Let's solve each problem step by step: --- ### **11. Rate at which the surface area decreases when the radius is 4 cm** Given: - The radius \( r \) decreases at rate \( \frac{dr}{dt} = -0.2 \) cm/min - Surface area of a sphere: \( SA = 4\pi r^2 \) - Find \( \frac{d(SA)}{dt} \) when \( r = 4 \) cm **Step 1: Differentiate the surface area with respect to time:** \[ \frac{d(SA)}{dt} = \frac{d}{dt}[4\pi r^2] = 8\pi r \frac{dr}{dt} \] **Step 2: Substitute \( r = 4 \) and \( \frac{dr}{dt} = -0.2 \):** \[ \frac{d(SA)}{dt} = 8\pi \times 4 \times (-0.2) \] \[ \frac{d(SA)}{dt} = 32\pi \times (-0.2) \] \[ \frac{d(SA)}{dt} = -6.4\pi \text{ cm}^2/\text{min} \] **Final Answer:** \[ \boxed{-6.4\pi \text{ cm}^2/\text{min}} \] or approximately \[ \boxed{-20.11 \text{ cm}^2/\text{min}} \] --- ### **12. Find \( a \) and \( b \) for the function \( f(x) = ax^3 + bx^2 - 5x + 9 \) given \( f(-1) = 12 \) and \( f'(-1) = 3 \)** **Step 1: Plug \( x = -1 \) into \( f(x) \):** \[ f(-1) = a(-1)^3 + b(-1)^2 - 5(-1) + 9 = -a + b + 5 + 9 = -a + b + 14 \] Set equal to 12: \[ -a + b + 14 = 12 \implies -a + b = -2 \tag{1} \] **Step 2: Find \( f'(x) \) and plug \( x = -1 \):** \[ f'(x) = 3ax^2 + 2bx - 5 \] \[ f'(-1) = 3a(-1)^2 + 2b(-1) - 5 = 3a - 2b - 5 \] Set equal to 3: \[ 3a - 2b - 5 = 3 \implies 3a - 2b = 8 \tag{2} \] **Step 3: Solve the system of equations:** From (1): \( -a + b = -2 \implies b = a - 2 \) Substitute into (2): \[ 3a - 2(a - 2) = 8 \implies 3a - 2a + 4 = 8 \implies a + 4 = 8 \implies a = 4 \] \[ b = a - 2 = 4 - 2 = 2 \] **Final Answer:** \[ \boxed{a = 4, \;\; b = 2} \] --- ### **13. Equation of lines through \((-4, -5)\) and tangent to \(y = x^2 + 1\)** Let the tangent point be \( (a, a^2 + 1) \). Slope of tangent = \( y' = 2a \). Equation of the tangent: \[ y - (a^2 + 1) = 2a(x - a) \] It passes through \( (-4, -5) \): \[ -5 - (a^2 + 1) = 2a(-4 - a) \] \[ -5 - a^2 - 1 = -8a - 2a^2 \] \[ -6 - a^2 = -8a - 2a^2 \] \[ -6 - a^2 + 2a^2 + 8a = 0 \] \[ a^2 + 8a - 6 = 0 \] Solve quadratic: \[ a = \frac{-8 \pm \sqrt{64 + 24}}{2} = \frac{-8 \pm \sqrt{88}}{2} = \frac{-8 \pm 2\sqrt{22}}{2} = -4 \pm \sqrt{22} \] So \( a_1 = -4 + \sqrt{22} \), \( a_2 = -4 - \sqrt{22} \) Equations: \[ \text{Slope } m_1 = 2a_1 = 2(-4 + \sqrt{22}) = -8 + 2\sqrt{22} \] \[ \text{Point } (a_1, a_1^2 + 1) \] So, tangent line 1: \[ y - (a_1^2 + 1) = m_1(x - a_1) \] Similarly for \( a_2 \). **Final Answers:** \[ \text{Line 1: } y - (a_1^2 + 1) = m_1(x - a_1), \quad a_1 = -4 + \sqrt{22} \] \[ \text{Line 2: } y - (a_2^2 + 1) = m_2(x - a_2), \quad a_2 = -4 - \sqrt{22} \] Or, explicitly (rounded to two decimals): \[ a_1 \approx -4 + 4.69 = 0.69,\quad m_1 \approx 2 \times 0.69 = 1.38 \] \[ y - (0.69^2 + 1) = 1.38(x - 0.69) \implies y - 1.48 = 1.38(x - 0.69) \] \[ a_2 \approx -4 - 4.69 = -8.69, \quad m_2 \approx 2 \times (-8.69) = -17.38 \] \[ y - ((-8.69)^2 + 1) = -17.38(x + 8.69) \implies y - 76.5 = -17.38(x + 8.69) \] --- ### **14. Positions and velocities when accelerations are equal** Given: \[ s_1 = 4t - t^2,\qquad s_2 = 5t^2 - t^3 \] **Step 1: Compute accelerations** \[ v_1 = \frac{ds_1}{dt} = 4 - 2t \implies a_1 = \frac{dv_1}{dt} = -2 \] \[ v_2 = \frac{ds_2}{dt} = 10t - 3t^2 \implies a_2 = \frac{dv_2}{dt} = 10 - 6t \] Set accelerations equal: \[ -2 = 10 - 6t \implies -6t = -12 \implies t = 2 \] **Step 2: Find positions and velocities at \( t = 2 \):** For particle 1: \[ s_1 = 4(2) - (2)^2 = 8 - 4 = 4 \] \[ v_1 = 4 - 2(2) = 4 - 4 = 0 \] For particle 2: \[ s_2 = 5(2)^2 - (2)^3 = 5 \times 4 - 8 = 20 - 8 = 12 \] \[ v_2 = 10 \times 2 - 3 \times (2)^2 = 20 - 12 = 8 \] **Final Answer:** At \( t = 2 \): - **Particle 1:** \( s_1 = 4 \), \( v_1 = 0 \) - **Particle 2:** \( s_2 = 12 \), \( v_2 = 8 \) --- If you need a detailed solution for any specific problem above, let me know!

Certainly! Here are the final answers for each problem: --- ### **11. Rate at which the surface area decreases when the radius is 4 cm** **Final Answer:** \(\boxed{-6.4\pi \text{ cm}^2/\text{min}}\) Approximately: \(\boxed{-20.11 \text{ cm}^2/\text{min}}\) --- ### **12. Values of \(a\) and \(b\) for the function \(f(x) = ax^3 + bx^2 - 5x + 9\)** **Final Answers:** \(\boxed{a = 4, \quad b = 2}\) --- ### **13. Equations of the two lines passing through \((-4, -5)\) and tangent to \( y = x^2 + 1 \)** Approximate coefficients rounded to two decimal places: - **Line 1:** \[ a_1 \approx 0.69,\quad m_1 \approx 1.38 \] Equation: \[ y - 1.48 = 1.38 (x - 0.69) \] - **Line 2:** \[ a_2 \approx -8.69,\quad m_2 \approx -17.38 \] Equation: \[ y - 76.5 = -17.38 (x + 8.69) \] --- ### **14. Positions and velocities when the accelerations are equal** At \( t = 2 \): - **Particle 1:** Position: \(\boxed{4}\) Velocity: \(\boxed{0}\) - **Particle 2:** Position: \(\boxed{12}\) Velocity: \(\boxed{8}\) --- Let me know if you need further clarification!