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. $$ \mathbf{r}(t)=3 \cos t \mathbf{i}+2 \sin t \mathbf{j},(3,0) $$

Short answer

The graph would show an ellipse centered at the origin with major axis of length 6 on the \(x\)-axis and minor axis of length 4 on the \(y\)-axis. The velocity vector at (3,0) would point directly upward and the acceleration vector would point to the left.

Step-by-step solution

Determine the path of the object

The parametric equations for \(x\) and \(y\) are given by \(x(t) = 3\cos(t)\) and \(y(t) = 2\sin(t)\). This describes an ellipse centered at the origin with radius 3 along the \(x\)-axis and 2 along the \(y\)-axis. So, draw an ellipse centered at the origin, with major axis on the \(x\)-axis of length 6 and minor axis on the \(y\)-axis of length 4.

Find the velocity and acceleration vectors

As the parametric equations show simple harmonic motion, the velocity \(\mathbf{v}(t)\) is the derivative of the position, and is given by \(\mathbf{v}(t) = -3\sin(t)\mathbf{i} + 2\cos(t)\mathbf{j}\). The acceleration \(\mathbf{a}(t)\) is the derivative of the velocity, and is given by \(\mathbf{a}(t) = -3\cos(t)\mathbf{i} -2\sin(t)\mathbf{j}\).

Sketching at the given point

The given point (3,0) corresponds to \(t=0\). For \(t=0\), we have \(\mathbf{v}(0) = 0\mathbf{i} + 2\mathbf{j}\) and \(\mathbf{a}(0) = -3\mathbf{i} - 0\mathbf{j}\). So, locate the point (3,0) on the ellipse, draw a small vector pointing straight up from this point for velocity, and a vector pointing to the left for acceleration.