Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • If the position of the particle is given by <img alt="y=e^{i \theta}+\frac{1}{2} e^{-i \theta}+\sin \theta" src="https://entr

If the position of the particle y is given by y=e^{i \theta}+\frac{1}{2} e^{-i \theta}+\sin \theta, then what is the rate of velocity at \theta=\pi / 2 -

 

Option: 1

-\left(e^{i \pi / 2}+\frac{1}{2} e^{-i \pi / 2}+1\right)


Option: 2

\left(e^{-i \pi / 2}+\frac{1}{2} e^{i \pi / 2}+1\right)


Option: 3

-\left(e^{-i \pi / 2}+\frac{1}{2} e^{i \pi / 2}\right)


Option: 4

-\left(e^{\pi i / 2}+\frac{1}{2} e^{-\pi i / 2}\right)


Answers (1)

best_answer

\text{position}(y)= e^{i \theta}+\frac{1}{2} e^{-i \theta}+\sin \theta \\
\text{velocity}(v)=\frac{d y}{d \theta} =i e^{i \theta}+\frac{1}{2} e^{-i \theta} \cdot(-i)+\cos \theta \\
v =i e^{i \theta}-\frac{i}{2} e^{-i \theta}+\cos \theta     .....(1)
rate of velocity (a)=\frac{d v}{d \theta} =i^2 e^{i \theta}-\frac{i}{2} e^{-i \theta}(-i)-\sin \theta \\
 =-1 e^{i \theta}+\frac{i^2}{2} e^{-i \theta}-\sin \theta \\
 =-e^{i \theta}-\frac{1}{2} e^{-i \theta}-\sin \theta    ......(2)
a=\frac{d v}{d \theta} at \theta=\pi / 2 will be -

\left.\frac{d v}{d \theta}\right|_{\text {at } \theta \rightarrow \pi / 2} =-e^{i \pi / 2}-\frac{1}{2} e^{-i \pi / 2}-\sin \pi / 2 \\
 =-e^{i \pi / 2}-\frac{1}{2} e^{-i \pi / 2}-01 \\
 =-\left[1+e^{i \pi / 2}+\frac{1}{2} e^{-i \pi / 2}\right]


 

Posted by

Ritika Kankaria

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

'PM Modi personally supervising the situation': Centre tells SC in NEET case
anam-khan

NEET Paper Leak Row: NTA informs SC of security overhaul, institutional reforms
anam-khan

Re-NEET 2026 City Intimation Slip LIVE: NTA MBBS admit card soon; direct link, login details, exam date

Latest Question