Exercises 5-8 give the position vectors of particles moving...

Exercises $5-8$ give the position vectors of particles moving along various curves in the $x y$ -plane. In each case, find the particle's velocity and acceleration vectors at the stated times and sketch them as vectors on the curve. \begin{equation} \mathbf{r}(t)=(\sin t) \mathbf{i}+(\cos t) \mathbf{j} ; \quad t=\pi / 4 \text { and } \pi / 2 \end{equation}

Exercises $5-8$ give the position vectors of particles moving along various curves in the $x y$ -plane. In each case, find the particle's velocity and acceleration vectors at the stated times and sketch them as vectors on the curve.
\begin{equation}
\mathbf{r}(t)=(\sin t) \mathbf{i}+(\cos t) \mathbf{j} ; \quad t=\pi / 4 \text { and } \pi / 2
\end{equation}

Key Concepts

Graphical Representation of Vectors in the Plane

This concept involves sketching the particle's trajectory along with its velocity and acceleration vectors at specific instances. Such visualizations aid in understanding the relationship between the position, velocity, and acceleration, and in interpreting the particle's motion dynamically in the xy-plane.

Parametric Representation of Particle Motion

This concept involves expressing the position of a moving particle using functions of a parameter, typically time. Instead of writing the coordinates as a single equation in x and y, both x and y are given as separate functions of t, allowing a dynamic description of the particle's path.

Differentiation of Vector Functions

Differentiation extends to vector functions by applying the derivative operation to each component separately. This process is essential for finding quantities like velocity and acceleration from the position vector, and it applies rules of differentiation, including those for trigonometric functions.

Velocity Vector

The velocity vector is obtained by differentiating the position vector with respect to time. It provides both the instantaneous speed and the direction of the particle's motion at any given time, thereby serving as a key element in understanding dynamics.

Acceleration Vector

Acceleration is the second derivative of the position vector with respect to time, or equivalently, the first derivative of the velocity vector. It describes how the particle's velocity changes over time, giving insight into the forces acting on the particle.

Transcript

00:01 In this problem, we are given a position vector, r of t, of a particle.

00:08 We want to find the particle's velocity and acceleration.

00:15 So our given position vector is parametrized by a variable t, and is given by the sine of t, i, plus the cos of t, j.

00:33 If we were to plot this in a cartesian system, we'll see that this corresponds to a circle.

00:44 Now to compute the velocity, we need to recall that the velocity corresponds to the derivative of the position, giving us the cos of ti minus the sine of tj.

01:08 Going even further, the acceleration corresponds to the derivative of the velocity or the second derivative of the position.

01:20 So let's differentiate our velocity vector again, and we obtain minus the sine of t i minus the cos of t j.

01:36 And we specifically want to plot these three vectors at the points t equal to pi over 4 and t equal to pi over 2.

01:48 So at t equal to pi over 4, we find that our position is equal to the square root of 2 2 over 2 times i plus j.

02:26 The velocity at this point will be similarly given by the square root of 2 over 2 times i minus j.

02:40 And lastly, the acceleration at pi over 2 is equal to the square root of 2 over 2 minus i minus j.

02:52 Now let's move on to what happens at t equal to pi over 2.

03:10 At this point, the position vector is equal to the vector i because the cos of pi over 2 is equal to 0, whereas the sin of pi over 2 is equal to 1.

03:30 The velocity will give us the vector minus j, and the acceleration yields minus i...

Please give Ace some feedback