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My problem states: Show that y: \begin{equation} y = e^{-t}sin(2t) \end{equation} is a maximum when \begin{equation} t = \frac{1}{2}\tan^{-1}(2) \end{equation} and determine this maximum value.

So basically i have to calculate first and second derivatives of function f(t) = y and find some relation with what i must prove. Well, i've almost done it i think. But not quite. Can someone help me by checking out my answer a bit and tell me where i'm wrong, what i should improve, anything that could help me really?

So first i calculate first derivative: \begin{equation} \frac{dy}{dt} = -e^{-t}\sin(2t) + 2e^{-t}\cos(2t) \end{equation} and i convert the result to a more simplified form \begin{equation} \sqrt{5}e^{-t}\cos(2t-0.464) \end{equation} which i equate to 0 to find t(which i can't find it here). And then i calculate the second derivative: \begin{equation} \frac{d^{2}y}{dt^{2}} = -\sqrt{5}e^{-t}\cos(2t-0.464)-2\sqrt{5}e^{-t}\sin(2t-0.464) \end{equation} which "simplifies" to: \begin{equation} 5e^{-t}\cos(2(t-\frac{tan^{-1}(2)}{2})-0.4636) \end{equation} I doubt i'm right so far but if i am then i'm close. However i don't see exactly how i can prove what is needed from here. If someone could help me a bit i would be most grateful. Thank you in advance.

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You would have been better off trying to solve for critical points directly in: $$\frac{dy}{dt} = -e^{-t}\sin(2t) + 2e^{-t}\cos(2t)$$ instead of first simplifying. $$0 = -e^{-t}\sin(2t) + 2e^{-t}\cos(2t) \implies e^{-t}\sin(2t) = 2e^{-t}\cos(2t)$$ and that gives $$\tan(2t) = 2$$

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    $\begingroup$ Ah i see. One question. Are we always allowed to divide by an expression that has the variable in it? For example you have divided by e^-t . I didn't know that was (always) possible. $\endgroup$ – Nikos Aug 3 '15 at 21:29
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    $\begingroup$ Yes with a caveat, $e^{-t}$ for each t is just a number, so as long as it isn't zero, you can divide both sides by it. If it is zero, you need to consider that case separately. However, $e^{-t}$ is always positive. $\endgroup$ – muaddib Aug 3 '15 at 21:31
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    $\begingroup$ Thank you very much. You helped me understand. $\endgroup$ – Nikos Aug 3 '15 at 21:33
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    $\begingroup$ @RestlessC0bra Great! $\endgroup$ – muaddib Aug 3 '15 at 21:34

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