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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

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