Problem 7 - 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 cycloid \(x=t-\sin t, y=1-\cos t\) $$ \mathbf{r}(t)=\left(4 \cos \frac{t}{2}\right) \mathbf{i}+\left(4 \sin \frac{t}{2}\right) \mathbf{j} ; \quad t=\pi \text { and } 3 \pi / 2 $$

Step-by-Step Solution

Answer

Velocity at \( t = \pi \) is \( -2\mathbf{i} \) and acceleration is \( -\mathbf{j} \); at \( t = \frac{3\pi}{2} \), velocity is \( -\sqrt{2}\mathbf{i} - \sqrt{2}\mathbf{j} \) and acceleration is \( \frac{\sqrt{2}}{2}\mathbf{i} - \frac{\sqrt{2}}{2}\mathbf{j} \).

Step 1: Understand the Position Vector

The position vector of the particle is given by \( \mathbf{r}(t)=\left(4 \cos \frac{t}{2}\right) \mathbf{i}+\left(4 \sin \frac{t}{2}\right) \mathbf{j} \). This represents the x-component and y-component of the vector in the i and j directions, respectively.

Step 2: Differentiate to Find the Velocity Vector

To find the velocity vector, we need to differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Differentiating gives:\[ \mathbf{v}(t) = \frac{d}{dt}\left(4 \cos \frac{t}{2}\right) \mathbf{i} + \frac{d}{dt}\left(4 \sin \frac{t}{2}\right) \mathbf{j} \]Applying the chain rule, we get:\[ \mathbf{v}(t) = -2 \sin \frac{t}{2} \mathbf{i} + 2 \cos \frac{t}{2} \mathbf{j} \]

Step 3: Evaluate Velocity at \( t = \pi \) and \( t = \frac{3\pi}{2} \)

Plug \( t = \pi \) into the velocity vector formula:\[ \mathbf{v}(\pi) = -2 \sin \frac{\pi}{2} \mathbf{i} + 2 \cos \frac{\pi}{2} \mathbf{j} = -2 \mathbf{i} + 0 \mathbf{j} \]Plug \( t = \frac{3\pi}{2} \) into the velocity vector formula:\[ \mathbf{v}\left(\frac{3\pi}{2}\right) = -2 \sin \frac{3\pi}{4} \mathbf{i} + 2 \cos \frac{3\pi}{4} \mathbf{j} = -2\left(\frac{\sqrt{2}}{2}\right) \mathbf{i} - 2\left(\frac{\sqrt{2}}{2}\right)\mathbf{j}= -\sqrt{2} \mathbf{i} - \sqrt{2}\mathbf{j} \]

Step 4: Differentiate to Find the Acceleration Vector

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

Step 5: Evaluate Acceleration at \( t = \pi \) and \( t = \frac{3\pi}{2} \)

Insert \( t = \pi \) into the acceleration vector:\[ \mathbf{a}(\pi) = - \cos \frac{\pi}{2} \mathbf{i} - \sin \frac{\pi}{2} \mathbf{j} = 0 \mathbf{i} - 1 \mathbf{j} = - \mathbf{j}\]Insert \( t = \frac{3\pi}{2} \) into the acceleration vector:\[ \mathbf{a}\left(\frac{3\pi}{2}\right) = - \cos \frac{3\pi}{4} \mathbf{i} - \sin \frac{3\pi}{4} \mathbf{j} = -\left(-\frac{\sqrt{2}}{2}\right) \mathbf{i} - \left(\frac{\sqrt{2}}{2}\right) \mathbf{j} = \frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j}\]

Step 6: Sketch the Vectors

Draw the position of the particle on the curve at \( t = \pi \) and \( t = \frac{3\pi}{2} \). At each position, sketch the corresponding velocity and acceleration vectors. - At \( t = \pi \), sketch \( \mathbf{v}(\pi) = -2 \mathbf{i} \) (a horizontal arrow pointing left) and \( \mathbf{a}(\pi) = - \mathbf{j} \) (a vertical arrow pointing down).- At \( t = \frac{3\pi}{2} \), sketch \( \mathbf{v}(\frac{3\pi}{2}) = -\sqrt{2} \mathbf{i} - \sqrt{2} \mathbf{j} \) (diagonal to the bottom left) and \( \mathbf{a}(\frac{3\pi}{2}) = \frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \) (diagonal to the top left).

Key Concepts

Velocity Vectors

Velocity vectors are crucial for understanding how fast and in which direction a particle moves along a path. In this exercise, we start with a position vector \( \mathbf{r}(t) = \left(4 \cos \frac{t}{2}\right) \mathbf{i} + \left(4 \sin \frac{t}{2}\right) \mathbf{j} \). This represents the location of the particle at time \( t \). To find the velocity vector, we differentiate the position vector with respect to time. When performing differentiation, we apply the chain rule, which gives us the velocity vector \( \mathbf{v}(t) = -2 \sin \frac{t}{2} \mathbf{i} + 2 \cos \frac{t}{2} \mathbf{j} \).

• The velocity component \( -2 \sin \frac{t}{2} \) in the \( \mathbf{i} \) direction shows the motion on the x-axis.
• The velocity component \( 2 \cos \frac{t}{2} \) in the \( \mathbf{j} \) direction indicates motion on the y-axis.

To find specific velocities at certain times, such as \( t = \pi \) and \( t = \frac{3\pi}{2} \), you substitute those values into \( \mathbf{v}(t) \). The results tell us how the particle moves at those specific times.

Acceleration Vectors

Acceleration vectors help us understand how the velocity of a particle changes with time. It is the rate at which the velocity vector itself changes. Once we have the velocity vector \( \mathbf{v}(t) = -2 \sin \frac{t}{2} \mathbf{i} + 2 \cos \frac{t}{2} \mathbf{j} \), we find the acceleration vector by differentiating the velocity vector with respect to time \( t \).Performing another differentiation while applying the chain rule yields the acceleration vector: \( \mathbf{a}(t) = - \cos \frac{t}{2} \mathbf{i} - \sin \frac{t}{2} \mathbf{j} \).

• The \( - \cos \frac{t}{2} \) in the \( \mathbf{i} \) direction shows how acceleration affects motion along the x-axis.
• The \( - \sin \frac{t}{2} \) in the \( \mathbf{j} \) direction affects motion along the y-axis.

By substituting specific values of \( t \) into \( \mathbf{a}(t) \), we can determine the acceleration at different moments, guiding us on how the particle's velocity changes at those times.

Differentiation

Differentiation is a fundamental mathematical process used to find rates of change, like velocity and acceleration. In the context of position vectors, differentiating helps us understand how a particle's position and motion evolve over time.When differentiating a position vector like \( \mathbf{r}(t) = \left(4 \cos \frac{t}{2}\right) \mathbf{i} + \left(4 \sin \frac{t}{2}\right) \mathbf{j} \), we get to compute the velocity vector \( \mathbf{v}(t) \). Another round of differentiation then provides us with the acceleration vector \( \mathbf{a}(t) \).

• Start with the function you want to differentiate.
• Use the chain rule, which is especially important when your function involves compositions as in trigonometric functions like \( \cos \) and \( \sin \).

Differentiation essentially helps in predicting future motion by analyzing current velocity and acceleration.

Parametric Equations

Parametric equations are a powerful way to express curves in the coordinate plane because they allow for separate equations to define the x and y coordinates in terms of a third variable, generally time \( t \). This approach is particularly useful for describing complex motion paths, such as cycloids.In this exercise, the position vector \( \mathbf{r}(t) = \left(4 \cos \frac{t}{2}\right) \mathbf{i} + \left(4 \sin \frac{t}{2}\right) \mathbf{j} \) uses trigonometric functions to describe the path of a particle over time.

• Parametric form allows for precise control and understanding of each coordinate’s evolution independently.
• The motion path can be easily visualized by sketching these equations over a range of \( t \) values.

Parametric equations are extremely valuable for sketching paths like circles, ellipses, and more complex curves, enabling a detailed analysis of each coordinate's behavior via differentiation and subsequently finding velocity and acceleration.