Problem 89 - Calculus 3

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 of the particle at \(t=\frac{\pi}{4}\)

Step-by-Step Solution

Answer

The acceleration is \(-\frac{\sqrt{2}}{2} \mathbf{i} - \sqrt{2} \mathbf{j}\) at \(t=\frac{\pi}{4}\).

Step 1: Differentiate r(t) to find velocity v(t)

Differentiate the position vector \( \mathbf{r}(t) = \cos t \ \mathbf{i} + 2 \sin t \ \mathbf{j} \) with respect to \( t \) to find the velocity vector \( \mathbf{v}(t) \). Using the derivatives of \( \cos t \) and \( \sin t \), we have:\[\mathbf{v}(t) = \frac{d}{dt}(\cos t) \ \mathbf{i} + \frac{d}{dt}(2 \sin t) \ \mathbf{j} = -\sin t \ \mathbf{i} + 2 \cos t \ \mathbf{j}.\]

Step 2: Differentiate v(t) to find acceleration a(t)

Differentiate the velocity vector \( \mathbf{v}(t) = -\sin t \ \mathbf{i} + 2 \cos t \ \mathbf{j} \) with respect to \( t \) to find the acceleration vector \( \mathbf{a}(t) \). This gives:\[\mathbf{a}(t) = \frac{d}{dt}(-\sin t) \ \mathbf{i} + \frac{d}{dt}(2 \cos t) \ \mathbf{j} = -\cos t \ \mathbf{i} - 2 \sin t \ \mathbf{j}.\]

Step 3: Evaluate a(t) at t=π/4

Substitute \( t = \frac{\pi}{4} \) into the acceleration vector \( \mathbf{a}(t) = -\cos t \ \mathbf{i} - 2 \sin t \ \mathbf{j} \). Using \( \cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \), we have:\[\mathbf{a}\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \ \mathbf{i} - 2 \left(\frac{\sqrt{2}}{2}\right) \ \mathbf{j} = -\frac{\sqrt{2}}{2} \ \mathbf{i} - \sqrt{2} \ \mathbf{j}.\]

Key Concepts

Understanding Differentiation

Differentiation is a fundamental concept in calculus that involves computing the derivative of a function. In simpler terms, it helps us determine the rate at which a quantity changes. For parametric equations, the process of differentiation is essential as it allows us to find the velocity and acceleration vectors.
With respect to the given exercise, the position vector \( \mathbf{r}(t) \) describes the position of a particle as it moves along a curve over time. By differentiating this vector with respect to time \( t \), we obtain the velocity vector \( \mathbf{v}(t) \).
In the exercise, we first differentiate \( \cos t \ \mathbf{i} + 2 \sin t \ \mathbf{j} \) to find the velocity. This involves taking the derivatives of \( \cos t \) and \( \sin t \), which are \(-\sin t\) and \(\cos t\) respectively. This step lays the groundwork for understanding the motion dynamics of the particle.

Exploring the Velocity Vector

The velocity vector is a vital concept when analyzing the motion of objects along parametric paths. It describes not just how fast an object is moving, but also its direction at any given point in time.
In our context, the velocity vector \( \mathbf{v}(t) \) is found by differentiating the position vector \( \mathbf{r}(t) \). After performing the differentiation, we arrive at \( \mathbf{v}(t) = -\sin t \ \mathbf{i} + 2 \cos t \ \mathbf{j} \). This equation clearly maps out how the particle's velocity changes as it travels along the elliptical path.

• \(-\sin t \ \mathbf{i}\) represents the component of velocity in the \( x \)-direction.
• \(2 \cos t \ \mathbf{j}\) gives the velocity's \( y \)-direction component.

Remember, the velocity vector helps paint a clear picture of both the speed and direction of the particle at any given instant along its path.

Deciphering the Acceleration Vector

Acceleration vectors are crucial in understanding how the velocity of a particle changes over time. In our exercise, the acceleration vector is determined by differentiating the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \).
After differentiating \(-\sin t \ \mathbf{i} + 2 \cos t \ \mathbf{j} \), we achieve the acceleration vector \( \mathbf{a}(t) = -\cos t \ \mathbf{i} - 2 \sin t \ \mathbf{j} \). This expression reveals how the velocity in each coordinate changes:

• \(-\cos t \ \mathbf{i}\) is the rate of change of the \( x \)-component of velocity over time.
• \(-2 \sin t \ \mathbf{j}\) shows how the \( y \)-component of velocity changes over time.

By evaluating the acceleration at specific points, such as \( t=\frac{\pi}{4} \), we gain deeper insight into the dynamics of the particle's motion.

Grasping Parametric Paths

Parametric paths offer a powerful way to describe the trajectories of particles in two or more dimensions using a set of equations dependent on a single parameter, usually time \( t \).
In this exercise, the path of the particle is given by the parametric equations \( x(t) = \cos t \) and \( y(t) = 2 \sin t \). Together, these equations map out the particle's path, specifically an ellipse, as \( t \) varies.
The beauty of parametric paths is in their flexibility, allowing us to easily describe complex motions and curves. They facilitate the analysis of motion by breaking down the object’s path into more manageable components, like velocity and acceleration vectors.

• The horizontal motion described by \( \cos t \ \mathbf{i} \) and the vertical movement via \( 2 \sin t \ \mathbf{j} \)
• Analyzing these separately aids in a thorough understanding of motion in each dimension.

Overall, mastering parametric paths provides insightful visualizations and enhances the comprehensive study of dynamic particles in physics and engineering.