Problem 54 - Calculus III

Question

A particle travels along the path of an ellipse with the equation \(\mathbf{r}(t)=\cos t \mathbf{i}+2 \sin t \mathbf{j}+0 \mathbf{k}\). Find the following:Acceleration

Step-by-Step Solution

Answer

The acceleration is \( \mathbf{a}(t) = -\cos t \ \mathbf{i} - 2 \sin t \ \mathbf{j} \).

Step 1: Differentiate to Find Velocity

To find acceleration, we first need to determine the velocity. Velocity \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)=\cos t\ \mathbf{i}+2 \sin t\ \mathbf{j}+0 \mathbf{k}\) with respect to time \(t\). Differentiating term-by-term: \[\mathbf{v}(t)=\frac{d}{dt}(\cos t) \ \mathbf{i}+\frac{d}{dt}(2 \sin t) \ \mathbf{j}+\frac{d}{dt}(0) \ \mathbf{k}= -\sin t\ \mathbf{i}+2\cos t\ \mathbf{j}+0\ \mathbf{k}.\]

Step 2: Differentiate Velocity to Find Acceleration

Acceleration \(\mathbf{a}(t)\) is the derivative of the velocity vector \(\mathbf{v}(t)=-\sin t\ \mathbf{i}+2 \cos t\ \mathbf{j}+0 \mathbf{k}\) with respect to \(t\). Differentiate each component: \\[\mathbf{a}(t)=\frac{d}{dt}(-\sin t)\ \mathbf{i}+\frac{d}{dt}(2 \cos t)\ \mathbf{j}+\frac{d}{dt}(0)\ \mathbf{k}= -\cos t\ \mathbf{i}-2 \sin t\ \mathbf{j}+0 \mathbf{k}.\]

Key Concepts

Velocity Vector

In vector calculus, a velocity vector is crucial for understanding the motion of particles. It represents both the direction and speed of a particle's movement. We determine the velocity vector by differentiating the position vector with respect to time.

• Velocity, in physics, defines how fast and in which direction something moves.
• The velocity vector is derived from the derivative of the position vector over time.

Visualizing this in our exercise, the particle travels along the path of an ellipse, described by \( \mathbf{r}(t) = \cos t\ \mathbf{i} + 2 \sin t\ \mathbf{j} + 0 \mathbf{k} \).To find the velocity vector \( \mathbf{v}(t) \), each component of \( \mathbf{r}(t) \) is differentiated:

• The derivative of \( \cos t \) in the \( \mathbf{i} \) direction is \( -\sin t \).
• The derivative of \( 2 \sin t \) in the \( \mathbf{j} \) direction is \( 2\cos t \).
• The \( \mathbf{k} \) component remains zero as its derivative is zero.

Thus, the velocity vector is \( \mathbf{v}(t) = -\sin t\ \mathbf{i} + 2 \cos t\ \mathbf{j} + 0 \mathbf{k} \). This vector will change over time, indicating how the velocity of the particle changes as it moves along the ellipse.

Ellipse Path

The path of the particle is described by the equation \( \mathbf{r}(t) = \cos t\ \mathbf{i} + 2 \sin t\ \mathbf{j} + 0 \mathbf{k} \). This equation represents an elliptical path, which requires understanding to solve related problems.

• Ellipses are closed, curved paths that resemble elongated circles.
• The general parametric equations for an ellipse are \( x = a \cos t \) and \( y = b \sin t \), where \( a \) and \( b \) are the ellipse's semi-major and semi-minor axes.

In this case:

• The \( \cos t \) and \( 2 \sin t \) components indicate the axes, with \( a = 1 \) and \( b = 2 \). This means that along the x-axis, the ellipse stretches 1 unit, and along the y-axis, it stretches 2 units.
• The particle moves around this ellipse as time changes, following a periodic and smooth path.

Understanding the geometry of the ellipse helps visualize how the particle travels and assists in analyzing its motion.

Differentiation

Differentiation is a fundamental concept in calculus, used to find how a function changes when its inputs change. In the context of motion, differentiation helps find velocity and acceleration from the position function.

• It calculates the rate of change, such as finding how position changes with time to find velocity.
• Differentiation involves basic rules, such as the derivative of \( \sin t \) being \( \cos t \), and the derivative of \( \cos t \) being \( -\sin t \).

Applying differentiation to \( \mathbf{r}(t) \), the position vector:

• First, we differentiate \( \mathbf{r}(t) \) to get the velocity vector \( \mathbf{v}(t) \).
• Then, we differentiate \( \mathbf{v}(t) \) to get the acceleration vector \( \mathbf{a}(t) \).
• For the provided problem, these calculations give \( \mathbf{a}(t) = -\cos t\ \mathbf{i} - 2 \sin t\ \mathbf{j} + 0 \mathbf{k} \), revealing how acceleration changes over time.

Differentiation thus serves as a bridge from understanding position to analyzing movement dynamics, offering a peek into the particle's changing speed and direction.