Question

The position vector \(r\) describes the path of an object moving in the \(x y\) -plane. Sketch a graph of the path and sketch the velocity and acceleration vectors at the given point. $$ \mathbf{r}(t)=\langle t-\sin t, 1-\cos t\rangle,(\pi, 2) $$

Short answer

The velocity vector at \( (\pi,2) \) is \( \langle 2,0 \rangle \), and the acceleration vector at this point is \( \langle 0,-1 \rangle \). These vectors are represented on the graph of the position function \( \mathbf{r}(t) = \langle t-\sin t, 1-\cos t \rangle \), with the velocity vector pointing to the right and the acceleration vector pointing down from the point \( (\pi,2) \).

Step-by-step solution

Calculate Velocity Vector

The velocity vector is the derivative of the position function. Using the derivative rules, the velocity vector \( \mathbf{v}(t) = \mathbf{r}'(t) \) is calculated as \( \langle 1-\cos t, \sin t \rangle \).

Calculate Acceleration Vector

The acceleration vector is the derivative of the velocity vector. Differentiating the velocity function gives the acceleration vector \( \mathbf{a}(t) = \mathbf{v}'(t) = \langle \sin t, \cos t \rangle \).

Find Velocity and Acceleration at t = π

Substituting \( t = \pi \) in both the velocity and acceleration functions gives the required vectors at the specified point. So, \( \mathbf{v}(\pi) = \langle 1 - \cos(\pi), \sin(\pi) \rangle = \langle 2,0 \rangle \) and \( \mathbf{a}(\pi) = \langle \sin(\pi), \cos(\pi) \rangle = \langle 0,-1 \rangle \).

Sketch the Graph

On a coordinate plane, plot the curve of \( \mathbf{r}(t) = \langle t-\sin t, 1-\cos t \rangle \). Then, draw the vectors \( \mathbf{v}(\pi) \) and \( \mathbf{a}(\pi) \) originating from the point \((\pi,2)\). The vector \(\mathbf{v}(\pi) = \langle 2,0 \rangle \) is a horizontal vector to the right, and the vector \(\mathbf{a}(\pi) = \langle 0,-1 \rangle \) is a vertical vector downward.