Given that y=e−2t(sin2t+cos2t)y = \mathrm{e}^{-2t}(\sin 2t + \cos 2t)y=e−2t(sin2t+cos2t), find dydt\frac{\mathrm{d}y}{\mathrm{d}t}dtdy. Simplify your answer.
Hence, find
∫e−2tsin2t dt=ae−2t(sin2t+cos2t)+C \int \mathrm{e}^{-2t} \sin 2t \, \mathrm{d}t = a \mathrm{e}^{-2t}(\sin 2t + \cos 2t) + C ∫e−2tsin2tdt=ae−2t(sin2t+cos2t)+Cwhere aaa is a rational number.
The displacement sss (in μm\mu\text{m}μm) of a micro-mechanical resonator at time ttt (in seconds) is modeled by s(t)=e−2tsin2ts(t) = \mathrm{e}^{-2t} \sin 2ts(t)=e−2tsin2t for t≥0t \ge 0t≥0. The areas of the finite regions bounded by the curve and the ttt-axis are denoted by A1,A2,…,An,…A_1, A_2, \dots, A_n, \dotsA1,A2,…,An,… where A1A_1A1 is the area of the region from t=0t=0t=0 to the first positive root.
(i) Find the exact value of the area A1A_1A1.
(ii) Show that An+1An=e−π\frac{A_{n+1}}{A_n} = \mathrm{e}^{-\pi}AnAn+1=e−π.
(iii) Show that the exact value of the total area enclosed between the curve and the ttt-axis for t≥0t \ge 0t≥0 is
1+e−π4(1−e−π) \frac{1 + \mathrm{e}^{-\pi}}{4(1 - \mathrm{e}^{-\pi})} 4(1−e−π)1+e−πor equivalently eπ+14(eπ−1)\frac{\mathrm{e}^{\pi} + 1}{4(\mathrm{e}^{\pi} - 1)}4(eπ−1)eπ+1
