Problem 9 - Calculus Early Transcendentals

Question

\(9-14\) Find the velocity, acceleration, and speed of a particle with the given position function. $$\mathbf{r}(t)=\left\langle t^{2}+1, t^{3}, t^{2}-1\right\rangle$$

Step-by-Step Solution

Answer

Velocity: \( \left\langle 2t, 3t^2, 2t \right\rangle \); Acceleration: \( \left\langle 2, 6t, 2 \right\rangle \); Speed: \( \sqrt{9t^4 + 8t^2} \)."

Step 1: Find the Velocity

The velocity of a particle is given by the derivative of the position vector with respect to time. So, we take the derivative of \(\mathbf{r}(t)=\left\langle t^{2}+1, t^{3}, t^{2}-1\right\rangle\):\[\mathbf{v}(t) = \frac{d}{dt}\left\langle t^{2}+1, t^{3}, t^{2}-1\right\rangle = \left\langle \frac{d}{dt}(t^2+1), \frac{d}{dt}(t^3), \frac{d}{dt}(t^2-1) \right\rangle = \left\langle 2t, 3t^2, 2t \right\rangle\]Thus, the velocity vector is \( \mathbf{v}(t) = \left\langle 2t, 3t^2, 2t \right\rangle \).

Step 2: Find the Acceleration

The acceleration of a particle is the derivative of the velocity vector with respect to time. Therefore, we take the derivative of \(\mathbf{v}(t) = \left\langle 2t, 3t^2, 2t \right\rangle\):\[\mathbf{a}(t) = \frac{d}{dt}\left\langle 2t, 3t^2, 2t \right\rangle = \left\langle \frac{d}{dt}(2t), \frac{d}{dt}(3t^2), \frac{d}{dt}(2t) \right\rangle = \left\langle 2, 6t, 2 \right\rangle\]So, the acceleration vector is \( \mathbf{a}(t) = \left\langle 2, 6t, 2 \right\rangle \).

Step 3: Find the Speed

Speed is the magnitude of the velocity vector. We calculate the magnitude of \( \mathbf{v}(t) = \left\langle 2t, 3t^2, 2t \right\rangle \) as follows:\[||\mathbf{v}(t)|| = \sqrt{(2t)^2 + (3t^2)^2 + (2t)^2} = \sqrt{4t^2 + 9t^4 + 4t^2} = \sqrt{9t^4 + 8t^2}\]Thus, the speed is \( \sqrt{9t^4 + 8t^2} \).

Key Concepts

Velocity

In calculus, velocity is a fundamental concept that relates to how fast the position of an object is changing with time. When you have a position function, such as a vector \(\mathbf{r}(t)\), the velocity is the derivative of this position function with respect to time. This means you are essentially measuring the rate of change of position, or how fast and in which direction the particle is moving at any given moment.

### Calculating Velocity
To calculate velocity, take the derivative of each component of the position vector. The given position function is \(\mathbf{r}(t) = \langle t^2+1, t^3, t^2-1 \rangle\). By differentiating each part with respect to time, you obtain:

• Derivative of \(t^2 + 1\) is \(2t\)
• Derivative of \(t^3\) is \(3t^2\)
• Derivative of \(t^2 - 1\) is \(2t\)

Thus, the velocity vector is given by \(\mathbf{v}(t) = \langle 2t, 3t^2, 2t \rangle\). This tells us not only the speed but also the direction of the particle's motion.

Acceleration

Acceleration in calculus describes how fast the velocity of an object is changing with respect to time. Just like velocity is the derivative of position, acceleration is the derivative of velocity. It provides insights into whether a particle is speeding up, slowing down, or changing direction.

### Calculating Acceleration
To determine the acceleration, differentiate the velocity vector \(\mathbf{v}(t) = \langle 2t, 3t^2, 2t \rangle\) with respect to time. By doing so, you compute:

• Derivative of \(2t\) is \(2\)
• Derivative of \(3t^2\) is \(6t\)
• Derivative of \(2t\) is \(2\)

The acceleration vector comes out to be \(\mathbf{a}(t) = \langle 2, 6t, 2 \rangle\). This vector tells us how the speed of the particle increases over time and the nature of its path.

Position Function

The position function in calculus describes where an object is located at any point over time. It's often given in vector form in physics and engineering, making it very versatile for describing motion in 2D, 3D, or more dimensions. For this problem, the position of a particle is described by the vector function \(\mathbf{r}(t) = \langle t^2+1, t^3, t^2-1 \rangle\).

### Understanding the Position Function
Each component of the position vector corresponds to a coordinate axis, defining the particle's place in space at any time \(t\):

• The first component \(t^2 + 1\) determines the position along the x-axis.
• The second component \(t^3\) is for the y-axis.
• The third component \(t^2 - 1\) indicates the z-axis.

By analyzing the position function, you can understand the initial position, trajectory, and dynamics of the particle’s movement in space. Calculating the derivative of this function gives you advanced insights, allowing transitions to velocity and acceleration calculations, which describe the state of motion at each time point.