Problem 18 - Thomas Calculus

Question

\({r}(t)\) is the position of a particle in space at time \(t .\) Find the angle between the velocity and acceleration vectors at time \(t=0 .\) \begin{equation} \mathbf{r}(t)=\frac{4}{9}(1+t)^{3 / 2} \mathbf{i}+\frac{4}{9}(1-t)^{3 / 2} \mathbf{j}+\frac{1}{3} t \mathbf{k} \end{equation}

Step-by-Step Solution

Answer

The vectors are orthogonal; the angle is 90 degrees.

Step 1: Compute the Velocity Vector

To find the velocity vector, take the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Calculate \( \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) \).The velocity vector is:\[\mathbf{v}(t) = \frac{d}{dt}\left( \frac{4}{9}(1+t)^{3/2} \right) \mathbf{i} + \frac{d}{dt}\left( \frac{4}{9}(1-t)^{3/2} \right) \mathbf{j} + \frac{d}{dt}\left( \frac{1}{3}t \right) \mathbf{k}\]After computing, we have:\[\mathbf{v}(t) = \frac{2}{3}(1+t)^{1/2} \mathbf{i} - \frac{2}{3}(1-t)^{1/2} \mathbf{j} + \frac{1}{3} \mathbf{k}\].

Step 2: Compute the Acceleration Vector

To find the acceleration vector, take the derivative of the velocity vector \( \mathbf{v}(t) \).Calculate \( \mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) \).The acceleration vector is:\[\mathbf{a}(t) = \frac{d}{dt}\left( \frac{2}{3}(1+t)^{1/2} \right) \mathbf{i} - \frac{d}{dt}\left( \frac{2}{3}(1-t)^{1/2} \right) \mathbf{j} + \frac{d}{dt}\left( \frac{1}{3} \right) \mathbf{k}\]After computing, we have:\[\mathbf{a}(t) = \frac{1}{3\sqrt{1+t}} \mathbf{i} + \frac{1}{3\sqrt{1-t}} \mathbf{j}\].

Step 3: Evaluate Velocity and Acceleration Vectors at t = 0

Substitute \( t = 0 \) into the expressions for \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \) that we obtained from Steps 1 and 2.Velocity vector at \( t = 0 \):\[\mathbf{v}(0) = \frac{2}{3} \mathbf{i} - \frac{2}{3} \mathbf{j} + \frac{1}{3} \mathbf{k} \].Acceleration vector at \( t = 0 \):\[\mathbf{a}(0) = \frac{1}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} \].

Step 4: Compute the Dot Product of Velocity and Acceleration

Calculate the dot product \( \mathbf{v}(0) \cdot \mathbf{a}(0) \) to help determine the angle between the two vectors at \( t = 0 \).\( \mathbf{v}(0) \cdot \mathbf{a}(0) = \left( \frac{2}{3} \right)\left( \frac{1}{3} \right) + \left( -\frac{2}{3} \right)\left( \frac{1}{3} \right) + \left( \frac{1}{3} \right)(0) = 0 \).

Step 5: Calculate the Magnitudes of the Vectors

Calculate the magnitudes of the velocity and acceleration vectors at \( t = 0 \).Magnitude of \( \mathbf{v}(0) \):\[| \mathbf{v}(0) | = \sqrt{\left( \frac{2}{3} \right)^2 + \left( -\frac{2}{3} \right)^2 + \left( \frac{1}{3} \right)^2} = 1 \].Magnitude of \( \mathbf{a}(0) \):\[| \mathbf{a}(0) | = \sqrt{\left( \frac{1}{3} \right)^2 + \left( \frac{1}{3} \right)^2} = \frac{1}{\sqrt{2}} \].

Step 6: Calculate the Angle Between the Vectors

Use the dot product and magnitudes to find the cosine of the angle \( \theta \) between the velocity and acceleration vectors.\[\cos \theta = \frac{\mathbf{v}(0) \cdot \mathbf{a}(0)}{| \mathbf{v}(0) | | \mathbf{a}(0) |}\].Since \( \mathbf{v}(0) \cdot \mathbf{a}(0) = 0 \),\[\cos \theta = 0 \].Thus, \( \theta = 90^\circ \). The angle between the vectors is \(90^\circ\), meaning they are orthogonal.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector calculus. It allows us to determine the angle between two vectors. Given two vectors, say \( \mathbf{u} \) and \( \mathbf{v} \), the dot product is calculated as:

• \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \)

Here, \( u_1, u_2, u_3 \) and \( v_1, v_2, v_3 \) are the components of \( \mathbf{u} \) and \( \mathbf{v} \) respectively.
The value of the dot product has a special significance.

• If it is zero, the vectors are orthogonal, implying a 90-degree angle between them.
• A positive value indicates an acute angle.
• A negative value shows an obtuse angle.

In this exercise, we calculated the dot product of the velocity and acceleration vectors at \( t = 0 \), finding it to be zero. This confirms their orthogonality, which we'll delve into next.

Orthogonality

Orthogonality is a concept used to describe when two vectors are perpendicular to each other. In a geometric sense, this means they form a right angle, or 90 degrees. This is crucial in many fields, including physics, engineering, and computer science.
The exercise we examined applied orthogonality to the velocity and acceleration vectors of a moving particle. At \( t = 0 \), the calculation showed that the dot product was zero:

• \( \mathbf{v}(0) \cdot \mathbf{a}(0) = 0 \)

This result confirms the vectors are orthogonal, reflecting the fact that the direction in which the particle is moving does not influence the acceleration at that particular time.
This specific condition is often met in scenarios where an object changes direction rapidly or experiences central force, such as circular motion. Keeping this in mind helps us understand complex motion and the interrelation between motion's direction and rate of change.

Derivatives of Vector Functions

In vector calculus, finding the derivative of a vector function is key to understanding dynamic systems. A vector function may represent the position of an object, and its derivatives represent its velocity and acceleration, which are crucial in physics and engineering.
The derivative of a vector function is taken component-wise. If \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \), its derivative \( \mathbf{r}'(t) \) (velocity) is:

• \( \mathbf{r}'(t) = x'(t) \mathbf{i} + y'(t) \mathbf{j} + z'(t) \mathbf{k} \)

Similarly, the second derivative (acceleration) is:

• \( \mathbf{r}''(t) = x''(t) \mathbf{i} + y''(t) \mathbf{j} + z''(t) \mathbf{k} \)

In the original exercise, we derived the velocity and acceleration vectors from the position vector by taking the first and second derivatives. This process helps us analyze motion in a spatial context, such as finding the angle between velocity and acceleration, which provides insights into how a particle's path and force interplay dynamically.