Problem 12 - Calculus Early Transcendentals

Question

\(9-14\) Find the velocity, acceleration, and speed of a particle with the given position function. $$\mathbf{r}(t)=t^{2} \mathbf{i}+\ln t \mathbf{j}+t \mathbf{k}$$

Step-by-Step Solution

Answer

Velocity: \(2t \mathbf{i} + \frac{1}{t} \mathbf{j} + 1 \mathbf{k}\). Acceleration: \(2 \mathbf{i} - \frac{1}{t^2} \mathbf{j}\). Speed: \(\sqrt{4t^2 + \frac{1}{t^2} + 1}\).

Step 1: Differentiate for Velocity

The position function is given as \( \mathbf{r}(t)=t^{2} \mathbf{i} + \ln t \mathbf{j} + t \mathbf{k} \). To find the velocity \( \mathbf{v}(t) \), we need to differentiate each component of \( \mathbf{r}(t) \) with respect to \( t \). The derivative of \( t^2 \) is \( 2t \), the derivative of \( \ln t \) is \( \frac{1}{t} \), and the derivative of \( t \) is \( 1 \). Thus, the velocity function is \( \mathbf{v}(t) = 2t \mathbf{i} + \frac{1}{t} \mathbf{j} + 1 \mathbf{k} \).

Step 2: Differentiate for Acceleration

To find the acceleration \( \mathbf{a}(t) \), we differentiate the velocity function \( \mathbf{v}(t) = 2t \mathbf{i} + \frac{1}{t} \mathbf{j} + 1 \mathbf{k} \). The derivative of \( 2t \) is \( 2 \), the derivative of \( \frac{1}{t} \) is \( -\frac{1}{t^2} \), and the derivative of \( 1 \) is \( 0 \). Therefore, the acceleration function is \( \mathbf{a}(t) = 2 \mathbf{i} - \frac{1}{t^2} \mathbf{j} + 0 \mathbf{k} \).

Step 3: Calculate the Speed

The speed of the particle is the magnitude of the velocity vector \( \mathbf{v}(t) = 2t \mathbf{i} + \frac{1}{t} \mathbf{j} + 1 \mathbf{k} \). The magnitude \(|\mathbf{v}(t)|\) is given by \( \sqrt{(2t)^2 + \left(\frac{1}{t}\right)^2 + 1^2} \). Simplifying, we get \( \sqrt{4t^2 + \frac{1}{t^2} + 1} \).

Key Concepts

velocity

In vector calculus, the concept of velocity is fundamental for understanding the motion of particles. Velocity is a vector quantity, which means it has both a magnitude and a direction. It represents the rate at which a particle's position changes with respect to time. To find the velocity of a particle given its position function, we differentiate each component of the position vector.

For example, consider the position function \( \mathbf{r}(t) = t^2 \mathbf{i} + \ln t \mathbf{j} + t \mathbf{k} \). We determine the velocity vector \( \mathbf{v}(t) \) by taking the derivative of each component:

• The derivative of \( t^2 \) is \( 2t \), identified as the \( \mathbf{i} \) component.
• The derivative of \( \ln t \) is \( \frac{1}{t} \), designated as the \( \mathbf{j} \) component.
• The derivative of \( t \) is \( 1 \), forming the \( \mathbf{k} \) component.

Therefore, the velocity function is expressed as:\[ \mathbf{v}(t) = 2t \mathbf{i} + \frac{1}{t} \mathbf{j} + 1 \mathbf{k} \]This equation indicates how fast and in what direction the particle is moving at any given time, \( t \). Understanding velocity is crucial as it forms the basis for deriving other motion aspects, like acceleration.

acceleration

Acceleration is another vector quantity closely related to velocity. It represents the rate at which the velocity of a particle changes over time. In simpler terms, it shows how quickly a particle is speeding up or slowing down and in which direction.

To find the acceleration of a particle, we differentiate its velocity vector with respect to time. Considering our velocity function\( \mathbf{v}(t) = 2t \mathbf{i} + \frac{1}{t} \mathbf{j} + 1 \mathbf{k} \), here's how we identify the acceleration:

• The derivative of \( 2t \) is \( 2 \), which is the constant \( \mathbf{i} \) component of the acceleration.
• The derivative of \( \frac{1}{t} \) is \( -\frac{1}{t^2} \), which forms the \( \mathbf{j} \) component.
• The \( 1 \) for the \( \mathbf{k} \) component results in the derivative \( 0 \), indicating no change in that direction.

So, the acceleration function is expressed as:\[ \mathbf{a}(t) = 2 \mathbf{i} - \frac{1}{t^2} \mathbf{j} + 0 \mathbf{k} \]This function helps us understand how the motion of the particle evolves by displaying the changes in velocity over time. By examining the acceleration vector, we can determine if a particle is gaining or losing speed in certain directions.

speed

Speed is a scalar quantity that only represents the magnitude of velocity. Unlike velocity, it does not include directional information. Speed tells us how fast a particle is moving regardless of the direction. To calculate speed, we take the magnitude of the velocity vector.

Given the velocity function\( \mathbf{v}(t) = 2t \mathbf{i} + \frac{1}{t} \mathbf{j} + 1 \mathbf{k} \), the speed \( |\mathbf{v}(t)| \) is derived by finding the square root of the sum of the squares of the components:

• \( (2t)^2 = 4t^2 \)
• \( \left(\frac{1}{t}\right)^2 = \frac{1}{t^2} \)
• \( 1^2 = 1 \)

Thus, the speed is given by:\[ |\mathbf{v}(t)| = \sqrt{4t^2 + \frac{1}{t^2} + 1} \]This scalar value quantifies how fast the particle is moving at any time "t". It's particularly important when the direction of motion is not a concern but we still need to understand the intensity of motion in a physical sense.