≠0, x=a(t+sin⁡t) and y=a(1−cos⁡t), then d2ydx2 at t=2π3 is

x=a(t+sin⁡t), y=a(1−cos⁡t)⇒dxdt=a(1+cos⁡t), dydt=asin⁡tdydx=dy/dtdx/dt=asin⁡ta(1+cos⁡t)=2sin⁡t/2cos⁡t/22cos2⁡t/2=tan⁡t2⇒d2ydx2=sec2⁡t2⋅12dtdx=12cos2⁡t2⋅1a(1+cos⁡t)=12acos2⁡t22cos2⁡t2=14acos4⁡t2⇒d2ydx2t=2π/3=14acos4⁡(π/3)=14a(1/2)4=4a.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $a \neq 0, x = a (t + \sin t) and y = a (1 - \cos t), then \frac{d^{2} y}{{dx}^{2}} at t = \frac{2 π}{3} is$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{4}{a}$

$\frac{1}{4 a}$

$4 a$

$\frac{a}{4}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} x = a (t + \sin t), y = a (1 - \cos t) \Rightarrow \frac{dx}{dt} = a (1 + \cos t), \frac{dy}{dt} = asin t \\ \frac{dy}{dx} = \frac{dy / dt}{dx / dt} = \frac{asin t}{a (1 + \cos t)} = \frac{2 \sin t / 2 \cos t / 2}{2 \cos^{2} t / 2} = \tan \frac{t}{2} \end{array}$

$\begin{array}{l} \Rightarrow \frac{d^{2} y}{{dx}^{2}} = \sec^{2} \frac{t}{2} \cdot \frac{1}{2} \frac{dt}{dx} = \frac{1}{2 \cos^{2} \frac{t}{2}} \cdot \frac{1}{a (1 + \cos t)} = \frac{1}{2 {acos}^{2} \frac{t}{2} 2 \cos^{2} \frac{t}{2}} = \frac{1}{4 {acos}^{4} \frac{t}{2}} \\ \Rightarrow {(\frac{d^{2} y}{{dx}^{2}})}_{t = 2 π / 3} = \frac{1}{4 {acos}^{4} (π / 3)} = \frac{1}{4 a (1 / 2)^{4}} = \frac{4}{a} . \end{array}$