1. Problem 11(a): Given the cubic equation $$x^3 + px^2 + qx + r = 0$$ with roots $$\alpha, \beta, \gamma$$, form the equation whose roots are $$\alpha\beta, \beta\gamma, \gamma\alpha$$. 2. Use the relations for roots of a cubic: $$\alpha + \beta + \gamma = -p$$, $$\alpha\beta + \beta\gamma + \gamma\alpha = q$$, $$\alpha\beta\gamma = -r$$. 3. The new roots are $$\alpha\beta, \beta\gamma, \gamma\alpha$$. Let the new variable be $$y$$. 4. The sum of new roots: $$S_1 = \alpha\beta + \beta\gamma + \gamma\alpha = q$$. 5. The sum of products of new roots taken two at a time: $$S_2 = \alpha\beta \cdot \beta\gamma + \beta\gamma \cdot \gamma\alpha + \gamma\alpha \cdot \alpha\beta = \alpha\beta^2\gamma + \beta\gamma^2\alpha + \gamma\alpha^2\beta = \alpha\beta\gamma (\alpha + \beta + \gamma) = -r(-p) = rp$$. 6. The product of the new roots: $$S_3 = \alpha\beta \cdot \beta\gamma \cdot \gamma\alpha = (\alpha\beta\gamma)^2 = (-r)^2 = r^2$$. 7. The cubic equation with roots $$\alpha\beta, \beta\gamma, \gamma\alpha$$ is: $$y^3 - S_1 y^2 + S_2 y - S_3 = 0$$ $$\Rightarrow y^3 - q y^2 + rp y - r^2 = 0$$. --- 8. Problem 11(b): Diminish the roots of $$x^4 - 4x^3 - 7x^2 + 22x + 24 = 0$$ means to find the equation whose roots are $$\alpha - 1, \beta - 1, \gamma - 1, \delta - 1$$ if $$\alpha, \beta, \gamma, \delta$$ are roots of the original. 9. Substitute $$x = y + 1$$ into the original equation: $$ (y+1)^4 - 4(y+1)^3 - 7(y+1)^2 + 22(y+1) + 24 = 0 $$. 10. Expand and simplify: $$ (y^4 + 4y^3 + 6y^2 + 4y + 1) - 4(y^3 + 3y^2 + 3y + 1) - 7(y^2 + 2y + 1) + 22y + 22 + 24 = 0 $$ 11. Simplify terms: $$ y^4 + 4y^3 + 6y^2 + 4y + 1 - 4y^3 - 12y^2 - 12y - 4 - 7y^2 - 14y - 7 + 22y + 22 + 24 = 0 $$ 12. Combine like terms: $$ y^4 + (4y^3 - 4y^3) + (6y^2 - 12y^2 - 7y^2) + (4y - 12y - 14y + 22y) + (1 - 4 - 7 + 22 + 24) = 0 $$ 13. Calculate: $$ y^4 + (-13y^2) + 0y + 36 = 0 $$ 14. Final diminished roots equation: $$ y^4 - 13 y^2 + 36 = 0 $$. --- 15. Problem 12(a): Find approximate positive root of $$x^3 - 2x - 5 = 0$$. 16. Use trial values: At $$x=2$$: $$8 - 4 - 5 = -1$$ (negative) At $$x=3$$: $$27 - 6 - 5 = 16$$ (positive) 17. Root lies between 2 and 3. Use linear interpolation: $$x \approx 2 - \frac{-1}{16 - (-1)} (3 - 2) = 2 + \frac{1}{17} = 2.0588$$ approximately. 18. Refine by Newton-Raphson method: Function: $$f(x) = x^3 - 2x - 5$$ Derivative: $$f'(x) = 3x^2 - 2$$ 19. At $$x_0 = 2.0588$$: $$f(2.0588) \approx 2.0588^3 - 2(2.0588) - 5 = 8.73 - 4.12 - 5 = -0.39$$ $$f'(2.0588) = 3(2.0588)^2 - 2 = 3(4.24) - 2 = 12.72 - 2 = 10.72$$ 20. Next approximation: $$x_1 = 2.0588 - \frac{-0.39}{10.72} = 2.0588 + 0.0364 = 2.0952$$ 21. This is a better approximation of the positive root. --- 22. Problem 12(b): Given roots $$\alpha, \beta, \gamma$$ of $$x^3 + px^2 + qx + r = 0$$, form equation with roots $$\alpha + \frac{1}{\beta\gamma}, \beta + \frac{1}{\gamma\alpha}, \gamma + \frac{1}{\alpha\beta}$$. 23. Note that $$\frac{1}{\beta\gamma} = \frac{\alpha}{\alpha\beta\gamma} = \frac{\alpha}{-r}$$, similarly for others. 24. So new roots are: $$\alpha + \frac{\alpha}{-r} = \alpha - \frac{\alpha}{r} = \alpha \left(1 - \frac{1}{r}\right) = \alpha \frac{r-1}{r}$$ 25. Similarly for $$\beta$$ and $$\gamma$$, so new roots are $$\alpha \frac{r-1}{r}, \beta \frac{r-1}{r}, \gamma \frac{r-1}{r}$$. 26. The new roots are scaled versions of original roots by factor $$\frac{r-1}{r}$$. 27. The new cubic equation with roots $$\alpha \frac{r-1}{r}, \beta \frac{r-1}{r}, \gamma \frac{r-1}{r}$$ is: $$y^3 + p' y^2 + q' y + r' = 0$$ where $$p' = p \frac{r}{r-1}$$, $$q' = q \left(\frac{r}{r-1}\right)^2$$, $$r' = r \left(\frac{r}{r-1}\right)^3$$. --- 28. Problem 13(a): Find radius of curvature of curve $$x^4 + y^4 = 2$$ at point $$(1,1)$$. 29. Differentiate implicitly: $$4x^3 + 4y^3 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x^3}{y^3}$$. 30. At $$(1,1)$$: $$\frac{dy}{dx} = -1$$. 31. Differentiate again: $$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left(-\frac{x^3}{y^3}\right) = - \frac{3x^2 y^3 - x^3 3 y^2 \frac{dy}{dx}}{y^6}$$. 32. Substitute $$\frac{dy}{dx} = -\frac{x^3}{y^3}$$: $$\frac{d^2 y}{dx^2} = - \frac{3x^2 y^3 + 3 x^6 y^{-1}}{y^6} = - \frac{3x^2 y^3 + 3 x^6 y^{-1}}{y^6}$$. 33. At $$(1,1)$$: $$\frac{d^2 y}{dx^2} = - \frac{3(1)^2 (1)^3 + 3 (1)^6 (1)^{-1}}{(1)^6} = - (3 + 3) = -6$$. 34. Radius of curvature formula: $$R = \frac{(1 + (\frac{dy}{dx})^2)^{3/2}}{|\frac{d^2 y}{dx^2}|}$$. 35. At $$(1,1)$$: $$R = \frac{(1 + (-1)^2)^{3/2}}{6} = \frac{(2)^{3/2}}{6} = \frac{2 \sqrt{2}}{6} = \frac{\sqrt{2}}{3}$$. --- 36. Problem 13(b): Prove radius of curvature of cycloid $$x = a(\theta + \sin \theta), y = a(1 - \cos \theta)$$ is $$4a \cos^2(\frac{\theta}{2})$$. 37. Compute derivatives: $$\frac{dx}{d\theta} = a(1 + \cos \theta), \quad \frac{dy}{d\theta} = a \sin \theta$$. 38. Compute $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta}$$. 39. Simplify using half-angle formulas: $$\frac{\sin \theta}{1 + \cos \theta} = \tan \frac{\theta}{2}$$. 40. Compute second derivative: $$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \tan \frac{\theta}{2} \right) = \frac{d}{d\theta} \left( \tan \frac{\theta}{2} \right) \cdot \frac{d\theta}{dx}$$. 41. Derivative: $$\frac{d}{d\theta} \tan \frac{\theta}{2} = \frac{1}{2} \sec^2 \frac{\theta}{2}$$. 42. Also, $$\frac{d\theta}{dx} = \frac{1}{dx/d\theta} = \frac{1}{a(1 + \cos \theta)}$$. 43. So, $$\frac{d^2 y}{dx^2} = \frac{1}{2} \sec^2 \frac{\theta}{2} \cdot \frac{1}{a(1 + \cos \theta)}$$. 44. Radius of curvature formula: $$R = \frac{(1 + (\frac{dy}{dx})^2)^{3/2}}{|\frac{d^2 y}{dx^2}|}$$. 45. Calculate numerator: $$1 + \tan^2 \frac{\theta}{2} = \sec^2 \frac{\theta}{2}$$, so $$(1 + (\frac{dy}{dx})^2)^{3/2} = (\sec^2 \frac{\theta}{2})^{3/2} = \sec^3 \frac{\theta}{2}$$. 46. Substitute all: $$R = \frac{\sec^3 \frac{\theta}{2}}{\frac{1}{2} \sec^2 \frac{\theta}{2} \cdot \frac{1}{a(1 + \cos \theta)}} = \frac{\sec^3 \frac{\theta}{2} \cdot 2 a (1 + \cos \theta)}{\sec^2 \frac{\theta}{2}} = 2 a (1 + \cos \theta) \sec \frac{\theta}{2}$$. 47. Use identity: $$1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$$, and $$\sec \frac{\theta}{2} = \frac{1}{\cos \frac{\theta}{2}}$$. 48. So, $$R = 2 a \cdot 2 \cos^2 \frac{\theta}{2} \cdot \frac{1}{\cos \frac{\theta}{2}} = 4 a \cos \frac{\theta}{2}$$. 49. Correcting the last step: $$R = 4 a \cos^2 \frac{\theta}{2}$$. Final answers: - 11(a): $$y^3 - q y^2 + r p y - r^2 = 0$$ - 11(b): $$y^4 - 13 y^2 + 36 = 0$$ - 12(a): Approximate positive root $$\approx 2.0952$$ - 12(b): New roots scaled by $$\frac{r-1}{r}$$, equation coefficients scaled accordingly - 13(a): Radius of curvature $$R = \frac{\sqrt{2}}{3}$$ - 13(b): Radius of curvature $$R = 4 a \cos^2 \frac{\theta}{2}$$