Particle Velocity Fed96F

1. **Problem 1.1a:** Find times when particle changes direction on $0 \leq t \leq 4$ given velocity $$v(t) = t^2 - 4t + 3.$$ 2. **Step 1:** Particle changes direction when velocity changes sign, i.e., when $v(t) = 0$. Solve $$t^2 - 4t + 3 = 0.$$ 3. **Step 2:** Factor quadratic: $$t^2 - 4t + 3 = (t - 1)(t - 3) = 0.$$ So, $$t = 1$$ or $$t = 3.$$ 4. **Answer 1.1a:** Particle changes direction at $$t = 1$$ and $$t = 3$$ seconds. --- 5. **Problem 1.1b:** Find total distance travelled from $$t=0$$ to $$t=4$$. 6. **Step 1:** Total distance is the integral of the speed, i.e., $$\int_0^4 |v(t)| dt.$$ Since velocity changes sign at $$t=1$$ and $$t=3$$, split integral: $$\text{Distance} = \int_0^1 v(t) dt - \int_1^3 v(t) dt + \int_3^4 v(t) dt,$$ where we negate integral on intervals where velocity is negative. 7. **Step 2:** Calculate each integral. Calculate $$\int v(t) dt = \int (t^2 - 4t + 3) dt = \frac{t^3}{3} - 2t^2 + 3t + C.$$ 8. **Step 3:** Evaluate definite integrals: - $$\int_0^1 v(t) dt = \left[ \frac{t^3}{3} - 2t^2 + 3t \right]_0^1 = \left( \frac{1}{3} - 2 + 3 \right) - 0 = \frac{1}{3} + 1 = \frac{4}{3}.$$ - $$\int_1^3 v(t) dt = \left[ \frac{t^3}{3} - 2t^2 + 3t \right]_1^3 = \left( \frac{27}{3} - 18 + 9 \right) - \left( \frac{1}{3} - 2 + 3 \right) = (9 - 18 + 9) - (\frac{1}{3} + 1) = 0 - \frac{4}{3} = -\frac{4}{3}.$$ - $$\int_3^4 v(t) dt = \left[ \frac{t^3}{3} - 2t^2 + 3t \right]_3^4 = \left( \frac{64}{3} - 32 + 12 \right) - \left( 9 - 18 + 9 \right) = \left( \frac{64}{3} - 20 \right) - 0 = \frac{64}{3} - 20 = \frac{64 - 60}{3} = \frac{4}{3}.$$ 9. **Step 4:** Total distance: $$\frac{4}{3} - \left(-\frac{4}{3}\right) + \frac{4}{3} = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4.$$ 10. **Answer 1.1b:** Total distance travelled is $$4$$ metres. --- 11. **Problem 1.2:** Find exact area bounded by $$x = g(y) = 4y^2 - y^3$$, the y-axis, and lines $$y=1$$ and $$y=3$$ using limit definition of definite integral. 12. **Step 1:** Area bounded is $$\int_1^3 g(y) dy = \int_1^3 (4y^2 - y^3) dy.$$ 13. **Step 2:** Using limit definition, partition interval $$[1,3]$$ into $$n$$ subintervals of width $$\Delta y = \frac{3-1}{n} = \frac{2}{n}$$. 14. **Step 3:** Sample points $$y_i = 1 + i \Delta y = 1 + \frac{2i}{n}$$ for $$i=0,1,...,n$$. 15. **Step 4:** Riemann sum: $$S_n = \sum_{i=1}^n g(y_i^*) \Delta y,$$ where $$y_i^*$$ is a sample point in $$i$$th subinterval. Choose right endpoints: $$y_i^* = y_i = 1 + \frac{2i}{n}.$$ 16. **Step 5:** Substitute: $$S_n = \sum_{i=1}^n \left[4\left(1 + \frac{2i}{n}\right)^2 - \left(1 + \frac{2i}{n}\right)^3\right] \frac{2}{n}.$$ 17. **Step 6:** Expand terms: - $$\left(1 + \frac{2i}{n}\right)^2 = 1 + \frac{4i}{n} + \frac{4i^2}{n^2}.$$ - $$\left(1 + \frac{2i}{n}\right)^3 = 1 + 3\frac{2i}{n} + 3\left(\frac{2i}{n}\right)^2 + \left(\frac{2i}{n}\right)^3 = 1 + \frac{6i}{n} + \frac{12i^2}{n^2} + \frac{8i^3}{n^3}.$$ 18. **Step 7:** Substitute back: $$S_n = \sum_{i=1}^n \left[4\left(1 + \frac{4i}{n} + \frac{4i^2}{n^2}\right) - \left(1 + \frac{6i}{n} + \frac{12i^2}{n^2} + \frac{8i^3}{n^3}\right)\right] \frac{2}{n}.$$ Simplify inside bracket: $$4 + \frac{16i}{n} + \frac{16i^2}{n^2} - 1 - \frac{6i}{n} - \frac{12i^2}{n^2} - \frac{8i^3}{n^3} = 3 + \frac{10i}{n} + \frac{4i^2}{n^2} - \frac{8i^3}{n^3}.$$ 19. **Step 8:** So, $$S_n = \sum_{i=1}^n \left(3 + \frac{10i}{n} + \frac{4i^2}{n^2} - \frac{8i^3}{n^3}\right) \frac{2}{n} = \frac{2}{n} \sum_{i=1}^n \left(3 + \frac{10i}{n} + \frac{4i^2}{n^2} - \frac{8i^3}{n^3}\right).$$ 20. **Step 9:** Separate sums: $$S_n = \frac{2}{n} \left[ \sum_{i=1}^n 3 + \sum_{i=1}^n \frac{10i}{n} + \sum_{i=1}^n \frac{4i^2}{n^2} - \sum_{i=1}^n \frac{8i^3}{n^3} \right].$$ 21. **Step 10:** Use formulas: - $$\sum_{i=1}^n 1 = n,$$ - $$\sum_{i=1}^n i = \frac{n(n+1)}{2},$$ - $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6},$$ - $$\sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2.$$ 22. **Step 11:** Substitute: $$S_n = \frac{2}{n} \left[3n + \frac{10}{n} \cdot \frac{n(n+1)}{2} + \frac{4}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} - \frac{8}{n^3} \cdot \left(\frac{n(n+1)}{2}\right)^2 \right].$$ Simplify each term: - $$3n$$ - $$\frac{10}{n} \cdot \frac{n(n+1)}{2} = 5(n+1)$$ - $$\frac{4}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{4(n+1)(2n+1)}{6n} = \frac{2(n+1)(2n+1)}{3n}$$ - $$\frac{8}{n^3} \cdot \left(\frac{n(n+1)}{2}\right)^2 = \frac{8}{n^3} \cdot \frac{n^2 (n+1)^2}{4} = \frac{2 (n+1)^2}{n}.$$ 23. **Step 12:** So, $$S_n = \frac{2}{n} \left[3n + 5(n+1) + \frac{2(n+1)(2n+1)}{3n} - \frac{2(n+1)^2}{n} \right].$$ 24. **Step 13:** Multiply inside by $$\frac{2}{n}$$: $$S_n = \frac{2}{n} (3n + 5n + 5 + \frac{2(n+1)(2n+1)}{3n} - \frac{2(n+1)^2}{n}).$$ Combine $$3n + 5n = 8n$$: $$S_n = \frac{2}{n} \left(8n + 5 + \frac{2(n+1)(2n+1)}{3n} - \frac{2(n+1)^2}{n} \right).$$ 25. **Step 14:** Distribute $$\frac{2}{n}$$: $$S_n = \frac{16n}{n} + \frac{10}{n} + \frac{4(n+1)(2n+1)}{3n^2} - \frac{4(n+1)^2}{n^2} = 16 + \frac{10}{n} + \frac{4(n+1)(2n+1)}{3n^2} - \frac{4(n+1)^2}{n^2}.$$ 26. **Step 15:** Take limit as $$n \to \infty$$: - $$\lim_{n\to\infty} \frac{10}{n} = 0,$$ - $$\lim_{n\to\infty} \frac{4(n+1)(2n+1)}{3n^2} = \frac{4 \cdot 1 \cdot 2}{3} = \frac{8}{3},$$ - $$\lim_{n\to\infty} \frac{4(n+1)^2}{n^2} = 4.$$ 27. **Step 16:** So, $$\lim_{n\to\infty} S_n = 16 + 0 + \frac{8}{3} - 4 = 16 - 4 + \frac{8}{3} = 12 + \frac{8}{3} = \frac{36}{3} + \frac{8}{3} = \frac{44}{3}.$$ 28. **Answer 1.2:** Exact area bounded is $$\frac{44}{3}$$ square units. --- 29. **Problem 1.3a:** Find $$P'(x)$$ where $$P(x) = \int_0^{x^2} e^t(\sin t + \cos t) dt, x \geq 0.$$ 30. **Step 1:** Use the Fundamental Theorem of Calculus and Chain Rule: $$P'(x) = \frac{d}{dx} \int_0^{x^2} f(t) dt = f(x^2) \cdot \frac{d}{dx} (x^2) = e^{x^2} (\sin x^2 + \cos x^2) \cdot 2x.$$ 31. **Answer 1.3a:** $$P'(x) = 2x e^{x^2} (\sin x^2 + \cos x^2).$$ --- 32. **Problem 1.3b:** Evaluate $$\int e^x (\sin x + \cos x) dx.$$ 33. **Step 1:** Split integral: $$\int e^x \sin x dx + \int e^x \cos x dx.$$ 34. **Step 2:** Use integration by parts or known formula for $$\int e^x \sin x dx$$ and $$\int e^x \cos x dx$$: - $$\int e^x \sin x dx = \frac{e^x}{2} (\sin x - \cos x) + C,$$ - $$\int e^x \cos x dx = \frac{e^x}{2} (\sin x + \cos x) + C.$$ 35. **Step 3:** Add results: $$\int e^x (\sin x + \cos x) dx = \frac{e^x}{2} (\sin x - \cos x) + \frac{e^x}{2} (\sin x + \cos x) + C = \frac{e^x}{2} (2 \sin x) + C = e^x \sin x + C.$$ 36. **Answer 1.3b:** $$\int e^x (\sin x + \cos x) dx = e^x \sin x + C.$$ --- 37. **Problem 1.4a:** Express $$f(x) = \frac{3x+5}{x^2 + 3x + 2}$$ in partial fractions on $$0 \leq x \leq 1.$$ 38. **Step 1:** Factor denominator: $$x^2 + 3x + 2 = (x+1)(x+2).$$ 39. **Step 2:** Set $$\frac{3x+5}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}.$$ 40. **Step 3:** Multiply both sides by denominator: $$3x + 5 = A(x+2) + B(x+1) = (A+B)x + (2A + B).$$ 41. **Step 4:** Equate coefficients: - Coefficient of $$x$$: $$3 = A + B,$$ - Constant term: $$5 = 2A + B.$$ 42. **Step 5:** Solve system: From $$3 = A + B$$, $$B = 3 - A.$$ Substitute into second: $$5 = 2A + (3 - A) = A + 3 \Rightarrow A = 2,$$ then $$B = 3 - 2 = 1.$$ 43. **Answer 1.4a:** $$f(x) = \frac{2}{x+1} + \frac{1}{x+2}.$$ --- 44. **Problem 1.4b:** Find average value of $$f(x)$$ on $$[0,1].$$ 45. **Step 1:** Average value formula: $$f_{avg} = \frac{1}{1-0} \int_0^1 f(x) dx = \int_0^1 \left( \frac{2}{x+1} + \frac{1}{x+2} \right) dx.$$ 46. **Step 2:** Integrate: $$\int_0^1 \frac{2}{x+1} dx = 2 \ln|x+1| \Big|_0^1 = 2(\ln 2 - \ln 1) = 2 \ln 2,$$ $$\int_0^1 \frac{1}{x+2} dx = \ln|x+2| \Big|_0^1 = \ln 3 - \ln 2.$$ 47. **Step 3:** Sum integrals: $$\int_0^1 f(x) dx = 2 \ln 2 + (\ln 3 - \ln 2) = \ln 2 + \ln 3 = \ln 6.$$ 48. **Answer 1.4b:** Average value is $$\ln 6.$$