Let's break down the solution as requested, step by step, with explanations after each step: --- ### **Step 1: Construct the point of the curve at t = 0.5 and the velocity vector** #### **Given points:** - \( P_0 = (400, 0) \) - \( P_1 = (400, 400) \) - \( P_2 = (0, 0) \) - \( P_3 = (0, 400) \) The cubic **Bezier curve** formula: \[ \vec{P}(t) = (1-t)^3\vec{P_0} + 3(1-t)^2 t \vec{P_1} + 3(1-t) t^2 \vec{P_2} + t^3 \vec{P_3} \] #### At \( t = 0.5 \): Calculate each coefficient: - \( (1-t)^3 = (0.5)^3 = 0.125 \) - \( 3(1-t)^2 t = 3 \cdot (0.5)^2 \cdot 0.5 = 3 \cdot 0.25 \cdot 0.5 = 0.375 \) - \( 3(1-t)t^2 = 3 \cdot 0.5 \cdot (0.5)^2 = 3 \cdot 0.5 \cdot 0.25 = 0.375 \) - \( t^3 = (0.5)^3 = 0.125 \) Now plug in the coordinates: \[ \begin{align*} \vec{P}(0.5) =&\ 0.125 \cdot (400,0) + 0.375 \cdot (400,400) \\ &+ 0.375 \cdot (0,0) + 0.125 \cdot (0,400) \\ =&\ (50,0) + (150,150) + (0,0) + (0,50) \\ =&\ (200, 200) \end{align*} \] **Explanation:** The point on the Bezier curve at \( t = 0.5 \) is **(200, 200)**, which is the exact center of the square. --- ### **Step 2: Calculate the velocity vector at \( t = 0.5 \) and determine points with a vertical tangent** The derivative of the cubic Bezier curve (velocity vector) is: \[ \vec{P}'(t) = 3(1-t)^2(\vec{P_1} - \vec{P_0}) + 6(1-t)t(\vec{P_2} - \vec{P_1}) + 3t^2(\vec{P_3} - \vec{P_2}) \] Plug in \( t = 0.5 \): - \( 3(1-t)^2 = 3 \cdot (0.5)^2 = 0.75 \) - \( 6(1-t)t = 6 \cdot 0.5 \cdot 0.5 = 1.5 \) - \( 3t^2 = 3 \cdot (0.5)^2 = 0.75 \) Now, calculate the differences: - \( \vec{P_1} - \vec{P_0} = (400,400) - (400,0) = (0,400) \) - \( \vec{P_2} - \vec{P_1} = (0,0) - (400,400) = (-400,-400) \) - \( \vec{P_3} - \vec{P_2} = (0,400) - (0,0) = (0,400) \) Now, substitute: \[ \vec{P}'(0.5) = 0.75 \cdot (0,400) + 1.5 \cdot (-400,-400) + 0.75 \cdot (0,400) \] \[ = (0,300) + (-600,-600) + (0,300) = ((0+0+0), (300-600+300)) = (-600, 0) \] **Explanation:** The velocity vector at \( t = 0.5 \) is **(-600, 0)**, which means the tangent is **horizontal** at the center of the curve (\( (200, 200) \)). For a **vertical tangent**, the x-component of the velocity must be zero. Let's solve for \( t \) where the x-component of \( \vec{P}'(t) \) is zero: - x-component of \( \vec{P}'(t) \): \[ x'(t) = 3(1-t)^2 (400-400) + 6(1-t)t (0-400) + 3t^2 (0-0) = 6(1-t)t \cdot (-400) \] Set \( x'(t) = 0 \): \[ 6(1-t)t \cdot (-400) = 0 \implies (1-t)t = 0 \implies t = 0 \text{ or } t = 1 \] At \( t = 0 \) and \( t = 1 \), the tangent is vertical. --- ## **Final Answers:** 1. **At \( t=0.5 \), the point on the Bezier curve is \((200,200)\) and the velocity vector is \((-600,0)\), which is horizontal.** 2. **The tangent is vertical at the endpoints (\( t = 0 \) and \( t = 1 \)), i.e., at \((400,0)\) and \((0,400)\).** 3. **The Bezier arc connects \((400,0) \to (400,400) \to (0,0) \to (0,400)\) and passes through the center (200,200) at \( t=0.5 \).** --- ### **b) Sketch for transformation lines \( x=200 \) and \( y=200 \):** - The lines \( x=200 \) and \( y=200 \) will intersect at the center (200,200). - The Bezier arc from part (a) should be drawn as a smooth curve from (400,0) to (0,400) using the given control points. - The frame is a square from (0,0) to (400,400). **(Draw the square, mark the lines \( x=200 \) and \( y=200 \), and overlay the Bezier arc as constructed.)**