1. Problem 1: Find the velocity of the end of the spring at $t=1$ s given $s = ae^{-kt} \sin 2\pi ft$ with $a=2$, $k=0.9$, $f=5$. 2. Differentiate $s$ with respect to $t$ to find velocity $v$: $$v = \frac{ds}{dt} = \frac{d}{dt}\left(ae^{-kt} \sin 2\pi ft\right).$$ 3. Use product rule: $$v = a\left(\frac{d}{dt}e^{-kt} \cdot \sin 2\pi ft + e^{-kt} \cdot \frac{d}{dt}\sin 2\pi ft\right).$$ 4. Calculate derivatives: $$\frac{d}{dt}e^{-kt} = -k e^{-kt}, \quad \frac{d}{dt}\sin 2\pi ft = 2\pi f \cos 2\pi ft.$$ 5. Substitute back: $$v = a \left(-k e^{-kt} \sin 2\pi ft + e^{-kt} 2\pi f \cos 2\pi ft\right) = a e^{-kt} \left(2\pi f \cos 2\pi ft - k \sin 2\pi ft\right).$$ 6. Plug in the values at $t=1$: $$v(1) = 2 e^{-0.9 \times 1} \left(2\pi \times 5 \cos 2\pi \times 5 \times 1 - 0.9 \sin 2\pi \times 5 \times 1\right).$$ 7. Note $\sin 10\pi = 0$ and $\cos 10\pi = 1$ (since $\sin n\pi = 0$ and $\cos n\pi = (-1)^n$, here $n=10$, even). 8. Evaluate: $$v(1) = 2 e^{-0.9} (2\pi \times 5 \times 1 - 0) = 2 e^{-0.9} (10\pi).$$ 9. Numerical approximation: $$e^{-0.9} \approx 0.40657, \quad v(1) \approx 2 \times 0.40657 \times 10\pi = 8.13 \pi \approx 25.53.$$ --- 10. Problem 2(d): Find rate of change of current $i = 10 \sin 2ft$ at $t=20$ ms, $f=150$ Hz. 11. Differentiate: $$\frac{di}{dt} = 10 \cdot 2f \cos 2ft = 20 f \cos(2ft).$$ 12. Substitute $f=150$ and $t=0.02$ s: $$\frac{di}{dt} = 20 \times 150 \times \cos(2 \times 150 \times 0.02) = 3000 \cos(6).$$ 13. Compute $\cos 6$ radians approximately: $$\cos 6 \approx 0.96017,$$ 14. Hence rate of change: $$\frac{di}{dt} \approx 3000 \times 0.96017 = 2880.5$$ amperes/second. --- 15. Problem 5(a): Compute $AB$ where $$A = \begin{pmatrix}2 & x \\ 3 & 1\end{pmatrix}, B = \begin{pmatrix}2 & 1 \\ 1 & 4\end{pmatrix}.$$ 16. Multiply: $$AB = \begin{pmatrix}2 \cdot 2 + x \cdot 1 & 2 \cdot 1 + x \cdot 4 \\ 3 \cdot 2 + 1 \cdot 1 & 3 \cdot 1 + 1 \cdot 4 \end{pmatrix} = \begin{pmatrix}4 + x & 2 + 4x \\ 6 + 1 & 3 + 4 \end{pmatrix} = \begin{pmatrix}4 + x & 2 + 4x \\ 7 & 7 \end{pmatrix}.$$ --- 17. Problem 5(b): Find $x$ given $B^T A^T = C$ with $$B^T = \begin{pmatrix}2 & 1 \\ 1 & 4\end{pmatrix}^T = \begin{pmatrix}2 & 1 \\ 1 & 4\end{pmatrix},$$ same as $B$, $$A^T= \begin{pmatrix}2 & 3 \\ x & 1\end{pmatrix},$$ $$C = \begin{pmatrix}3x+2 & 7 \\ 7 - x & 7 \end{pmatrix}.$$ 18. Multiply $B^T A^T$: $$B^T A^T = \begin{pmatrix}2 & 1 \\ 1 & 4\end{pmatrix} \begin{pmatrix}2 & 3 \\ x & 1\end{pmatrix} = \begin{pmatrix}2 \cdot 2 + 1 \cdot x & 2 \cdot 3 + 1 \cdot 1 \\ 1 \cdot 2 + 4 \cdot x & 1 \cdot 3 + 4 \cdot 1 \end{pmatrix} = \begin{pmatrix}4 + x & 7 \\ 2 + 4x & 7 \end{pmatrix}.$$ 19. Set equal to $C$: $$\begin{cases}4 + x = 3x + 2 \\ 7 = 7 \\ 2 + 4x = 7 - x \\ 7 = 7 \end{cases}$$ 20. From first, $4 + x = 3x + 2 \implies 2 = 2x \implies x = 1$. 21. Check with third: $2 + 4(1) = 7 - 1 \implies 6 = 6$. Correct. So $x=1$. --- 22. Problem 5(c): Solve system using determinants: $$\begin{cases}2I_1 + 3I_2 - 4I_3 = 26 \\ I_1 - 5I_2 - 3I_3 = -87 \\ -7I_1 + 2I_2 + 6I_3 = 12 \end{cases}$$ 23. Coefficient matrix $D$: $$\begin{vmatrix} 2 & 3 & -4 \\ 1 & -5 & -3 \\ -7 & 2 & 6 \end{vmatrix}.$$ 24. Compute determinant $\Delta$: $$\Delta = 2 \begin{vmatrix} -5 & -3 \\ 2 & 6 \end{vmatrix} -3 \begin{vmatrix} 1 & -3 \\ -7 & 6 \end{vmatrix} -4 \begin{vmatrix} 1 & -5 \\ -7 & 2 \end{vmatrix}.$$ 25. Calculate minors: $$M_1 = (-5)(6) - (-3)(2) = -30 + 6 = -24,$$ $$M_2 = 1 \times 6 - (-3)(-7) = 6 - 21 = -15,$$ $$M_3 = 1 \times 2 - (-5)(-7) = 2 - 35 = -33.$$ 26. So, $$\Delta = 2(-24) - 3(-15) - 4(-33) = -48 + 45 + 132 = 129.$$ 27. For $I_1$, replace first column with constants: $$D_1 = \begin{vmatrix} 26 & 3 & -4 \\ -87 & -5 & -3 \\ 12 & 2 & 6 \end{vmatrix}.$$ 28. Calculate: $$= 26 \begin{vmatrix} -5 & -3 \\ 2 & 6 \end{vmatrix} - 3 \begin{vmatrix} -87 & -3 \\ 12 & 6 \end{vmatrix} -4 \begin{vmatrix} -87 & -5 \\ 12 & 2 \end{vmatrix}.$$ 29. Minor matrices: $$= 26(-24) - 3(-87 \cdot 6 - (-3) \cdot 12) -4 (-87 \cdot 2 - (-5) \cdot 12)$$ $$= 26(-24) - 3(-522 + 36) -4(-174 + 60)$$ $$= -624 - 3(-486) - 4(-114) = -624 + 1458 + 456 = 1290.$$ 30. Then $I_1 = \frac{1290}{129} = 10$. 31. For $I_2$, replace second column: $$D_2 = \begin{vmatrix} 2 & 26 & -4 \\ 1 & -87 & -3 \\ -7 & 12 & 6 \end{vmatrix}.$$ 32. Calculate: $$= 2 \begin{vmatrix} -87 & -3 \\ 12 & 6 \end{vmatrix} - 26 \begin{vmatrix} 1 & -3 \\ -7 & 6 \end{vmatrix} -4 \begin{vmatrix} 1 & -87 \\ -7 & 12 \end{vmatrix}.$$ 33. Minors: $$= 2(-87 \times 6 - (-3) \times 12) - 26(1 \times 6 - (-3) \times -7) -4 (1 \times 12 - (-87)(-7))$$ $$= 2(-522 + 36) - 26(6 - 21) - 4(12 - 609)$$ $$= 2(-486) - 26(-15) - 4(-597) = -972 + 390 + 2388 = 1806.$$ 34. $I_2 = \frac{1806}{129} = 14$. 35. For $I_3$, replace third column: $$D_3 = \begin{vmatrix} 2 & 3 & 26 \\ 1 & -5 & -87 \\ -7 & 2 & 12 \end{vmatrix}.$$ 36. Calculate: $$= 2 \begin{vmatrix} -5 & -87 \\ 2 & 12 \end{vmatrix} - 3 \begin{vmatrix} 1 & -87 \\ -7 & 12 \end{vmatrix} + 26 \begin{vmatrix} 1 & -5 \\ -7 & 2 \end{vmatrix}.$$ 37. Minors: $$= 2(-5 \times 12 - (-87) \times 2) - 3(1 \times 12 - (-87)(-7)) + 26(1 \times 2 - (-5)(-7))$$ $$= 2(-60 + 174) - 3(12 - 609) + 26(2 - 35)$$ $$= 2(114) -3(-597) + 26(-33) = 228 + 1791 - 858 = 1161.$$ 38. $I_3 = \frac{1161}{129} = 9$. --- 39. Problem 6(a): Net return decreases by 10% annually, starting at 400. Use formula for infinite decreasing geometric series: $$S = \frac{a}{1-r}$$ where $a=400$, $r = 1 - 0.1 = 0.9$. 40. Calculate: $$S = \frac{400}{1 - 0.9} = \frac{400}{0.1} = 4000.$$ --- 41. Problem 6(b): 6 speeds from 50 rpm to 750 rpm in geometric progression. 42. Let speeds be $a, ar, ar^2, ar^3, ar^4, ar^5$ with $a=50$, last term $ar^5=750$. 43. Solve for $r$: $$750 = 50 r^5 \implies r^5 = 15 \implies r = \sqrt[5]{15}.$$ 44. Approximate: $$r \approx 1.7171.$$ 45. Calculate all speeds: $$50, 50 \times 1.7171 = 86, 86 \times 1.7171 = 148, 148 \times 1.7171 = 254, 254 \times 1.7171 = 436, 436 \times 1.7171 = 750.$$ Speeds to nearest whole number: 50, 86, 148, 254, 436, 750.