1. **Problem (a):** Given the curve $$y = 4x^2 + \frac{1}{x^2} - 8,$$ find the rate of change of $x$ when the $y$-coordinate decreases at 5 units per second and $x=2$. 2. Differentiate $y$ with respect to $t$: $$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}.$$ Calculate $\frac{dy}{dx}$ using the given $y$: $$\frac{dy}{dx} = 8x - \frac{2}{x^3}.$$ 3. Substitute $x=2$: $$\frac{dy}{dx}\Big|_{x=2} = 8(2) - \frac{2}{2^3} = 16 - \frac{2}{8} = 16 - 0.25 = 15.75.$$ 4. Given that $\frac{dy}{dt} = -5$, solve for $\frac{dx}{dt}$: $$-5 = 15.75 \cdot \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{-5}{15.75} = -\frac{20}{63}.$$ 5. **Answer (a):** The rate at which $x$ changes when $x=2$ is $$\frac{dx}{dt} = -\frac{20}{63}.$$ --- 6. **Problem (b):** Find stationary points of $y = 4x^2 + \frac{1}{x^2} - 8$ where $\frac{dy}{dx} = 0$. 7. Set derivative to zero: $$8x - \frac{2}{x^3} = 0 \implies 8x = \frac{2}{x^3} \implies 8x^4 = 2 \implies x^4 = \frac{1}{4}.$$ 8. Solve for $x$: $$x = \pm \frac{1}{\sqrt{2}}.$$ 9. Find $y$ coordinates: $$y = 4x^2 + \frac{1}{x^2} - 8 = 4 \cdot \frac{1}{2} + 2 -8 = 2 + 2 -8 = -4.$$ 10. Stationary points are $$\left( \frac{1}{\sqrt{2}}, -4 \right) \text{ and } \left( -\frac{1}{\sqrt{2}}, -4 \right).$$ 11. Find second derivative: $$\frac{d^2y}{dx^2} = 8 + \frac{6}{x^4}.$$ 12. Evaluate at stationary points: For $x= \pm \frac{1}{\sqrt{2}}$, $$\frac{d^2y}{dx^2} = 8 + 6 \cdot 4 = 8 + 24 = 32 > 0.$$ 13. Since second derivative is positive, both points are local minima. --- 14. **Problem 2:** For curve $$y = 2x + \frac{12}{x^2},$$ find the tangent at $(-2,-1)$. 15. Find $dy/dx$: $$\frac{dy}{dx} = 2 - \frac{24}{x^3}.$$ 16. Evaluate slope at $x=-2$: $$m = 2 - \frac{24}{(-2)^3} = 2 - \frac{24}{-8} = 2 + 3 = 5.$$ 17. Use point-slope form: $$y - (-1) = 5(x - (-2)) \implies y + 1 = 5(x + 2).$$ 18. Simplify to $y = mx + c$: $$y = 5x + 10 - 1 = 5x + 9.$$ --- 19. **Problem 3 (a):** For curve $$y = \frac{9}{2x - 5} + 2x - 5,$$ find coordinates of stationary points. 20. Compute derivative using quotient or product rule: $$\frac{dy}{dx} = -\frac{18}{(2x - 5)^2} + 2.$$ 21. Set derivative to zero to find stationary points: $$-\frac{18}{(2x - 5)^2} + 2 = 0 \implies 2 = \frac{18}{(2x - 5)^2} \implies (2x - 5)^2 = 9.$$ 22. Solve: $$2x - 5 = \pm 3.$$ 23. Cases: - If $2x -5 = 3 \implies 2x = 8 \implies x=4.$ - If $2x -5 = -3 \implies 2x=2 \implies x=1.$ 24. Find corresponding $y$ values: For $x=4$, $$y = \frac{9}{3} + 8 - 5 = 3 + 3 = 6.$$ For $x=1$, $$y = \frac{9}{-3} + 2 - 5 = -3 - 3 = -6.$$ 25. Stationary points: $(4,6)$ and $(1,-6)$. --- 26. **Problem 3 (b):** Find second derivative and nature of stationary points. 27. Differentiate $\frac{dy}{dx}$ again: $$\frac{d^2y}{dx^2} = \frac{2 \cdot 18 \cdot 2}{(2x - 5)^3} = \frac{72}{(2x - 5)^3}.$$ 28. Evaluate at $x=4$: $$\frac{d^2y}{dx^2} = \frac{72}{3^3} = \frac{72}{27} = \frac{8}{3} > 0,$$ so $(4,6)$ is a minimum. 29. Evaluate at $x=1$: $$\frac{d^2y}{dx^2} = \frac{72}{(-3)^3} = \frac{72}{-27} = -\frac{8}{3} < 0,$$ so $(1,-6)$ is a maximum. --- 30. **Problem 4 (a):** For curve $$y = x^3 + ax^2 + bx + 5,$$ with stationary point $(1,9)$, find $a$ and $b$. 31. Use point condition: $$9 = 1 + a + b + 5 \implies a + b = 3.$$ 32. Derivative: $$\frac{dy}{dx} = 3x^2 + 2ax + b.$$ 33. At $x=1$, stationary point means derivative zero: $$0 = 3 + 2a + b.$$ 34. From previous, $b = 3 - a$, substitute: $$0 = 3 + 2a + 3 - a = 6 + a \implies a = -6.$$ 35. Then $b = 3 - (-6) = 9.$ 36. **Answer (a):** $a = -6$, $b = 9$. --- 37. **Problem 4 (b):** Find other stationary points. 38. Substitute $a,b$ into derivative: $$\frac{dy}{dx} = 3x^2 - 12x + 9.$$ 39. Set derivative zero: $$3x^2 - 12x + 9 = 0 \implies x^2 - 4x + 3 = 0.$$ 40. Solve quadratic: $$(x - 3)(x - 1) = 0 \implies x = 3, 1.$$ 41. The other stationary point is at $x=3$. Find $y$: $$y = 27 - 54 + 27 + 5 = 5.$$ 42. Coordinates: $(3,5)$. --- 43. **Problem 4 (c):** Rate of change of $x$ when $y$ increases at 6 units/sec at $x=5$. 44. Compute $\frac{dy}{dx}$ using known $a$, $b$: $$\frac{dy}{dx} = 3x^2 - 12x + 9.$$ 45. Evaluate at $x=5$: $$\frac{dy}{dx} = 75 - 60 + 9 = 24.$$ 46. Use related rates: $$\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} \implies 6 = 24 \cdot \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{6}{24} = \frac{1}{4}.$$ --- **Final Answers:** (a) $\frac{dx}{dt} = -\frac{20}{63}$ when $x=2$ on first curve. (b) Stationary points at $\left(\pm \frac{1}{\sqrt{2}}, -4\right)$ are minima. 2. Tangent to $y=2x + \frac{12}{x^2}$ at $(-2,-1)$ is $y=5x+9$. 3. Stationary points of given rational curve: $(4,6)$ (min) and $(1,-6)$ (max). 4. For cubic curve, $a=-6$, $b=9$, second stationary point: $(3,5)$, and at $x=5$, $\frac{dx}{dt} = \frac{1}{4}$.