1. Problem: Find the radius of curvature for $y=2x^2$ at the point $(1,2)$. Step 1: Compute the first derivative $y' = \frac{dy}{dx} = 4x$. Step 2: Compute the second derivative $y'' = \frac{d^2y}{dx^2} = 4$. Step 3: Use the radius of curvature formula: $$R = \frac{(1 + (y')^2)^{3/2}}{|y''|}$$ At $x=1$, $y' = 4(1) = 4$. Step 4: Calculate: $$R = \frac{(1 + 4^2)^{3/2}}{4} = \frac{(1 + 16)^{3/2}}{4} = \frac{17^{3/2}}{4} = \frac{17 \times \sqrt{17}}{4}$$ 2. Problem: Find the integral $\int (1 + 3x) dx$. Step 1: Integrate term by term: $$\int 1 dx = x$$ $$\int 3x dx = \frac{3x^2}{2}$$ Step 2: Combine and add constant of integration $C$: $$\int (1 + 3x) dx = x + \frac{3x^2}{2} + C$$ 3. Problem: Differentiate $y = (1 + 6x)^{10}$. Step 1: Use chain rule: $$\frac{dy}{dx} = 10(1 + 6x)^9 \times 6 = 60(1 + 6x)^9$$ 4. Problem: Differentiate $$y = 1 - \frac{x^2}{2} + \frac{x^4}{4} - \frac{x^6}{6} + \frac{x^8}{8}$$ Step 1: Differentiate term by term: $$\frac{dy}{dx} = 0 - x + x^3 - x^5 + x^7$$ 5. Problem: Find $\frac{dy}{dx}$ for $y = (5x + 6)(3x + 8)$. Step 1: Use product rule: $$\frac{dy}{dx} = (5)(3x + 8) + (5x + 6)(3) = 15x + 40 + 15x + 18 = 30x + 58$$ 6. Problem: If $y = a x^2 + b x^{-1/2}$, prove that $$2x^2 \frac{d^2y}{dx^2} = x \frac{dy}{dx} + 2y$$ Step 1: Compute first derivative: $$\frac{dy}{dx} = 2ax - \frac{b}{2} x^{-3/2}$$ Step 2: Compute second derivative: $$\frac{d^2y}{dx^2} = 2a + \frac{3b}{4} x^{-5/2}$$ Step 3: Calculate left side: $$2x^2 \frac{d^2y}{dx^2} = 2x^2 \left(2a + \frac{3b}{4} x^{-5/2}\right) = 4a x^2 + \frac{3b}{2} x^{-1/2}$$ Step 4: Calculate right side: $$x \frac{dy}{dx} + 2y = x \left(2ax - \frac{b}{2} x^{-3/2}\right) + 2 \left(a x^2 + b x^{-1/2}\right) = 2a x^2 - \frac{b}{2} x^{-1/2} + 2a x^2 + 2b x^{-1/2} = 4a x^2 + \frac{3b}{2} x^{-1/2}$$ Step 5: Both sides are equal, hence proved. 7. Problem: Find critical points of $y = 8x^3 + 81x^2 - 42x - 8$. Step 1: Find derivative: $$\frac{dy}{dx} = 24x^2 + 162x - 42$$ Step 2: Set derivative to zero: $$24x^2 + 162x - 42 = 0$$ Step 3: Simplify by dividing by 6: $$4x^2 + 27x - 7 = 0$$ Step 4: Use quadratic formula: $$x = \frac{-27 \pm \sqrt{27^2 - 4 \times 4 \times (-7)}}{2 \times 4} = \frac{-27 \pm \sqrt{729 + 112}}{8} = \frac{-27 \pm \sqrt{841}}{8} = \frac{-27 \pm 29}{8}$$ Step 5: Solutions: $$x_1 = \frac{2}{8} = \frac{1}{4}, \quad x_2 = \frac{-56}{8} = -7$$ 8. Problem: Find radius of curvature for $f(x) = 5x^3 - x + 14$ at $x=2$. Step 1: Compute first derivative: $$f'(x) = 15x^2 - 1$$ Step 2: Compute second derivative: $$f''(x) = 30x$$ Step 3: Evaluate at $x=2$: $$f'(2) = 15(4) - 1 = 60 - 1 = 59$$ $$f''(2) = 30(2) = 60$$ Step 4: Use radius of curvature formula: $$R = \frac{(1 + (f')^2)^{3/2}}{|f''|} = \frac{(1 + 59^2)^{3/2}}{60} = \frac{(1 + 3481)^{3/2}}{60} = \frac{3482^{3/2}}{60} = \frac{3482 \times \sqrt{3482}}{60}$$ 9. Problem: Find radius of curvature for $f(x) = 4x^2 + 3x - 7$ at $x=4$. Step 1: Compute first derivative: $$f'(x) = 8x + 3$$ Step 2: Compute second derivative: $$f''(x) = 8$$ Step 3: Evaluate at $x=4$: $$f'(4) = 8(4) + 3 = 32 + 3 = 35$$ Step 4: Use radius of curvature formula: $$R = \frac{(1 + (f')^2)^{3/2}}{|f''|} = \frac{(1 + 35^2)^{3/2}}{8} = \frac{(1 + 1225)^{3/2}}{8} = \frac{1226^{3/2}}{8} = \frac{1226 \times \sqrt{1226}}{8}$$