Let's solve the problem step by step as requested: --- ### **Step 1: Construct the point of the Bezier curve at \( t = 0.5 \) and find the velocity vector** #### **Given:** - \( P_0 = (400, 0) \) - \( P_1 = (400, 400) \) - \( P_2 = (0, 0) \) - \( P_3 = (0, 400) \) Bezier curve formula: \[ \vec{P}(t) = (1-t)^3 \vec{P}_0 + 3(1-t)^2 t \vec{P}_1 + 3(1-t) t^2 \vec{P}_2 + t^3 \vec{P}_3 \] #### **For \( t = 0.5 \):** Calculate each coefficient: - \( (1-t)^3 = (0.5)^3 = 0.125 \) - \( 3(1-t)^2 t = 3 \times (0.5)^2 \times 0.5 = 3 \times 0.25 \times 0.5 = 0.375 \) - \( 3(1-t) t^2 = 3 \times 0.5 \times (0.5)^2 = 3 \times 0.5 \times 0.25 = 0.375 \) - \( t^3 = (0.5)^3 = 0.125 \) Now, calculate \( \vec{P}(0.5) \): \[ \vec{P}(0.5) = 0.125 \cdot (400, 0) + 0.375 \cdot (400, 400) + 0.375 \cdot (0, 0) + 0.125 \cdot (0, 400) \] Break into \( x \) and \( y \) components: **x-component:** \[ x = 0.125 \cdot 400 + 0.375 \cdot 400 + 0.375 \cdot 0 + 0.125 \cdot 0 = (0.125 + 0.375) \cdot 400 = 0.5 \cdot 400 = 200 \] **y-component:** \[ y = 0.125 \cdot 0 + 0.375 \cdot 400 + 0.375 \cdot 0 + 0.125 \cdot 400 = (0.375 + 0.125) \cdot 400 = 0.5 \cdot 400 = 200 \] **So,** \[ \boxed{\vec{P}(0.5) = (200, 200)} \] **Velocity vector:** Take the derivative of the Bezier formula with respect to \( t \): \[ \vec{P}'(t) = 3(1-t)^2 (\vec{P}_1 - \vec{P}_0) + 6(1-t)t (\vec{P}_2 - \vec{P}_1) + 3t^2 (\vec{P}_3 - \vec{P}_2) \] Plug in \( t = 0.5 \): - \( (1-t)^2 = (0.5)^2 = 0.25 \) - \( 6(1-t)t = 6 \times 0.5 \times 0.5 = 1.5 \) - \( t^2 = (0.5)^2 = 0.25 \) \[ \vec{P}'(0.5) = 3 \cdot 0.25 \cdot ((400,400) - (400,0)) + 1.5 \cdot ((0,0) - (400,400)) + 3 \cdot 0.25 \cdot ((0,400) - (0,0)) \] Now calculate: - \( (400,400)-(400,0) = (0,400) \) - \( (0,0)-(400,400) = (-400,-400) \) - \( (0,400)-(0,0) = (0,400) \) So: \[ = 3 \cdot 0.25 \cdot (0,400) + 1.5 \cdot (-400,-400) + 3 \cdot 0.25 \cdot (0,400) \] \[ = (0,300) + (-600,-600) + (0,300) \] \[ = (0,300) + (0,300) = (0,600) \] \[ = (0,600) + (-600,-600) = (-600, 0) \] **So,** \[ \boxed{\vec{P}'(0.5) = (-600, 0)} \] **Explanation (Step 1 conclusion):** - The point at \( t=0.5 \) on the Bezier curve is at **(200, 200)**. - The velocity vector at this point is **(-600, 0)**, indicating a purely horizontal direction (to the left). --- ### **Step 2: Calculate the points at vertical tangent** For a vertical tangent, the x-component of the velocity vector should be zero, i.e., \( \frac{dx}{dt} = 0 \). Recall, \[ \frac{dx}{dt} = 3(1-t)^2 (x_1 - x_0) + 6(1-t)t (x_2 - x_1) + 3t^2 (x_3 - x_2) \] Plug in the \( x \) values: - \( x_0 = 400 \) - \( x_1 = 400 \) - \( x_2 = 0 \) - \( x_3 = 0 \) \[ \frac{dx}{dt} = 3(1-t)^2 (400-400) + 6(1-t)t (0-400) + 3t^2 (0-0) = 0 + 6(1-t)t(-400) + 0 = -2400t(1-t) \] Set \( \frac{dx}{dt} = 0 \): \[ -2400t(1-t) = 0 \implies t = 0 \quad \text{or} \quad t = 1 \] At \( t = 0 \): \( \vec{P}(0) = (400, 0) \) (start point) At \( t = 1 \): \( \vec{P}(1) = (0, 400) \) (end point) **Explanation (Step 2 conclusion):** - The only points where the tangent is vertical are at the endpoints: **(400, 0)** and **(0, 400)**. --- ## **Final Answers** 1. **Point at \( t=0.5 \):** \( (200, 200) \) 2. **Velocity vector at this point:** \( (-600, 0) \) 3. **Points with vertical tangent:** \( (400, 0) \) and \( (0, 400) \) --- Let me know if you want to proceed to the sketching or the next sub-question!