Let's solve each question step by step. --- ### **6.1 (a) Calculate the rate of change of vertical position at \( t = 3 \) seconds.** **Given:** \[ v(t) = \sin\left(\frac{t}{\sqrt{t^2 + 1}}\right) \] We need to find \(\frac{dv}{dt}\) at \(t = 3\). **Step 1: Let \( u = \frac{t}{\sqrt{t^2 + 1}} \), then \( v(t) = \sin(u) \).** **Step 2: Differentiate using chain rule:** \[ \frac{dv}{dt} = \cos(u) \cdot \frac{du}{dt} \] **Step 3: Find \(\frac{du}{dt}\):** \[ u = \frac{t}{\sqrt{t^2 + 1}} \] Let \( y = t \), \( z = \sqrt{t^2 + 1} \) \[ \frac{d}{dt}\left(\frac{y}{z}\right) = \frac{z \cdot \frac{dy}{dt} - y \cdot \frac{dz}{dt}}{z^2} \] \[ \frac{dy}{dt} = 1, \quad \frac{dz}{dt} = \frac{1}{2}(t^2 + 1)^{-1/2} \cdot 2t = \frac{t}{\sqrt{t^2 + 1}} \] \[ \frac{du}{dt} = \frac{\sqrt{t^2 + 1} (1) - t \left( \frac{t}{\sqrt{t^2 + 1}} \right)}{t^2 + 1} = \frac{\sqrt{t^2 + 1} - \frac{t^2}{\sqrt{t^2 + 1}}}{t^2 + 1} = \frac{t^2 + 1 - t^2}{(t^2 + 1)\sqrt{t^2 + 1}} = \frac{1}{(t^2 + 1)^{3/2}} \] **Step 4: Substitute back:** \[ \frac{dv}{dt} = \cos\left( \frac{t}{\sqrt{t^2 + 1}} \right) \cdot \frac{1}{(t^2 + 1)^{3/2}} \] **Step 5: Calculate at \( t = 3 \):** \[ u = \frac{3}{\sqrt{3^2 + 1}} = \frac{3}{\sqrt{10}} \approx 0.949 \] \[ (t^2 + 1)^{3/2} = (9 + 1)^{3/2} = 10^{3/2} = 10 \cdot \sqrt{10} \approx 31.623 \] \[ \cos(u) = \cos(0.949) \approx 0.582 \] \[ \frac{dv}{dt} \approx \frac{0.582}{31.623} \approx 0.0184 \] **Final Answer (rounded to three decimals):** \[ \boxed{0.018} \] --- ### **6.1 (b) Explain what this rate means in context.** **Answer:** This rate represents how quickly the drone's vertical position is changing at \( t = 3 \) seconds. A positive value means the drone is ascending (moving upward), and the magnitude tells us how fast this change is happening at that instant. --- ### **6.1 (c) At what time will the drone momentarily stop ascending or descending?** **We want:** When \(\frac{dv}{dt} = 0\). \[ \frac{dv}{dt} = \cos\left(\frac{t}{\sqrt{t^2 + 1}}\right) \cdot \frac{1}{(t^2 + 1)^{3/2}} = 0 \] The second factor is never zero, so: \[ \cos\left(\frac{t}{\sqrt{t^2 + 1}}\right) = 0 \] \[ \frac{t}{\sqrt{t^2 + 1}} = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \] For the **smallest positive \( t \)**, set \( n = 0 \): \[ \frac{t}{\sqrt{t^2 + 1}} = \frac{\pi}{2} \] Let \( x = t \): \[ \frac{x}{\sqrt{x^2 + 1}} = \frac{\pi}{2} \] Solve for \( x \): 1. Cross-multiply: \[ x = \frac{\pi}{2} \sqrt{x^2 + 1} \] 2. Square both sides: \[ x^2 = \frac{\pi^2}{4} (x^2 + 1) \] \[ 4x^2 = \pi^2 (x^2 + 1) \] \[ 4x^2 = \pi^2 x^2 + \pi^2 \] \[ 4x^2 - \pi^2 x^2 = \pi^2 \] \[ x^2 (4 - \pi^2) = \pi^2 \] \[ x^2 = \frac{\pi^2}{4 - \pi^2} \] \[ x = \frac{\pi}{\sqrt{4 - \pi^2}} \] Calculate numerically (\(\pi \approx 3.142\)): \[ 4 - \pi^2 \approx 4 - 9.870 = -5.870 \] So, \[ x^2 = \frac{9.870}{-5.870} = -1.682 \] Since negative, this equation has **no real solution** for \( n = 0 \). Try \( n = 1 \): \[ \frac{t}{\sqrt{t^2 + 1}} = \frac{3\pi}{2} \] Let \( y = t \): \[ y = \frac{3\pi}{2} \sqrt{y^2 + 1} \] \[ y^2 = \frac{9\pi^2}{4}(y^2 + 1) \] \[ 4y^2 = 9\pi^2(y^2 + 1) \] \[ 4y^2 = 9\pi^2 y^2 + 9\pi^2 \] \[ 4y^2 - 9\pi^2 y^2 = 9\pi^2 \] \[ y^2 (4 - 9\pi^2) = 9\pi^2 \] \[ y^2 = \frac{9\pi^2}{4 - 9\pi^2} \] \[ 4 - 9\pi^2 \approx 4 - 9 \times 9.870 = 4 - 88.83 = -84.83 \] \[ y^2 = \frac{9 \times 9.87}{-84.83} = \frac{88.83}{-84.83} = -1.047 \] Still negative. Try negative \( n \), for the first positive solution. Since the argument of cosine is always less than 1, the function never reaches \(\frac{\pi}{2}\) so the answer is that the drone never momentarily stops ascending or descending after takeoff for real \( t \), **unless the sine function output is restricted**. **Thus:** For this function, the rate of change never actually reaches zero for real \( t \) (since \( \frac{t}{\sqrt{t^2 + 1}} < 1 \) for all real \( t \)), so the drone never momentarily stops ascending or descending. --- ### **6.2 Calculate \( f'(1) \) if \( f(x) = \frac{x^3 \ln x}{3x^2 - 1} \)** Use the quotient rule: \[ f'(x) = \frac{(3x^2 \ln x + x^2) (3x^2 - 1) - x^3 \ln x \cdot 6x}{(3x^2 - 1)^2} \] Let’s compute step by step: Let \( u = x^3 \ln x \), \( v = 3x^2 - 1 \): - \( u' = 3x^2 \ln x + x^2 \) - \( v' = 6x \) So, \[ f'(x) = \frac{u'v - uv'}{v^2} \] \[ = \frac{(3x^2 \ln x + x^2)(3x^2 - 1) - x^3 \ln x \cdot 6x}{(3x^2 - 1)^2} \] Now substitute \( x = 1 \): - \( \ln 1 = 0 \) - \( x^2 = 1 \) - \( 3x^2 - 1 = 2 \) - \( x^3 = 1 \) Plug in: \[ u' = 3 \cdot 1^2 \cdot 0 + 1^2 = 0 + 1 = 1 \] \[ u = 1^3 \cdot 0 = 0 \] \[ v = 2 \] \[ v' = 6 \cdot 1 = 6 \] So, \[ f'(1) = \frac{1 \cdot 2 - 0 \cdot 6}{2^2} = \frac{2}{4} = 0.5 \] --- ## **Final Answers** ### 6.1 (a) \[ \boxed{0.018} \] ### 6.1 (b) > This rate represents how quickly the drone's vertical position is changing at \( t = 3 \) seconds. A positive value means the drone is ascending, and the magnitude tells how fast. ### 6.1 (c) > The drone will **never momentarily stop ascending or descending** for real \( t \), as the rate of change never reaches zero. ### 6.2 \[ \boxed{0.5} \] --- Let me know if you need detailed working for any step!