Let's break down the problem step by step: --- **Given:** - The curve: \( x = \frac{y^3}{3} \) - Vertical motion: \( y = \frac{t^2}{4} \) - Find **velocity** (\(\vec{v}\)), **acceleration** (\(\vec{a}\)), and **radius of curvature** (\(R\)) at \( t = 2 \) seconds. --- ## Step 1: Find \(y\) and \(x\) as functions of \(t\) Given: \( y = \frac{t^2}{4} \) So, at any time \( t \), \[ y(t) = \frac{t^2}{4} \] For \( x \): \[ x = \frac{y^3}{3} \] Substitute \( y(t) \): \[ x(t) = \frac{1}{3} \left( \frac{t^2}{4} \right)^3 = \frac{1}{3} \cdot \frac{t^6}{64} = \frac{t^6}{192} \] So, \[ x(t) = \frac{t^6}{192} \] --- ## Step 2: Compute derivatives (velocity components) ### (a) \( \dot{y} = \frac{dy}{dt} \) \[ \frac{dy}{dt} = \frac{d}{dt} \left( \frac{t^2}{4} \right) = \frac{2t}{4} = \frac{t}{2} \] ### (b) \( \dot{x} = \frac{dx}{dt} \) \[ \frac{dx}{dt} = \frac{d}{dt} \left( \frac{t^6}{192} \right) = \frac{6t^5}{192} = \frac{t^5}{32} \] --- ## Step 3: Velocity at \( t = 2 \) seconds **Plug \( t = 2 \) into the above:** - \( \dot{y} = \frac{2}{2} = 1 \) - \( \dot{x} = \frac{(2)^5}{32} = \frac{32}{32} = 1 \) **So,** \[ \vec{v} = (\dot{x}, \dot{y}) = (1, 1) \] **Magnitude:** \[ |\vec{v}| = \sqrt{(\dot{x})^2 + (\dot{y})^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \] --- ## Step 4: Acceleration components ### (a) \( \ddot{y} = \frac{d^2y}{dt^2} \) \[ \frac{d^2y}{dt^2} = \frac{d}{dt} \left( \frac{t}{2} \right) = \frac{1}{2} \] ### (b) \( \ddot{x} = \frac{d^2x}{dt^2} \) \[ \frac{d^2x}{dt^2} = \frac{d}{dt} \left( \frac{t^5}{32} \right) = \frac{5 t^4}{32} \] At \( t = 2 \): \[ \ddot{x} = \frac{5 \times (2)^4}{32} = \frac{5 \times 16}{32} = \frac{80}{32} = 2.5 \] **So,** \[ \vec{a} = (\ddot{x}, \ddot{y}) = (2.5, 0.5) \] **Magnitude:** \[ |\vec{a}| = \sqrt{(2.5)^2 + (0.5)^2} = \sqrt{6.25 + 0.25} = \sqrt{6.5} \approx 2.55 \] --- ## Step 5: Radius of Curvature (\(R\)) at \( t = 2 \) For parametric curves \( x(t), y(t) \): \[ R = \frac{\left( \dot{x}^2 + \dot{y}^2 \right)^{3/2}}{|\dot{x} \ddot{y} - \dot{y} \ddot{x}|} \] At \( t = 2 \): - \( \dot{x} = 1 \) - \( \dot{y} = 1 \) - \( \ddot{x} = 2.5 \) - \( \ddot{y} = 0.5 \) Compute numerator: \[ (\dot{x}^2 + \dot{y}^2)^{3/2} = (1^2 + 1^2)^{3/2} = (2)^{3/2} = 2^{1.5} = 2 \sqrt{2} \] Compute denominator: \[ |\dot{x} \ddot{y} - \dot{y} \ddot{x}| = |(1)(0.5) - (1)(2.5)| = |0.5 - 2.5| = 2 \] So: \[ R = \frac{2 \sqrt{2}}{2} = \sqrt{2} \approx 1.414 \] --- ## **Final Answers (boxed):** ### 1. **Velocity at \( t = 2 \) s:** \[ \boxed{ \vec{v} = (1, 1) \quad \text{magnitude: } \sqrt{2} } \] ### 2. **Acceleration at \( t = 2 \) s:** \[ \boxed{ \vec{a} = (2.5, 0.5) \quad \text{magnitude: } \approx 2.55 } \] ### 3. **Radius of curvature at \( t = 2 \) s:** \[ \boxed{ R = \sqrt{2} \approx 1.41 } \] --- **Summary Table:** | Quantity | Value at \( t = 2 \) | |--------------------|--------------------------| | Velocity | \( (1, 1),\; |\vec{v}| = \sqrt{2} \) | | Acceleration | \( (2.5, 0.5),\; |\vec{a}| \approx 2.55 \) | | Radius of curvature| \( R = \sqrt{2} \approx 1.41 \) |