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. $$ \mathbf{r}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j},(\sqrt{2}, \sqrt{2}) $$

Short answer

The graph is a circle, and at the point \(\sqrt{2}, \sqrt{2}\) the velocity vector points to the direction \(-\sqrt{2} \mathbf{i} + \sqrt{2} \mathbf{j}\) and the acceleration vector points to the direction \(-\sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j}\). Both vectors are drawn as arrows extending from the given point in their respective directions.

Step-by-step solution

Draw the graph of the motion

Start by drawing the graph of the function \(r(t)\) which describes the motion of the object. The function \(r(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j}\) shows that this motion is a circle on the plane. The coordinates \(\sqrt{2}, \sqrt{2} \) is the given point plotted on the graph.

Find the velocity vector

The velocity vector \(\mathbf{v}(t)\) of the motion is obtained by taking the first derivative of the position vector function \(r(t)\). So for the given function, \(\mathbf{v}(t) = -2 \sin(t) \mathbf{i} + 2 \cos(t) \mathbf{j}\).At the point \(\sqrt{2}, \sqrt{2} \), we find the corresponding time \(t\) by equating the parameters of \(r(t)\) with the coordinates. Solving \(2 \cos(t) = \sqrt{2} \) gives \(t = \frac{\pi}{4}\). Substituting \( \frac{\pi}{4}\) into \(\mathbf{v}(t)\), we get the velocity vector \(-\sqrt{2} \mathbf{i} + \sqrt{2} \mathbf{j}\) which is sketched from the given point.

Find the acceleration vector

The acceleration vector \(\mathbf{a}(t)\), is the derivative of the velocity vector. Differentiating \(\mathbf{v}(t)\) gives \(\mathbf{a}(t) = -2 \cos(t) \mathbf{i} - 2 \sin(t) \mathbf{j} \). Substituting \( \frac{\pi}{4}\) into \(\mathbf{a}(t)\), we get the acceleration vector \(-\sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j}\) which is also sketched from the given point.