Problem 8
Question
Find the velocity, acceleration, and speed of a particle with the given position function. $$ \mathbf{r}(t)=\langle\sin t, t, \cos t\rangle . \text { The graph is shown here: } $$
Step-by-Step Solution
Verified Answer
Velocity: \( \langle \cos t, 1, -\sin t \rangle \), Acceleration: \( \langle -\sin t, 0, -\cos t \rangle \), Speed: \( \sqrt{2} \).
1Step 1: Differentiate to find velocity
The velocity vector is the derivative of the position vector with respect to time. So, differentiate each component of \( \mathbf{r}(t) = \langle \sin t, t, \cos t \rangle \). The derivative of \( \sin t \) is \( \cos t \), the derivative of \( t \) is \( 1 \), and the derivative of \( \cos t \) is \( -\sin t \). Thus, the velocity vector is \( \mathbf{v}(t) = \langle \cos t, 1, -\sin t \rangle \).
2Step 2: Differentiate velocity to find acceleration
The acceleration vector is the derivative of the velocity vector with respect to time. Differentiate each component of \( \mathbf{v}(t) = \langle \cos t, 1, -\sin t \rangle \). The derivative of \( \cos t \) is \( -\sin t \), the derivative of \( 1 \) is \( 0 \), and the derivative of \( -\sin t \) is \( -\cos t \). Hence, the acceleration vector is \( \mathbf{a}(t) = \langle -\sin t, 0, -\cos t \rangle \).
3Step 3: Calculate the speed of the particle
Speed is the magnitude of the velocity vector. Given \( \mathbf{v}(t) = \langle \cos t, 1, -\sin t \rangle \), the speed is calculated as the square root of the sum of the squares of its components: \[ \text{Speed} = \sqrt{(\cos t)^2 + 1^2 + (-\sin t)^2} = \sqrt{\cos^2 t + 1 + \sin^2 t} \]. Since \( \cos^2 t + \sin^2 t = 1 \), the speed simplifies to \( \sqrt{2} \).
Key Concepts
Position FunctionCalculus DifferentiationSpeed Calculation
Position Function
In the realm of calculus, the position function is crucial for describing the movement of a particle in space. It defines the location of a particle at any given time. The position function is typically described as \( \mathbf{r}(t) \), where each component represents a spatial dimension.
In our example, the position function is given as \( \mathbf{r}(t) = \langle \sin t, t, \cos t \rangle \). This function implies that:
Understanding this function is foundational for determining other aspects like velocity and acceleration, which are derived from the position function.
In our example, the position function is given as \( \mathbf{r}(t) = \langle \sin t, t, \cos t \rangle \). This function implies that:
- \( \sin t \) dictates motion along the x-axis.
- \( t \) represents motion along the y-axis.
- \( \cos t \) controls movement along the z-axis.
Understanding this function is foundational for determining other aspects like velocity and acceleration, which are derived from the position function.
Calculus Differentiation
Differentiation is a core principle in calculus used to analyze how functions change. When you differentiate the position function \( \mathbf{r}(t) \), you obtain the velocity function \( \mathbf{v}(t) \), which informs us about how fast and in what direction the particle is moving at any given point in time.
For our problem, differentiating the position function \( \mathbf{r}(t) = \langle \sin t, t, \cos t \rangle \) involves taking the derivative of each component:
This component-wise differentiation process is equally crucial when moving from velocity to acceleration, illustrating how the particle's speed evolves.
By deriving these functions, you gain deeper insights into the particle's motion, making differentiation an indispensable tool.
For our problem, differentiating the position function \( \mathbf{r}(t) = \langle \sin t, t, \cos t \rangle \) involves taking the derivative of each component:
- The derivative of \( \sin t \) is \( \cos t \).
- The derivative of \( t \) is \( 1 \).
- The derivative of \( \cos t \) is \( -\sin t \).
This component-wise differentiation process is equally crucial when moving from velocity to acceleration, illustrating how the particle's speed evolves.
By deriving these functions, you gain deeper insights into the particle's motion, making differentiation an indispensable tool.
Speed Calculation
Speed is a scalar quantity that represents how fast an object is moving, regardless of direction. To find the speed of a particle from its velocity vector, one computes the magnitude of that vector.
In our example, the velocity vector is \( \mathbf{v}(t) = \langle \cos t, 1, -\sin t \rangle \). Its magnitude is calculated by finding the square root of the sum of the squares of its components:\[ \text{Speed} = \sqrt{(\cos t)^2 + 1^2 + (-\sin t)^2} = \sqrt{\cos^2 t + 1 + \sin^2 t} \]
Using the trigonometric identity \( \cos^2 t + \sin^2 t = 1 \), the speed simplifies to \( \sqrt{2} \).
This simplified calculation is very common in physics problems, allowing fast assessment of motion intensity without full vectorial analysis.
Understanding speed calculation empowers you to quantify motion effectively, grounding your understanding of the underlying kinetics.
In our example, the velocity vector is \( \mathbf{v}(t) = \langle \cos t, 1, -\sin t \rangle \). Its magnitude is calculated by finding the square root of the sum of the squares of its components:\[ \text{Speed} = \sqrt{(\cos t)^2 + 1^2 + (-\sin t)^2} = \sqrt{\cos^2 t + 1 + \sin^2 t} \]
Using the trigonometric identity \( \cos^2 t + \sin^2 t = 1 \), the speed simplifies to \( \sqrt{2} \).
This simplified calculation is very common in physics problems, allowing fast assessment of motion intensity without full vectorial analysis.
Understanding speed calculation empowers you to quantify motion effectively, grounding your understanding of the underlying kinetics.
Other exercises in this chapter
Problem 7
Find the velocity, acceleration, and speed of a particle with the given position function. $$ \mathbf{r}(t)=\left\langle e^{t}, e^{-t}\right\rangle $$
View solution Problem 7
Finding the Principal Unit Normal Vector and Binormal Vector For each of the following vector-valued functions, find the principal unit normal vector. Then, if
View solution Problem 8
Compute the derivatives of the vector-valued functions.\(\mathbf{r}(t)=\tan (2 t) \mathbf{i}+\sec (2 t) \mathbf{j}+\sin ^{2}(t) \mathbf{k}\)
View solution Problem 8
Find the unit normal vector for the vector-valued function \(\mathbf{r}(t)=\left(t^{2}-3 t\right) \mathbf{i}+(4 t+1) \mathbf{j}\) and evaluate it at \(t=2\).
View solution