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A dog moving on a floor with velocity \left(V_0\right) which is represented as -

 \mathrm{V_0(t)=\frac{1}{2} e^t+\frac{3}{2} e^{it}} , where i is a complex number. what will be the ratio of change of velocity at t=2 sec

Option: 1

\mathrm{e^2+3 c^{2 i}}


Option: 2

\mathrm{\frac{1}{2} e^2+3 e^{2 i}}


Option: 3

\mathrm{\frac{1}{2}\left(e^2+3 e^{2 i}\right)}


Option: 4

\mathrm{e^2+2 e^{3 i}}


Answers (1)

best_answer

Given, \mathrm{v_0(t)=\frac{1}{2} e^t+\frac{3}{2} e^{i t}}

Rate of charge of velocity \mathrm{\frac{d V(t)}{d t}=?}

Now, 

\mathrm{\begin{aligned} \frac{d v_0(t)}{d t} & =\frac{d}{d t}\left[\frac{1}{2} e^t+\frac{3}{2} e^{i t}\right] \\ & =\frac{1}{2} e^t+\frac{3}{2} e^{i t} \cdot i \\ \frac{d v_0(t)}{d t} & =\frac{1}{2} e^t+\frac{3 i}{2} e^{i t} \\ \left.\frac{d v_0(t)}{d t}\right|_{\text {at } t=2 \mathrm{sec}} & =\frac{1}{2} e^2+\frac{3 i}{2} e^{2 i} \\ & =\frac{1}{2}\left[e^2+3 e^{2 i}\right] \end{aligned}}\mathrm{\begin{aligned} \frac{d v_0(t)}{d t} & =\frac{d}{d t}\left[\frac{1}{2} e^t+\frac{3}{2} e^{i t}\right] \\ & =\frac{1}{2} e^t+\frac{3}{2} e^{i t} \cdot i \\ \frac{d v_0(t)}{d t} & =\frac{1}{2} e^t+\frac{3 i}{2} e^{i t} \\ \left.\frac{d v_0(t)}{d t}\right|_{\text {at } t=2 \mathrm{sec}} & =\frac{1}{2} e^2+\frac{3 i}{2} e^{2 i} \\ & =\frac{1}{2}\left[e^2+3 e^{2 i}\right] \end{aligned}}

Posted by

seema garhwal

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