1. Problem 7: A street light is mounted at the top of a 15-ft pole. A man 6 ft tall walks away from the pole at 5 ft/s. Find how fast the tip of his shadow is moving when he is 40 ft from the pole. 2. Let $x$ be the distance of the man from the pole, and $s$ be the length of his shadow. 3. Using similar triangles, the ratio of heights to shadows is $\frac{15}{x+s} = \frac{6}{s}$. 4. Cross-multiplied: $15s = 6(x+s)$, so $15s = 6x + 6s$. 5. Rearranged: $15s - 6s = 6x \Rightarrow 9s = 6x \Rightarrow s = \frac{2}{3}x$. 6. Differentiate both sides with respect to time $t$: $\frac{ds}{dt} = \frac{2}{3} \frac{dx}{dt}$. 7. Given $\frac{dx}{dt} = 5$ ft/s, so $\frac{ds}{dt} = \frac{2}{3} \times 5 = \frac{10}{3}$ ft/s. 8. The tip of the shadow is at distance $x + s$, so its rate of change is $\frac{d}{dt}(x + s) = \frac{dx}{dt} + \frac{ds}{dt} = 5 + \frac{10}{3} = \frac{25}{3}$ ft/s. 9. Final answer: The tip of the shadow moves at $\frac{25}{3} \approx 8.33$ ft/s when the man is 40 ft from the pole. --- 10. Problem 8: A ball is thrown upward with velocity 80 ft/s; height after $t$ seconds is $s = 80t - 16t^2$. 11. (a) Maximum height occurs when velocity $v = s' = 0$. 12. Compute $s' = 80 - 32t$. 13. Set $s' = 0 \Rightarrow 80 - 32t = 0 \Rightarrow t = \frac{80}{32} = 2.5$ seconds. 14. Maximum height: $s(2.5) = 80(2.5) - 16(2.5)^2 = 200 - 16 \times 6.25 = 200 - 100 = 100$ ft. 15. (b) Velocity when ball is 96 ft on way up and down. 16. Solve $80t - 16t^2 = 96 \Rightarrow 16t^2 - 80t + 96 = 0$. 17. Divide by 16: $t^2 - 5t + 6 = 0$. 18. Factor: $(t - 2)(t - 3) = 0 \Rightarrow t = 2$ or $3$ seconds. 19. Velocity at $t=2$: $v = 80 - 32(2) = 80 - 64 = 16$ ft/s (upward). 20. Velocity at $t=3$: $v = 80 - 32(3) = 80 - 96 = -16$ ft/s (downward). --- 21. Problem 9: Cost function $C(x) = 1200 + 12x - 0.1x^2 + 0.0005x^3$. 22. (a) Marginal cost $C'(x) = 12 - 0.2x + 0.0015x^2$. 23. (b) Calculate $C'(200) = 12 - 0.2(200) + 0.0015(200)^2 = 12 - 40 + 0.0015 imes 40000 = 12 - 40 + 60 = 32$. 24. Interpretation: At 200 yards, the cost to produce one more yard is approximately 32 dollars. 25. (c) Cost of 201st yard is $C(201) - C(200)$. 26. Calculate $C(201) = 1200 + 12(201) - 0.1(201)^2 + 0.0005(201)^3$. 27. $201^2 = 40401$, $201^3 = 201 imes 40401 = 8,120,601$. 28. So $C(201) = 1200 + 2412 - 4040.1 + 0.0005 imes 8,120,601 = 1200 + 2412 - 4040.1 + 4060.3 = 4632.2$. 29. Similarly, $C(200) = 1200 + 2400 - 4000 + 0.0005 imes 8,000,000 = 1200 + 2400 - 4000 + 4000 = 4600$. 30. Cost of 201st yard = $4632.2 - 4600 = 32.2$ dollars, close to marginal cost 32. --- 31. Problem 10: Graphs with vertical tangents. 32. (a) $y = x^{5/3} - 5x^{2/3}$. 33. Derivative: $y' = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3}$. 34. Vertical tangent where $y' \to \pm \infty$, i.e., where denominator zero: at $x=0$ because of $x^{-1/3}$. 35. Confirm limit: as $x \to 0^+$, $y' \to -\infty$, vertical tangent at $x=0$. 36. (b) $y = x^{1/5}$. 37. Derivative: $y' = \frac{1}{5} x^{-4/5}$. 38. Vertical tangent at $x=0$ since $y' \to \infty$. 39. (c) $y = x^{2/3} - (x-1)^{1/3}$. 40. Derivative: $y' = \frac{2}{3} x^{-1/3} - \frac{1}{3} (x-1)^{-2/3}$. 41. Vertical tangents at $x=0$ and $x=1$ due to negative exponents. --- 42. Problem 11: Find derivative of $$f(t) = \frac{\sqrt{t}(3t + t^5 - 10) + (t^3 - e^t)}{t^2}.$$ 43. Rewrite numerator: $\sqrt{t} = t^{1/2}$, so numerator is $t^{1/2}(3t + t^5 - 10) + t^3 - e^t$. 44. Simplify numerator: $t^{1/2} \times 3t = 3t^{3/2}$, $t^{1/2} \times t^5 = t^{11/2}$, $t^{1/2} \times (-10) = -10 t^{1/2}$. 45. Numerator: $3t^{3/2} + t^{11/2} - 10 t^{1/2} + t^3 - e^t$. 46. Use quotient rule: $f'(t) = \frac{(N') t^2 - N (2t)}{t^4}$ where $N$ is numerator. 47. Compute $N'$: $\frac{d}{dt} 3t^{3/2} = \frac{9}{2} t^{1/2}$, $\frac{d}{dt} t^{11/2} = \frac{11}{2} t^{9/2}$, $\frac{d}{dt} (-10 t^{1/2}) = -5 t^{-1/2}$, $\frac{d}{dt} t^3 = 3 t^2$, $\frac{d}{dt} (-e^t) = -e^t$. 48. So $N' = \frac{9}{2} t^{1/2} + \frac{11}{2} t^{9/2} - 5 t^{-1/2} + 3 t^2 - e^t$. 49. Final derivative: $$f'(t) = \frac{\left(\frac{9}{2} t^{1/2} + \frac{11}{2} t^{9/2} - 5 t^{-1/2} + 3 t^2 - e^t\right) t^2 - \left(3 t^{3/2} + t^{11/2} - 10 t^{1/2} + t^3 - e^t\right) 2 t}{t^4}.$$ --- 50. Problem 12: Find first, second, third derivatives of $$f(x) = 4 e^{2x} - \sin(3x) + 5 x^3 - \ln(2x).$$ 51. First derivative: $f'(x) = 4 \times 2 e^{2x} - 3 \cos(3x) + 15 x^2 - \frac{1}{2x} = 8 e^{2x} - 3 \cos(3x) + 15 x^2 - \frac{1}{2x}$. 52. Second derivative: $f''(x) = 8 \times 2 e^{2x} + 9 \sin(3x) + 30 x + \frac{1}{2 x^2} = 16 e^{2x} + 9 \sin(3x) + 30 x + \frac{1}{2 x^2}$. 53. Third derivative: $f'''(x) = 16 \times 2 e^{2x} + 27 \cos(3x) + 30 - \frac{1}{x^3} = 32 e^{2x} + 27 \cos(3x) + 30 - \frac{1}{x^3}$. --- 54. Problem 13: (a) $y = \sqrt{x^2 - 1} + x$, $x > 1$. 55. Derivative: $y' = \frac{1}{2 \sqrt{x^2 - 1}} \times 2x + 1 = \frac{x}{\sqrt{x^2 - 1}} + 1$. (b) $y = \ln \ln (x^2 + 4) - x \frac{x}{2} = \ln \ln (x^2 + 4) - \frac{x^2}{2}$. 56. Derivative: $y' = \frac{1}{\ln(x^2 + 4)} \times \frac{1}{x^2 + 4} \times 2x - x = \frac{2x}{(x^2 + 4) \ln(x^2 + 4)} - x$. --- Final answers provided with detailed steps for all problems 7 to 13.