Problem 50 - 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:Speed of the particle at \(t=\frac{\pi}{4}\)

Step-by-Step Solution

Answer

Speed at \( t = \frac{\pi}{4} \) is \( \sqrt{\frac{5}{2}} \).

Step 1: Differentiate the Position Vector

The speed of a particle is the magnitude of its velocity vector. First, find the velocity vector by taking the derivative of the position vector \( \mathbf{r}(t) = \cos t \mathbf{i} + 2 \sin t \mathbf{j} + 0 \mathbf{k} \) with respect to \( t \). This gives us: \[ \mathbf{v}(t) = \frac{d}{dt}\left(\cos t\right) \mathbf{i} + \frac{d}{dt}\left(2 \sin t\right) \mathbf{j} \] which simplifies to \( \mathbf{v}(t) = -\sin t \mathbf{i} + 2 \cos t \mathbf{j} \).

Step 2: Evaluate the Velocity Vector at Given Time

Substitute \( t = \frac{\pi}{4} \) into the velocity vector found: \[ \mathbf{v}\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) \mathbf{i} + 2 \cos\left(\frac{\pi}{4}\right) \mathbf{j} \] Simplifying using \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \) and \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \), we have: \[ \mathbf{v}\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \mathbf{i} + \sqrt{2} \mathbf{j} \].

Step 3: Compute the Speed from the Velocity Magnitude

The speed is the magnitude of the velocity vector. Use the formula for the magnitude of a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) which is \( \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2} \). Substitute the components from \( \mathbf{v}\left(\frac{\pi}{4}\right) \): \[ \|\mathbf{v}\left(\frac{\pi}{4}\right)\| = \sqrt{\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(\sqrt{2}\right)^2} = \sqrt{\frac{2}{4} + 2} = \sqrt{\frac{1}{2} + 2} \].

Step 4: Simplify the Result to Find the Speed

Continuing from the previous calculation: \[ \|\mathbf{v}\left(\frac{\pi}{4}\right)\| = \sqrt{\frac{1}{2} + 2} = \sqrt{\frac{1}{2} + \frac{4}{2}} = \sqrt{\frac{5}{2}} \]. Hence, the speed of the particle at \( t = \frac{\pi}{4} \) is \( \sqrt{\frac{5}{2}} \).

Key Concepts

Position Vector

A position vector plays a crucial role in determining the location of a particle in a multi-dimensional space at any given time. In this exercise, the position vector is represented as \( \mathbf{r}(t) = \cos t \mathbf{i} + 2 \sin t \mathbf{j} + 0 \mathbf{k} \). This mathematical expression tells us how the particle moves along an elliptical path, with each component indicating the particle's position along the respective axis.

The \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) are unit vectors pointing in the direction of the x-axis, y-axis, and z-axis respectively. However, since the \(\mathbf{k}\) component is zero, the particle's path is restricted to the xy-plane, making the motion effectively two-dimensional. This setup is common in problems involving circular or elliptical paths in physics and engineering.

Understanding position vectors helps us track the motion of particles without physically observing each move. This vector highlights the particle's "current" position on its traveling path at any time \(t\). Recognizing these components has practical applications in mechanics, robotics, and many branches of physics.

Velocity Vector

The velocity vector is derived by differentiating the position vector with respect to time. In essence, it represents how fast and in which direction an object is moving at a given point in time. For this problem, by differentiating the position vector \( \mathbf{r}(t) = \cos t \mathbf{i} + 2 \sin t \mathbf{j} + 0 \mathbf{k} \), the velocity vector is calculated as \( \mathbf{v}(t) = -\sin t \mathbf{i} + 2 \cos t \mathbf{j} \).

Each component of the velocity vector \(-\sin t \mathbf{i}\) and \(2 \cos t \mathbf{j}\) represents the rates of change of position along the x-axis and y-axis respectively. In physical terms, this means:

• The \(-\sin t \mathbf{i}\) component indicates the horizontal velocity component: how the x-coordinate of the particle changes over time.
• The \(2 \cos t \mathbf{j}\) component indicates the vertical velocity component: how the y-coordinate changes.

A time-specific velocity vector gives insights into the instantaneous speed and direction of the particle at any moment, crucial for trajectory analysis and understanding dynamic motion.

Magnitude of a Vector

The magnitude of a vector, sometimes called its "length" or "norm," provides a scalar quantity that describes how long the vector is, irrespective of its direction. For the purpose of this exercise, we use the magnitude to find the speed of the particle. The speed is essentially the magnitude of the velocity vector.

Using the formula for the magnitude of a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), it is computed as \(\|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2}\). When we calculated the magnitude of the velocity vector \(\mathbf{v}(t) = -\sin t \mathbf{i} + 2 \cos t \mathbf{j} \) at \( t = \frac{\pi}{4} \), the components \(a\) and \(b\) are substituted accordingly:\[ \|\mathbf{v}(\frac{\pi}{4})\| = \sqrt{\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(\sqrt{2}\right)^2} \]
This process concludes with obtaining the speed by calculation.

With understanding the magnitude, you're equipped to measure the intensity of vectors, essential in various real-world applications like physics simulations, computer graphics, or any field requiring vector analysis.

Parametric Equations of Motion

Parametric equations allow us to describe motion in a rather elegant and comprehensive way by setting x, y, and z as functions of a third variable, often time \(t\). In this problem, both position and velocity of the particle are articulated through parametric equations.

For the exercise, we have the parametric equation of the path as \(\mathbf{r}(t) = \cos t \mathbf{i} + 2 \sin t \mathbf{j}\). This expression gives us the coordinates of the particle at any time \(t\), showing that the position depends directly on that specific time.

Parametric equations provide several advantages:

• They simplify complex trajectories by describing motion in terms of simple equations for each component.
• They make it easier to calculate related quantities, for instance, derivatives like acceleration when described using parametric equations of motion.
• They are adaptable to various coordinate systems, making them a universal tool in mathematics and physics.

Understanding parametric equations enriches your toolset for solving problems involving moving objects, not just in academics but in any applied physics or engineering role.