Problem 17 - Thomas Calculus

Question

In Exercises \(15-18, \mathbf{r}(t)\) is the position of a particle in space at time \(t .\) Find the angle between the velocity and acceleration vectors at time \(t=0 .\) $$ \mathbf{r}(t)=\left(\ln \left(t^{2}+1\right)\right) \mathbf{i}+\left(\tan ^{-1} t\right) \mathbf{j}+\sqrt{t^{2}+1} \mathbf{k} $$

Step-by-Step Solution

Answer

The angle is 90 degrees.

Step 1: Find the Velocity Vector

First, determine the velocity vector, \( \mathbf{v}(t) \), by differentiating the position vector \( \mathbf{r}(t) \) with respect to \( t \). Given \( \mathbf{r}(t)=\left(\ln \left(t^{2}+1\right)\right) \mathbf{i}+\left(\tan^{-1} t\right) \mathbf{j}+\sqrt{t^{2}+1} \mathbf{k} \), calculate its derivative: - \( \frac{d}{dt} \ln(t^2+1) = \frac{2t}{t^2+1} \)- \( \frac{d}{dt} \tan^{-1}t = \frac{1}{t^2+1} \)- \( \frac{d}{dt} \sqrt{t^2+1} = \frac{t}{\sqrt{t^2+1}} \)Thus, \[ \mathbf{v}(t) = \frac{2t}{t^2+1}\,\mathbf{i} + \frac{1}{t^2+1} \,\mathbf{j} + \frac{t}{\sqrt{t^2+1}}\, \mathbf{k} \].

Step 2: Find the Acceleration Vector

Next, find the acceleration vector, \( \mathbf{a}(t) \), by differentiating the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). Differentiate each component of \( \mathbf{v}(t) \): - \( \frac{d}{dt} \left( \frac{2t}{t^2+1} \right) = \frac{2(t^2+1) - 2t(2t)}{(t^2+1)^2} = \frac{2 - 2t^2}{(t^2+1)^2} \)- \( \frac{d}{dt} \left( \frac{1}{t^2+1} \right) = \frac{-2t}{(t^2+1)^2} \)- \( \frac{d}{dt} \left( \frac{t}{\sqrt{t^2+1}} \right) = \frac{\sqrt{t^2+1} - \frac{t^2}{\sqrt{t^2+1}}}{t^2+1} = \frac{1}{(t^2+1)^{3/2}} \)So, \[ \mathbf{a}(t) = \frac{2 - 2t^2}{(t^2+1)^2}\, \mathbf{i} + \frac{-2t}{(t^2+1)^2} \, \mathbf{j} + \frac{1}{(t^2+1)^{3/2}}\, \mathbf{k} \].

Step 3: Evaluate Velocity and Acceleration at t=0

Evaluate the velocity and acceleration vectors at \( t=0 \):- \( \mathbf{v}(0) = \frac{2(0)}{0^2+1} \mathbf{i} + \frac{1}{0^2+1} \mathbf{j} + \frac{0}{\sqrt{0^2+1}} \mathbf{k} = \mathbf{j} \)- \( \mathbf{a}(0) = \frac{2 - 0}{(0^2+1)^2} \mathbf{i} + \frac{-2(0)}{(0^2+1)^2} \mathbf{j} + \frac{1}{(0^2+1)^{3/2}} \mathbf{k} = 2\,\mathbf{i} + \mathbf{k} \)

Step 4: Find the Dot Product of the Vectors

Find the dot product \( \mathbf{v}(0) \cdot \mathbf{a}(0) \). \( \mathbf{v}(0) = \mathbf{j} \) and \( \mathbf{a}(0) = 2\,\mathbf{i} + \mathbf{k} \), so their dot product is:\( \mathbf{j} \cdot (2\,\mathbf{i} + \mathbf{k}) = 0 \).

Step 5: Magnitudes of Velocity and Acceleration

Calculate the magnitudes of \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \):- \( |\mathbf{v}(0)| = \sqrt{0^2 + 1^2 + 0^2} = 1 \)- \( |\mathbf{a}(0)| = \sqrt{2^2 + 0^2 + 1^2} = \sqrt{5} \)

Step 6: Find the Angle Between Vectors

Use the dot product formula to find the angle \( \theta \) between \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \):\[ \cos \theta = \frac{\mathbf{v}(0) \cdot \mathbf{a}(0)}{|\mathbf{v}(0)| \cdot |\mathbf{a}(0)|} = \frac{0}{1 \cdot \sqrt{5}} = 0 \]Since \( \cos \theta = 0 \), \( \theta = 90^\circ \).

Key Concepts

Velocity Vector

The velocity vector represents the rate of change of an object's position over time. It tells us how fast and in what direction the object is moving at any given moment.
For a position vector \( \mathbf{r}(t) \), the velocity vector \( \mathbf{v}(t) \) is obtained by differentiating \( \mathbf{r}(t) \) with respect to time \( t \).
This process is known as differentiation, and it allows us to analyze the instantaneous change in position.
In the exercise given, the position vector \( \mathbf{r}(t) = \left(\ln \left(t^{2}+1\right)\right) \mathbf{i}+\left(\tan^{-1} t\right) \mathbf{j}+\sqrt{t^{2}+1} \mathbf{k} \) leads to the velocity vector

• \( \mathbf{v}(t) = \frac{2t}{t^2+1}\,\mathbf{i} + \frac{1}{t^2+1} \,\mathbf{j} + \frac{t}{\sqrt{t^2+1}}\, \mathbf{k} \)

At \( t=0 \), \( \mathbf{v}(0) \) simplifies to \( \mathbf{j} \), showing that the velocity is in the direction of the \( \mathbf{j} \) unit vector, or simply the y-axis.

Acceleration Vector

Acceleration is the rate at which an object's velocity changes over time. It can affect both the speed and direction of the velocity. To find the acceleration vector \( \mathbf{a}(t) \), we differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \).
This second differentiation provides a mathematical representation of how quickly velocity changes, reflecting the dynamic nature of motion.
For the given velocity vector \( \mathbf{v}(t) = \frac{2t}{t^2+1}\,\mathbf{i} + \frac{1}{t^2+1} \,\mathbf{j} + \frac{t}{\sqrt{t^2+1}}\, \mathbf{k} \), the differentiation yields

• \( \mathbf{a}(t) = \frac{2 - 2t^2}{(t^2+1)^2}\, \mathbf{i} + \frac{-2t}{(t^2+1)^2} \, \mathbf{j} + \frac{1}{(t^2+1)^{3/2}}\, \mathbf{k} \)

Evaluating at \( t=0 \), the acceleration vector simplifies to \( 2\,\mathbf{i} + \mathbf{k} \), which indicates influence in the x and z-axes.

Dot Product

The dot product of two vectors is a way to multiply them, producing a scalar (a single number) which tells us how much of one vector goes in the direction of the other.
It is found by multiplying corresponding components of the vectors and summing up those products.
The formula for the dot product \( \mathbf{u} \cdot \mathbf{v} \) is:

• \( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \)

In our exercise, the velocity vector \( \mathbf{v}(0) = \mathbf{j} \) and the acceleration vector \( \mathbf{a}(0) = 2\,\mathbf{i} + \mathbf{k} \) yield a dot product

• \( \mathbf{j} \cdot (2\,\mathbf{i} + \mathbf{k}) = 0 \)

This result indicates orthogonality or perpendicularly between the vectors, crucial for determining angles.

Angle Between Vectors

The angle between two vectors can be computed using the dot product and the magnitudes of the vectors. The angle is important as it describes the geometric relationship between them.
The formula to find the cosine of the angle \( \theta \) is:

• \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{a}}{|\mathbf{v}| \cdot |\mathbf{a}|} \)

In the given problem, the dot product is zero, which implies:

• \( \cos \theta = \frac{0}{1 \cdot \sqrt{5}} = 0 \)

This calculation leads to \( \theta = 90^\circ \), showing that the velocity and acceleration vectors are perpendicular to each other. This serves as a classic example of trigonometric application in vector calculus.