Problem 8 - Calculus Early Transcendentals

Question

\(3-8\) Find the velocity, acceleration, and speed of a particle with the given position function. Sketch the path of the particle and draw the velocity and acceleration vectors for the specified value of \(t .\) $$\mathbf{r}(t)=t \mathbf{i}+2 \cos t \mathbf{j}+\sin t \mathbf{k}, \quad t=0$$

Step-by-Step Solution

Answer

Velocity at \( t=0 \) is \( \mathbf{i} + \mathbf{k} \), acceleration is \( -2\mathbf{j} \), speed is \( \sqrt{2} \).

Step 1: Find Velocity Function

Velocity is the first derivative of the position function \( \mathbf{r}(t) \) with respect to time \( t \). Given \( \mathbf{r}(t) = t \mathbf{i} + 2\cos(t) \mathbf{j} + \sin(t) \mathbf{k} \), differentiate each component: \[ \mathbf{v}(t) = \frac{d}{dt}(t) \mathbf{i} + \frac{d}{dt}(2\cos(t)) \mathbf{j} + \frac{d}{dt}(\sin(t)) \mathbf{k} = \mathbf{i} - 2\sin(t) \mathbf{j} + \cos(t) \mathbf{k} \]

Step 2: Calculate Velocity at t=0

Substitute \( t = 0 \) into the velocity function:\[\mathbf{v}(0) = \mathbf{i} - 2\sin(0) \mathbf{j} + \cos(0) \mathbf{k} = \mathbf{i} + \mathbf{k}\]

Step 3: Find Acceleration Function

Acceleration is the derivative of the velocity function \( \mathbf{v}(t) \). Differentiate:\[\mathbf{a}(t) = \frac{d}{dt}(\mathbf{i}) - \frac{d}{dt}(2\sin(t) \mathbf{j}) + \frac{d}{dt}(\cos(t) \mathbf{k}) = -2\cos(t) \mathbf{j} - \sin(t) \mathbf{k}\]

Step 4: Calculate Acceleration at t=0

Substitute \( t = 0 \) into the acceleration function:\[\mathbf{a}(0) = -2\cos(0) \mathbf{j} - \sin(0) \mathbf{k} = -2\mathbf{j}\]

Step 5: Determine Speed at t=0

Speed is the magnitude of the velocity vector \( \mathbf{v}(t) \). Evaluate at \( t = 0 \):\[\text{speed} = \|\mathbf{v}(0)\| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}\]

Step 6: Sketch Path and Vectors

The path is a 3D helix determined by the parametric equations \( x = t \), \( y = 2\cos(t) \), \( z = \sin(t) \). At \( t=0 \), plot position as (0,2,0), velocity vector \( \mathbf{v}(0) = \mathbf{i} + \mathbf{k} \), acceleration vector \( \mathbf{a}(0) = -2\mathbf{j} \). The velocity vector points diagonally in the xz-plane, and the acceleration vector points downwards along y-axis.

Key Concepts

Velocity

Velocity is an important concept in calculus and physical sciences alike. It describes how fast an object is moving and in what direction. In more technical terms, velocity is the rate of change of the position function with respect to time.
Given the position function \( \mathbf{r}(t) = t \mathbf{i} + 2\cos(t) \mathbf{j} + \sin(t) \mathbf{k} \), we find the velocity by differentiating each component with respect to \( t \). This gives us the function:

• \( \, \frac{d}{dt}(t) \mathbf{i} = \mathbf{i} \) - which represents constant motion in the x-direction.
• \( \, \frac{d}{dt}(2\cos(t)) \mathbf{j} = -2\sin(t) \mathbf{j} \) - which is variable motion dictated by the sine function in the y-direction.
• \( \, \frac{d}{dt}(\sin(t)) \mathbf{k} = \cos(t) \mathbf{k} \) - which shows periodic motion in the z-direction.

Plug these into the velocity vector \( \mathbf{v}(t) = \mathbf{i} - 2\sin(t) \mathbf{j} + \cos(t) \mathbf{k} \).
To find the velocity at a specific time, such as \( t=0 \), substitute in the values and evaluate the expression. So we have:
\( \mathbf{v}(0) = \mathbf{i} + \mathbf{k} \).
This velocity vector tells us the object is moving in a positive direction along both the x- and z-axes.

Acceleration

Acceleration is the fancy term for how much the velocity of an object changes with time. Essentially, it's just the derivative of the velocity function. For our position function problem, the acceleration function is found by taking the derivative of \( \mathbf{v}(t) = \mathbf{i} - 2\sin(t) \mathbf{j} + \cos(t) \mathbf{k} \).
Here's how the differentiation works for each component:

• \( \frac{d}{dt}(\mathbf{i}) = 0 \) - because any constant term disappears in differentiation.
• \( \frac{d}{dt}(-2\sin(t) \mathbf{j}) = -2\cos(t) \mathbf{j} \) - cosine creeps in, changing the scale of motion in the y-direction.
• \( \frac{d}{dt}(\cos(t) \mathbf{k}) = -\sin(t) \mathbf{k} \) - presents a varying negative direction for z.

This simplifies to the acceleration vector:\[ \mathbf{a}(t) = -2\cos(t) \mathbf{j} - \sin(t) \mathbf{k} \].
At \( t=0 \), evaluate \( \mathbf{a}(0) = -2\mathbf{j} \).
Here, the acceleration vector points only in the negative y-direction. This means that, at this moment, the object experiences a downward pull along the y-axis.

Position Function

The position function is the starting point in our journey through calculus for motion analysis. It provides the coordinates of a particle as it moves through space over time. In our exercise, this function is represented as a vector:
\( \mathbf{r}(t) = t \mathbf{i} + 2\cos(t) \mathbf{j} + \sin(t) \mathbf{k} \).
Here, \( t \) is time, and this vector breaks down motion into components along the x, y, and z directions, using the standard unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) respectively.

• \( t \mathbf{i} \) - linear progress along the x-axis at an increasing rate.
• \( 2\cos(t) \mathbf{j} \) - signifies oscillation back and forth in the y-axis - think of it as swinging.
• \( \sin(t) \mathbf{k} \) - denotes up-and-down periodic motion in the z-axis.

This function essentially charts out a 3D helical path over time.
At any specific moment, like \( t=0 \), it determines the particle's location with spelling out the exact 3D coordinates.
Understanding this foundational concept is key to mastering how velocity and acceleration work down the line!