Problem 2 - Calculus III

Question

Given \(\mathbf{r}(t)=\left(3 t^{2}-2\right) \mathbf{i}+(2 t-\sin (t)) \mathbf{j}\), find the acceleration vector of a particle moving along the curve in the preceding exercise.

Step-by-Step Solution

Answer

The acceleration vector is \( 6\mathbf{i} + \sin(t)\mathbf{j} \).

Step 1: Understand the Problem

We need to find the acceleration vector of a particle moving along a curve defined by the position vector \( \mathbf{r}(t) = (3t^2 - 2)\mathbf{i} + (2t - \sin(t))\mathbf{j} \). First, we will find the velocity vector by differentiating the position vector, and then differentiate the velocity vector to find the acceleration vector.

Step 2: Differentiate to Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). Taking the derivative: \[ \mathbf{v}(t) = \frac{d}{dt}[(3t^2 - 2)\mathbf{i} + (2t - \sin(t))\mathbf{j}] = (6t)\mathbf{i} + (2 - \cos(t))\mathbf{j}. \]

Step 3: Differentiate to Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). Taking the derivative: \[ \mathbf{a}(t) = \frac{d}{dt}[(6t)\mathbf{i} + (2 - \cos(t))\mathbf{j}] = 6\mathbf{i} + \sin(t)\mathbf{j}. \]

Step 4: Conclusion

Thus, the acceleration vector \( \mathbf{a}(t) \) is \( 6\mathbf{i} + \sin(t)\mathbf{j} \).

Key Concepts

Understanding the Position Vector

A position vector is a vital concept in describing the location of a particle in space. It expresses the particle's current position relative to a specified origin. In a 2D or 3D space, the position vector is typically written as \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \), where \( x(t), y(t), \) and \( z(t) \) are functions of time \( t \). Here, the functions represent the coordinates of the particle at any time \( t \).

In this specific exercise, the position vector \( \mathbf{r}(t) = (3t^2 - 2)\mathbf{i} + (2t - \sin(t))\mathbf{j} \) defines the curve along which the particle moves. Each component of the vector is a function of time, allowing us to determine the particle's location in the plane at any given instant.

• The term \( 3t^2 - 2 \) indicates how the \( x \)-coordinate changes over time.
• The term \( 2t - \sin(t) \) reveals the movement in the \( y \)-direction.

Understanding this structure is key to analyzing motion because it provides the starting point to derive other vectors such as velocity and acceleration. In real life, this could represent anything from a car traveling on a road to a satellite orbiting Earth.

Velocity Vector and its Derivation

The velocity vector of a particle describes how its position changes over time. It is the first derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Mathematically, if we denote the velocity vector by \( \mathbf{v}(t) \), it can be expressed as: \[ \mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t) = \frac{d}{dt}[(3t^2 - 2)\mathbf{i} + (2t - \sin(t))\mathbf{j}] = (6t)\mathbf{i} + (2 - \cos(t))\mathbf{j} \]
This computation effectively involves:

• Differentiating each component of the position vector separately.
• The \( x \)-component \( (3t^2 - 2) \) turns into \( (6t) \), signifying the change rate of the particle's horizontal position.
• The \( y \)-component \( (2t - \sin(t)) \) becomes \( (2 - \cos(t)) \), illustrating the rate of change in the vertical direction.

The velocity vector is crucial because it provides both the speed and direction of the particle. Just as a car’s speedometer tells speed, the direction of this vector indicates which way the particle is moving.

Diving into Differentiation to Find Acceleration

Differentiation is a fundamental tool in calculus used to determine the rate of change of a function. In motion analysis, we apply this technique to the velocity vector to obtain the acceleration vector, which shows how the velocity itself is changing over time. This second derivative provides insight into the dynamical forces acting on the particle.

For our particle, the acceleration vector \( \mathbf{a}(t) \) can be found by differentiating the velocity vector \( \mathbf{v}(t) \) with respect to \( t \):\[ \mathbf{a}(t) = \frac{d}{dt}[(6t)\mathbf{i} + (2 - \cos(t))\mathbf{j}] = 6\mathbf{i} + \sin(t)\mathbf{j} \]

• The derivative of \( 6t \) simply becomes \( 6 \), indicating a constant horizontal acceleration.
• For the derivative of \( (2 - \cos(t)) \), we get \( \sin(t) \) as the change in the \( y \)-component over time.

Thus, the final acceleration vector shows both constant and variable components, denoting diverse forces’ impacts on the particle's motion over time. In more complex scenarios, this might reflect various real-world influences such as gravity,forces, or other dynamic constraints on the object’s path.