Problem 21
Question
In Exercises \(19-22, \mathbf{r}(t)\) 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
Verified Answer
The angle between the velocity and acceleration vectors at time \( t=0 \) is 90 degrees.
1Step 1: Find the velocity vector \( \mathbf{v}(t) \)
The velocity vector \( \mathbf{v}(t) \) is the derivative of the position function \( \mathbf{r}(t) \) with respect to time \( t \).Compute the derivative of each component of \( \mathbf{r}(t) \):1. \( \frac{d}{dt} \left( \ln(t^2 + 1) \right) = \frac{2t}{t^2 + 1} \)2. \( \frac{d}{dt} \left( \tan^{-1} t \right) = \frac{1}{t^2 + 1} \)3. \( \frac{d}{dt} \left( \sqrt{t^2 + 1} \right) = \frac{t}{\sqrt{t^2 + 1}} \)Thus, \[ \mathbf{v}(t) = \frac{2t}{t^2 + 1} \mathbf{i} + \frac{1}{t^2 + 1} \mathbf{j} + \frac{t}{\sqrt{t^2 + 1}} \mathbf{k} \].
2Step 2: Find the acceleration vector \( \mathbf{a}(t) \)
The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity function \( \mathbf{v}(t) \) with respect to time \( t \).Compute the derivative of each component of \( \mathbf{v}(t) \):1. \( \frac{d}{dt} \left( \frac{2t}{t^2 + 1} \right) = \frac{2(t^2 + 1) - 4t^2}{(t^2 + 1)^2} = \frac{-2t^2 + 2}{(t^2 + 1)^2} \)2. \( \frac{d}{dt} \left( \frac{1}{t^2 + 1} \right) = \frac{-2t}{(t^2 + 1)^2} \)3. \( \frac{d}{dt} \left( \frac{t}{\sqrt{t^2 + 1}} \right) = \frac{1}{\sqrt{t^2 + 1}} - \frac{t^2}{(t^2 + 1)^{3/2}} = \frac{1}{(t^2 + 1)^{3/2}} \)Thus, \[ \mathbf{a}(t) = \frac{-2t^2 + 2}{(t^2 + 1)^2} \mathbf{i} + \frac{-2t}{(t^2 + 1)^2} \mathbf{j} + \frac{1}{(t^2 + 1)^{3/2}} \mathbf{k} \].
3Step 3: Evaluate \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \)
Substitute \( t = 0 \) into the expressions for the velocity and acceleration vectors.1. \( \mathbf{v}(0) = \left( \frac{2(0)}{0^2 + 1} \right) \mathbf{i} + \left( \frac{1}{0^2 + 1} \right) \mathbf{j} + \left( \frac{0}{\sqrt{0^2 + 1}} \right) \mathbf{k} = 0 \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k} = \mathbf{j} \).2. \( \mathbf{a}(0) = \left( \frac{-2(0)^2 + 2}{(0^2 + 1)^2} \right) \mathbf{i} + \left( \frac{-2(0)}{(0^2 + 1)^2} \right) \mathbf{j} + \left( \frac{1}{(0^2 + 1)^{3/2}} \right) \mathbf{k} = 2 \mathbf{i} + 0 \mathbf{j} + 1 \mathbf{k} \).
4Step 4: Find the dot product \( \mathbf{v}(0) \cdot \mathbf{a}(0) \)
Calculate the dot product of \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \).\[ \mathbf{v}(0) \cdot \mathbf{a}(0) = (0)(2) + (1)(0) + (0)(1) = 0 \].
5Step 5: Find magnitudes \( \| \mathbf{v}(0) \| \) and \( \| \mathbf{a}(0) \| \)
Calculate the magnitudes of \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \).1. \( \| \mathbf{v}(0) \| = \sqrt{0^2 + 1^2 + 0^2} = 1 \).2. \( \| \mathbf{a}(0) \| = \sqrt{2^2 + 0^2 + 1^2} = \sqrt{5} \).
6Step 6: Find the angle \( \theta \) between \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \)
Use the dot product formula to find the angle \( \theta \):\[ \mathbf{v}(0) \cdot \mathbf{a}(0) = \| \mathbf{v}(0) \| \| \mathbf{a}(0) \| \cos \theta \].Substitute the known values:\[ 0 = 1 \times \sqrt{5} \times \cos \theta \].Solve for \( \theta \):\[ \cos \theta = 0 \] implies \( \theta = \frac{\pi}{2} \) or 90 degrees.
Key Concepts
Velocity VectorAcceleration VectorDot ProductAngle Between Vectors
Velocity Vector
In vector calculus, a velocity vector represents the rate of change of an object's position with respect to time. Think of it as a vector that points in the direction of motion and whose magnitude tells us how fast the object is moving.
When calculating the velocity vector from a given position function \( \mathbf{r}(t) \), you simply take the derivative of each component of the vector with respect to time \( t \). This process gives you a new vector, \( \mathbf{v}(t) \), representing the instantaneous velocity of the particle in space.
For example, if you have a position vector like \( \mathbf{r}(t) = (\ln(t^2 + 1)) \mathbf{i} + (\tan^{-1} t) \mathbf{j} + \sqrt{t^2 + 1} \mathbf{k} \), you differentiate each component to get the velocity vector \( \mathbf{v}(t) = \frac{2t}{t^2 + 1} \mathbf{i} + \frac{1}{t^2 + 1} \mathbf{j} + \frac{t}{\sqrt{t^2 + 1}} \mathbf{k} \). This vector describes how the particle's position is changing at any given moment in time.
When calculating the velocity vector from a given position function \( \mathbf{r}(t) \), you simply take the derivative of each component of the vector with respect to time \( t \). This process gives you a new vector, \( \mathbf{v}(t) \), representing the instantaneous velocity of the particle in space.
For example, if you have a position vector like \( \mathbf{r}(t) = (\ln(t^2 + 1)) \mathbf{i} + (\tan^{-1} t) \mathbf{j} + \sqrt{t^2 + 1} \mathbf{k} \), you differentiate each component to get the velocity vector \( \mathbf{v}(t) = \frac{2t}{t^2 + 1} \mathbf{i} + \frac{1}{t^2 + 1} \mathbf{j} + \frac{t}{\sqrt{t^2 + 1}} \mathbf{k} \). This vector describes how the particle's position is changing at any given moment in time.
Acceleration Vector
Just as the velocity vector tells us about the rate of change of position, the acceleration vector describes the rate of change of velocity. It's like the velocity vector's even more dynamic sibling because it provides insight into how the velocity of an object changes over time.
To find the acceleration vector \( \mathbf{a}(t) \), you need to differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time. This operation will yield another vector that tells us not only the direction in which the velocity is changing but also the rate of this change.
In our example, taking the derivative of the velocity vector \( \mathbf{v}(t) = \frac{2t}{t^2 + 1} \mathbf{i} + \frac{1}{t^2 + 1} \mathbf{j} + \frac{t}{\sqrt{t^2 + 1}} \mathbf{k} \), results in the acceleration vector \( \mathbf{a}(t) = \frac{-2t^2 + 2}{(t^2 + 1)^2} \mathbf{i} + \frac{-2t}{(t^2 + 1)^2} \mathbf{j} + \frac{1}{(t^2 + 1)^{3/2}} \mathbf{k} \). This vector shows how the velocity's magnitude or direction is changing at any moment.
To find the acceleration vector \( \mathbf{a}(t) \), you need to differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time. This operation will yield another vector that tells us not only the direction in which the velocity is changing but also the rate of this change.
In our example, taking the derivative of the velocity vector \( \mathbf{v}(t) = \frac{2t}{t^2 + 1} \mathbf{i} + \frac{1}{t^2 + 1} \mathbf{j} + \frac{t}{\sqrt{t^2 + 1}} \mathbf{k} \), results in the acceleration vector \( \mathbf{a}(t) = \frac{-2t^2 + 2}{(t^2 + 1)^2} \mathbf{i} + \frac{-2t}{(t^2 + 1)^2} \mathbf{j} + \frac{1}{(t^2 + 1)^{3/2}} \mathbf{k} \). This vector shows how the velocity's magnitude or direction is changing at any moment.
Dot Product
The dot product, also known as the scalar product, is a way to multiply two vectors that results in a scalar. Unlike vector multiplication that can yield another vector, the dot product gives us a single number which has several useful properties.
This operation is particularly handy for finding angles and projections. For two vectors \( \mathbf{v} \) and \( \mathbf{a} \), their dot product is computed as \( \mathbf{v} \cdot \mathbf{a} = v_1a_1 + v_2a_2 + v_3a_3 \) if they are in three-dimensional space. The dot product is zero if the two vectors are perpendicular.
In our exercise, finding the dot product of \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \) gives us \( 0\times2 + 1\times0 + 0\times1 = 0 \). Since the product is zero, it confirms that \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \) are perpendicular to each other.
This operation is particularly handy for finding angles and projections. For two vectors \( \mathbf{v} \) and \( \mathbf{a} \), their dot product is computed as \( \mathbf{v} \cdot \mathbf{a} = v_1a_1 + v_2a_2 + v_3a_3 \) if they are in three-dimensional space. The dot product is zero if the two vectors are perpendicular.
In our exercise, finding the dot product of \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \) gives us \( 0\times2 + 1\times0 + 0\times1 = 0 \). Since the product is zero, it confirms that \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \) are perpendicular to each other.
Angle Between Vectors
Finding the angle between vectors is crucial for understanding their spatial relationship. The angle \( \theta \) between two vectors can be determined using the dot product and magnitudes of the vectors with the formula: \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{a}}{\|\mathbf{v}\| \|\mathbf{a}\|} \).
To determine this angle, first, find the dot product of the two vectors and their respective magnitudes. Then, solve for \( \theta \). If the dot product is zero, \( \cos \theta = 0 \), meaning \( \theta = \frac{\pi}{2} \), indicating that the vectors are perpendicular.
In the example, \( \mathbf{v}(0) \cdot \mathbf{a}(0) = 0 \), the magnitudes are \( \| \mathbf{v}(0) \| = 1 \) and \( \| \mathbf{a}(0) \| = \sqrt{5} \), leading to \( \cos \theta = 0 \). Thus, the angle \( \theta \) between the velocity and acceleration vectors at \( t=0 \) is \( 90 \) degrees or \( \frac{\pi}{2} \). This signifies that they are exactly perpendicular at this point in time.
To determine this angle, first, find the dot product of the two vectors and their respective magnitudes. Then, solve for \( \theta \). If the dot product is zero, \( \cos \theta = 0 \), meaning \( \theta = \frac{\pi}{2} \), indicating that the vectors are perpendicular.
In the example, \( \mathbf{v}(0) \cdot \mathbf{a}(0) = 0 \), the magnitudes are \( \| \mathbf{v}(0) \| = 1 \) and \( \| \mathbf{a}(0) \| = \sqrt{5} \), leading to \( \cos \theta = 0 \). Thus, the angle \( \theta \) between the velocity and acceleration vectors at \( t=0 \) is \( 90 \) degrees or \( \frac{\pi}{2} \). This signifies that they are exactly perpendicular at this point in time.
Other exercises in this chapter
Problem 21
Show that \(\kappa\) and \(\tau\) are both zero for the line $$\mathbf{r}(t)=\left(x_{0}+A t\right) \mathbf{i}+\left(y_{0}+B t\right) \mathbf{j}+\left(z_{0}+C t
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Distance along a line Show that if \(\mathbf{u}\) is a unit vector, then the arc length parameter along the line \(\mathbf{r}(t)=P_{0}+t \mathbf{u}\) from the p
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A particle traveling in a straight line is located at the point \((1,-1,2)\) and has speed 2 at time \(t=0 .\) The particle moves toward the point \((3,0,3)\) w
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Find an equation for the circle of curvature of the curve \(\mathbf{r}(t)=\) \((2 \ln t) \mathbf{i}-[t+(1 / t)] \mathbf{j}, e^{-2} \leq t \leq e^{2},\) at the p
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