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Differential calculus plays a crucial role in understanding motion in polar coordinates. It helps us find the rate of change for various quantities, which is essential for velocity and acceleration calculations. Here, we're working with the differentiation of two specific functions: the radial distance function \( r = 2e^t \) and the angular function \( \theta = 8t^2 \).

To determine how these quantities change with time, we differentiate them with respect to \( t \). For the radial distance, the derivative \( \frac{dr}{dt} = 2e^t \) gives the rate at which the distance from the origin is changing. For the angle, \( \frac{d\theta}{dt} = 16t \) shows how fast the particle is moving along its angular path. These derivatives are fundamental to understanding the particle's instantaneous motion.

Velocity in polar coordinates can be split into two components: radial and transverse.

• Radial velocity is the rate of change of the distance from the origin, represented by \( \frac{dr}{dt} \).
• Transverse velocity, or angular velocity, is given by \( r \frac{d\theta}{dt} \), describing how fast the particle is moving along its circular path.

In this exercise, we compute the radial velocity using the derivative \( \frac{dr}{dt} = 2e^t \) and evaluate it at \( t = 1 \) s to find \( 2e^1 \) ft/s. For the angular velocity, substituting \( r = 2e^1 \) and \( \frac{d\theta}{dt} = 16 \) gives the transverse velocity as \( 2e \times 16 \) ft/s. These components provide a complete picture of the velocity vector in polar coordinates at any given time.

Acceleration in polar coordinates is divided into radial and transverse components, much like velocity.

• The radial acceleration component is given by the formula \( \frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2 \). It indicates how quickly the radial distance is changing with respect to time.
• The transverse acceleration component is represented by \( r \frac{d^2\theta}{dt^2} + 2 \frac{dr}{dt} \frac{d\theta}{dt} \).

The radial acceleration requires the second derivative of the radial distance, while the transverse requires both the derivative of velocity and radius. In our scenario, differentiating \( 2e^t \) again and evaluating it provides the necessary information for these calculations when \( t = 1 \) s, helping us understand the particle's changing state of motion.

Kinematics is the branch of physics that deals with motion without considering the forces that cause it. In polar coordinates, kinematics can become more complex as it involves analyzing motion in terms of radius and angle.

Understanding kinematics in this way helps in visualizing the path of a particle and predicting its future position and motion when parameters like velocity and acceleration are known.

• By knowing the particle's \( r(t) \) and \( \theta(t) \, \) you can determine its trajectory.
• Additionally, velocity and acceleration components calculated using calculus provide insights into speed and directional changes over time.

Overall, studying kinematics in polar coordinates allows us to approach problems involving curvilinear motion, which is common in many real-world applications.