Problem 6 - 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. Motion on the circle \(x^{2}+y^{2}=16\) $$ \mathbf{r}(t)=\left(4 \cos \frac{t}{2}\right) \mathbf{i}+\left(4 \sin \frac{t}{2}\right) \mathbf{j} ; \quad t=\pi \text { and } 3 \pi / 2 $$

Step-by-Step Solution

Answer

Find velocity and acceleration vectors at \(t = \pi\) and \(t = \frac{3\pi}{2}\) using derivatives; then sketch them.

Step 1: Understand the Position Vector

The position vector given is \(\mathbf{r}(t) = \left(4 \cos \frac{t}{2}\right) \mathbf{i} + \left(4 \sin \frac{t}{2}\right) \mathbf{j}\). This describes a curve in the plane and we need to calculate the velocity and acceleration at specific times \(t = \pi\) and \(t = \frac{3\pi}{2}\).

Step 2: Compute the Velocity Vector

Velocity is the derivative of the position vector \(\mathbf{r}(t)\) with respect to time \(t\). We use differentiation:\[\mathbf{v}(t) = \frac{d}{dt}\left(4 \cos \frac{t}{2}\right) \mathbf{i} + \frac{d}{dt}\left(4 \sin \frac{t}{2}\right) \mathbf{j} = -2 \sin \frac{t}{2} \mathbf{i} + 2 \cos \frac{t}{2} \mathbf{j}\].

Step 3: Evaluate the Velocity at Specified Times

Substitute \(t = \pi\) into \(\mathbf{v}(t)\):\[\mathbf{v}(\pi) = -2 \sin \frac{\pi}{2} \mathbf{i} + 2 \cos \frac{\pi}{2} \mathbf{j} = -2\mathbf{i}\]Substitute \(t = \frac{3\pi}{2}\) into \(\mathbf{v}(t)\):\[\mathbf{v}\left(\frac{3\pi}{2}\right) = -2 \sin \frac{3\pi}{4} \mathbf{i} + 2 \cos \frac{3\pi}{4} \mathbf{j} = -\sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j}\].

Step 4: Compute the Acceleration Vector

Acceleration is the derivative of the velocity vector \(\mathbf{v}(t)\). Differentiate again:\[\mathbf{a}(t) = \frac{d}{dt}\left(-2 \sin \frac{t}{2}\right) \mathbf{i} + \frac{d}{dt}\left(2 \cos \frac{t}{2}\right) \mathbf{j} = -\cos \frac{t}{2} \mathbf{i} - \sin \frac{t}{2} \mathbf{j}\].

Step 5: Evaluate the Acceleration at Specified Times

Substitute \(t = \pi\) into \(\mathbf{a}(t)\):\[\mathbf{a}(\pi) = - \cos \frac{\pi}{2} \mathbf{i} - \sin \frac{\pi}{2} \mathbf{j} = -\mathbf{j}\]Substitute \(t = \frac{3\pi}{2}\) into \(\mathbf{a}(t)\):\[\mathbf{a}\left(\frac{3\pi}{2}\right) = -\cos \frac{3\pi}{4} \mathbf{i} - \sin \frac{3\pi}{4} \mathbf{j} = \frac{1}{\sqrt{2}} \mathbf{i} - \frac{1}{\sqrt{2}} \mathbf{j}\].

Step 6: Sketch the Vectors

Plot the curve \(x^2 + y^2 = 16\), which is a circle with radius 4. At \(t = \pi\), plot the velocity vector \(-2\mathbf{i}\) and acceleration vector \(-\mathbf{j}\) at the position \((0, -4)\). At \(t = \frac{3\pi}{2}\), plot the velocity vector \(-\sqrt{2}\mathbf{i} - \sqrt{2}\mathbf{j}\) and the acceleration vector \(\frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}\) at position \((-\sqrt{2}, -\sqrt{2})\).

Key Concepts

Differentiation

Differentiation is a key concept in calculus, particularly when dealing with velocity and acceleration vectors. It involves finding the rate at which a function changes at any given point. In the context of motion, we often differentiate position vectors to obtain velocity vectors, revealing how fast and in which direction a point is moving. This is crucial for understanding motion along curves, like the circle described in the given exercise.

When differentiating a function, you are essentially finding the slope of the tangent line at a point on a curve. This slope represents the instantaneous rate of change. For example, if our position vector is given by \[ \mathbf{r}(t) = \left(4 \cos \frac{t}{2}\right) \mathbf{i} + \left(4 \sin \frac{t}{2}\right) \mathbf{j} \] we differentiate each component with respect to time \( t \), resulting in the velocity vector:

• \( \mathbf{v}(t) = \frac{d}{dt}\left(4 \cos \frac{t}{2}\right) \mathbf{i} + \frac{d}{dt}\left(4 \sin \frac{t}{2}\right) \mathbf{j} \)
• \( \mathbf{v}(t) = -2 \sin \frac{t}{2} \mathbf{i} + 2 \cos \frac{t}{2} \mathbf{j} \)

So, differentiation allows us to transition from knowing where an object is, to understanding how it moves.

Position Vector

The position vector is a fundamental way to describe the location of a point in space relative to an origin. It is often used in physics and engineering to denote where a particle or object is situated within a defined coordinate system. In the given exercise, the position vector is \[ \mathbf{r}(t) = \left(4 \cos \frac{t}{2}\right) \mathbf{i} + \left(4 \sin \frac{t}{2}\right) \mathbf{j} \], which describes motion along a circular path.

This vector combines two components, each representing a dimension in the plane:

• \( 4 \cos \frac{t}{2} \) is the horizontal component, indicating how far along the x-axis the point is.
• \( 4 \sin \frac{t}{2} \) is the vertical component, showing the distance along the y-axis.

The radius of the circle is consistently 4, since the addition of the squares of these components equals 16, following the equation \( x^2 + y^2 = 16 \). The position vector serves as the starting point for determining dynamics parameters like velocity and acceleration.

Parametric Equations

Parametric equations express the coordinates of the points that make up a geometric object as functions of a variable, typically time \( t \). They are especially useful in capturing the motion of particles, as they allow each coordinate to vary independently. In our scenario, the motion of the particle is described by parametric equations through the position vector \[ \mathbf{r}(t) = \left(4 \cos \frac{t}{2}\right) \mathbf{i} + \left(4 \sin \frac{t}{2}\right) \mathbf{j} \].

This form is beneficial as it offers a clear depiction of motion across dimensions:

• The x-coordinate as a function of \( t \): \( 4 \cos \frac{t}{2} \)
• The y-coordinate as a function of \( t \): \( 4 \sin \frac{t}{2} \)

These equations effectively describe how the particle traverses the defined circle \( x^2 + y^2 = 16 \), letting us explore the trajectory in relation to time. It simplifies the analysis of the motion by considering separate contributions from each spatial dimension.

Calculus

Calculus is the branch of mathematics focusing on change. It offers tools for comprehending the behavior of functions. Fundamental to calculus are differentiation and integration, which tackle rates of change and areas under curves, respectively. The exercise in question extensively employs calculus concepts to extract dynamic information from a position vector.

By differentiating, we obtained the velocity \[ \mathbf{v}(t) = -2 \sin \frac{t}{2} \mathbf{i} + 2 \cos \frac{t}{2} \mathbf{j} \]. This is a classic application of derivatives to measure how position changes over time. Further differentiation of velocity gives acceleration, providing a deeper understanding of the forces and influences acting on a particle.

• This results in the acceleration vector: \( \mathbf{a}(t) = -\cos \frac{t}{2} \mathbf{i} - \sin \frac{t}{2} \mathbf{j} \)

Calculus not only aids in predicting how particles will move, but also in understanding broader phenomena by applying these principles across different contexts of motion and behavior.