Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the velocity \(\mathbf{v}\), acceleration \(\mathbf{a}\), and speed \(s\) at the indicated time \(t=t_{1}\). \mathbf{r}(t)=t \mathbf{i}+(t-1)^{2} \mathbf{j}+(t-3)^{3} \mathbf{k} ; t_{1}=0

Step-by-Step Solution

Answer

Velocity: \( \mathbf{i} - 2 \mathbf{j} + 27 \mathbf{k} \); Acceleration: \( 2 \mathbf{j} - 18 \mathbf{k} \); Speed: \( \sqrt{734} \).

Step 1: Differentiate the Position Vector

To find the velocity vector \( \mathbf{v}(t) \), we need to differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \). The position vector is given by \( \mathbf{r}(t) = t \mathbf{i} + (t-1)^2 \mathbf{j} + (t-3)^3 \mathbf{k} \). Differentiating each component separately, we get:\[\mathbf{v}(t) = \frac{d}{dt}(t \mathbf{i}) + \frac{d}{dt}((t-1)^2 \mathbf{j}) + \frac{d}{dt}((t-3)^3 \mathbf{k})\]This results in:\[\mathbf{v}(t) = \mathbf{i} + 2(t-1) \mathbf{j} + 3(t-3)^2 \mathbf{k}\]

Step 2: Calculate Velocity at Indicated Time

Substitute \( t_1 = 0 \) into the velocity expression obtained in the previous step:\[\mathbf{v}(0) = \mathbf{i} + 2(0-1) \mathbf{j} + 3(0-3)^2 \mathbf{k}\]Calculating each term, we get:\[\mathbf{v}(0) = \mathbf{i} - 2 \mathbf{j} + 27 \mathbf{k}\]

Step 3: Differentiate the Velocity Vector

To find the acceleration vector \( \mathbf{a}(t) \), differentiate the velocity vector \( \mathbf{v}(t) = \mathbf{i} + 2(t-1) \mathbf{j} + 3(t-3)^2 \mathbf{k} \) with respect to time \( t \). Differentiate each component separately:\[\mathbf{a}(t) = \frac{d}{dt}(\mathbf{i}) + \frac{d}{dt}(2(t-1) \mathbf{j}) + \frac{d}{dt}(3(t-3)^2 \mathbf{k})\]This results in:\[\mathbf{a}(t) = 0 \mathbf{i} + 2 \mathbf{j} + 6(t-3) \mathbf{k}\]

Step 4: Calculate Acceleration at Indicated Time

Substitute \( t_1 = 0 \) into the acceleration expression obtained in the previous step:\[\mathbf{a}(0) = 0 \mathbf{i} + 2 \mathbf{j} + 6(0-3) \mathbf{k}\]This simplifies to:\[\mathbf{a}(0) = 2 \mathbf{j} - 18 \mathbf{k}\]

Step 5: Calculate the Speed at Indicated Time

The speed \( s(t) \) is the magnitude of the velocity vector \( \mathbf{v}(t) \). Calculating the magnitude of \( \mathbf{v}(0) = \mathbf{i} - 2 \mathbf{j} + 27 \mathbf{k} \), we have:\[ s(0) = \sqrt{1^2 + (-2)^2 + 27^2} = \sqrt{1 + 4 + 729} = \sqrt{734}\]

Step 6: Summary of the Solution

The velocity vector at \( t = 0 \) is \( \mathbf{v}(0) = \mathbf{i} - 2 \mathbf{j} + 27 \mathbf{k} \), the acceleration vector is \( \mathbf{a}(0) = 2 \mathbf{j} - 18 \mathbf{k} \), and the speed is \( \sqrt{734} \).

Key Concepts

Differentiation

Differentiation is a powerful tool in calculus that allows us to find the rate at which a quantity changes. When we talk about differentiating a position vector \( \mathbf{r}(t) \), we're determining how the position changes over time to find the velocity vector \( \mathbf{v}(t) \). By taking the derivative of each component of \( \mathbf{r}(t) \), we can individually calculate the rate of change in each direction.
In the given problem, the position vector \( \mathbf{r}(t) = t \mathbf{i} + (t-1)^2 \mathbf{j} + (t-3)^3 \mathbf{k} \) is differentiated to find the velocity vector by calculating \( \frac{d}{dt} \). For example:

• \( \frac{d}{dt}(t \mathbf{i}) = \mathbf{i} \)
• \( \frac{d}{dt}((t-1)^2 \mathbf{j}) = 2(t-1) \mathbf{j} \)
• \( \frac{d}{dt}((t-3)^3 \mathbf{k}) = 3(t-3)^2 \mathbf{k} \)

This gives us the velocity vector \( \mathbf{v}(t) = \mathbf{i} + 2(t-1) \mathbf{j} + 3(t-3)^2 \mathbf{k} \). Differentiation enables us to understand and calculate these changes efficiently.

Position Vector

The position vector \( \mathbf{r}(t) \) describes the position of a point or object in space as a function of time. It is composed of components along the coordinate axes, typically denoted \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) for three-dimensional space. The vector provides a complete description of where the object is at any given \( t \).
In the exercise, the position vector is given in the form \( \mathbf{r}(t) = t \mathbf{i} + (t-1)^2 \mathbf{j} + (t-3)^3 \mathbf{k} \). Each part of this vector corresponds to movement or positioning along one of the axes:

• The \( \mathbf{i} \) component represents linear movement along the x-axis.
• The \( \mathbf{j} \) component, \( (t-1)^2 \), represents a quadratic trajectory along the y-axis.
• The \( \mathbf{k} \) component, \( (t-3)^3 \), indicates a cubic trajectory along the z-axis.

Understanding the position vector is crucial as a starting point for calculating velocity and acceleration.

Calculus

Calculus is the branch of mathematics that studies continuous change, seamlessly linking concepts like differentiation and integration. It's central to understanding motion and dynamics in physics.
In this problem, we employ differential calculus to analyze motion. By differentiating the position vector function with respect to time, we can determine the velocity, which tells us how fast and in what direction the object is moving. Further differentiation of the velocity vector helps us find the acceleration, revealing how the velocity itself changes over time. Calculus enables the step-by-step transformation from position to velocity to acceleration by using derivatives. It allows us to calculate instantaneous rates of change and understand the behavior of moving objects thoroughly.

Magnitude

Magnitude refers to the size or length of a vector. For a velocity vector, the magnitude represents the speed of the object, i.e., how fast it is moving regardless of the direction.
To calculate the magnitude of a vector like \( \mathbf{v}(0) = \mathbf{i} - 2 \mathbf{j} + 27 \mathbf{k} \), we use the formula:\[ s(0) = \sqrt{1^2 + (-2)^2 + 27^2} = \sqrt{1 + 4 + 729} = \sqrt{734}\] The term inside the square root aggregates the squares of each component, reflecting Pythagorean theorem's extension into three dimensions. Magnitude is essential to determine the scalar speed from vector quantities, providing a quick measure of how much ground is being covered over time. Understanding how to compute magnitudes aids in interpreting the complete picture of motion described by velocity vectors.