Problem 5 - Thomas Calculus

Question

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}

Step-by-Step Solution

Answer

Velocity at \( t=\pi/4 \) is \(\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j}\), at \( t=\pi/2 \) is \(-\mathbf{j}\). Acceleration at \( t=\pi/4 \) is \(-\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j}\), at \( t=\pi/2 \) is \(-\mathbf{i}\).

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) of a particle is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Given that \( \mathbf{r}(t) = \sin t \mathbf{i} + \cos t \mathbf{j} \), we differentiate each component:\[ \frac{d}{dt}(\sin t) = \cos t \]\[ \frac{d}{dt}(\cos t) = -\sin t \]Thus, the velocity vector is \( \mathbf{v}(t) = \cos t \mathbf{i} - \sin t \mathbf{j} \).

Step 2: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). We'll differentiate each component of \( \mathbf{v}(t) = \cos t \mathbf{i} - \sin t \mathbf{j} \):\[ \frac{d}{dt}(\cos t) = -\sin t \]\[ \frac{d}{dt}(-\sin t) = -\cos t \]Thus, the acceleration vector is \( \mathbf{a}(t) = -\sin t \mathbf{i} - \cos t \mathbf{j} \).

Step 3: Calculate Velocity at \( t = \pi/4 \) and \( \pi/2 \)

Substitute \( t = \pi/4 \) into the velocity vector \( \mathbf{v}(t) = \cos t \mathbf{i} - \sin t \mathbf{j} \):\[ \mathbf{v}(\pi/4) = \cos(\pi/4) \mathbf{i} - \sin(\pi/4) \mathbf{j} = \frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \]For \( t = \pi/2 \):\[ \mathbf{v}(\pi/2) = \cos(\pi/2) \mathbf{i} - \sin(\pi/2) \mathbf{j} = 0 \mathbf{i} - 1 \mathbf{j} = -\mathbf{j} \]

Step 4: Calculate Acceleration at \( t = \pi/4 \) and \( \pi/2 \)

Substitute \( t = \pi/4 \) into the acceleration vector \( \mathbf{a}(t) = -\sin t \mathbf{i} - \cos t \mathbf{j} \):\[ \mathbf{a}(\pi/4) = -\sin(\pi/4) \mathbf{i} - \cos(\pi/4) \mathbf{j} = -\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \]For \( t = \pi/2 \):\[ \mathbf{a}(\pi/2) = -\sin(\pi/2) \mathbf{i} - \cos(\pi/2) \mathbf{j} = -1 \mathbf{i} - 0 \mathbf{j} = -\mathbf{i} \]

Step 5: Sketch the Vectors

At \( t = \pi/4 \), the velocity vector \( \frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \) and the acceleration vector \( -\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \) form equal magnitude vectors in different directions. At \( t = \pi/2 \), the velocity vector points downward \(-\mathbf{j}\) and the acceleration vector points leftward \(-\mathbf{i}\) at the end of the circular path.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and operations on them. It is a key component in physics and engineering because it helps us understand physical phenomena.
Vector calculus involves concepts like vector addition, scalar multiplication, and differentiation of vector functions.

• It allows us to find velocity and acceleration from position vectors, which is crucial in mechanics.
• It includes understanding gradient, divergence, and curl, essential in electromagnetism and fluid dynamics.

By using vector calculus, we can effectively describe changes in physical systems and predict the future states of moving objects.

Position Vectors

A position vector describes the location of a point in space. In the plane, we often represent position vectors using the Cartesian coordinate system. For a moving particle, the position vector can change with time.
The general form of a position vector \(\mathbf{r}(t)\) in two dimensions can be expressed as:

• \(\mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j}\), where \(x(t)\) and \(y(t)\) are functions of time.

These functions define the path of the particle. In the given exercise, the position vector \(\mathbf{r}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j}\) indicates that the particle moves in a circular path.

Velocity and Acceleration Vectors

Velocity and acceleration vectors are derived from the position vector. The velocity vector \(\mathbf{v}(t)\) provides the speed and direction of the moving particle, while the acceleration vector \(\mathbf{a}(t)\) shows how the velocity changes over time.
To find these vectors:

• The velocity vector is the derivative of the position vector: \(\mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t)\).
• The acceleration vector is the derivative of the velocity vector: \(\mathbf{a}(t) = \frac{d}{dt}\mathbf{v}(t)\).

This process, called differentiation, allows us to understand and predict the particle's future motion by examining its current state.

Differentiation

Differentiation is a mathematical process used to compute the derivative of a function. It measures how a function changes as its input changes. In vector calculus, differentiation is used to find velocity and acceleration from position vectors.
For a function \(f(t)\), its derivative \(f'(t)\) shows the rate of change at any point. In the example:

• The derivative of \(\sin t\) is \(\cos t\), used to find the velocity vector component along \(\mathbf{i}\).
• The derivative of \(\cos t\) is \(-\sin t\), used for the component along \(\mathbf{j}\).

Differentiation gives us valuable information to analyze particles' motion and understand dynamic systems.