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Chapter 3: Problem 80

The position vector for a particle is \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k}\). The graph is shown here: Find the velocity vector at any time.

Short Answer

Expert verified
Velocity vector \( \mathbf{v}(t) = 1 \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).

Step by step solution

01

Understanding the Position Vector

The position vector is given as \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \). This vector describes the particle's position in 3D space as a function of time \( t \). The components are \( t \mathbf{i} \), \( t^2 \mathbf{j} \), and \( t^3 \mathbf{k} \).
02

Differentiating the Position Vector

To find the velocity vector, we differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \). This is because velocity is the rate of change of position with respect to time.
03

Derivative of the First Component

The derivative of \( t \) with respect to \( t \) is 1. So, the first component of the velocity vector is \( \frac{d}{dt}(t) = 1 \).
04

Derivative of the Second Component

The derivative of \( t^2 \) with respect to \( t \) is \( 2t \). So, the second component of the velocity vector is \( \frac{d}{dt}(t^2) = 2t \).
05

Derivative of the Third Component

The derivative of \( t^3 \) with respect to \( t \) is \( 3t^2 \). So, the third component of the velocity vector is \( \frac{d}{dt}(t^3) = 3t^2 \).
06

Constructing the Velocity Vector

Combining all the differentiated components, the velocity vector \( \mathbf{v}(t) \) is \( \mathbf{v}(t) = 1 \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Position Vector
In the world of vector calculus, the **position vector** is an essential concept. It provides information about the location of a particle in space over time. Given by the function \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \), this vector has several components:
  • \( t \mathbf{i} \): Represents movement along the x-axis.
  • \( t^2 \mathbf{j} \): Illustrates the movement along the y-axis.
  • \( t^3 \mathbf{k} \): Describes the movement along the z-axis.
As time \( t \) changes, the position vector traces a path in three-dimensional space. Understanding this helps conceptualize how particles move and helps in predicting future positions.
Velocity Vector
The **velocity vector** is crucial as it describes how the position of the particle changes over time. Derived from differentiating the position vector, it gives us a snapshot of the particle's direction and speed at any given moment. The velocity vector is calculated as:
  • First component: Derivative of \( t \) is 1. Resulting in velocity \( \, \mathbf{i} \).
  • Second component: Derivative of \( t^2 \) is \( 2t \). Resulting in velocity \( 2t \mathbf{j} \).
  • Third component: Derivative of \( t^3 \) is \( 3t^2 \). Resulting in velocity \( 3t^2 \mathbf{k} \).
The full velocity vector is \( \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).This means the particle's speed and direction are changing, providing dynamic insights into its trajectory.
Differentiation
**Differentiation** is a fundamental tool in calculus and is essential for finding rates of change. In the context of vector calculus:
  • It helps in determining velocities from position vectors.
  • It allows us to understand how each component of the position vector changes with respect to time.
For example, differentiating \( t \), \( t^2 \), and \( t^3 \), gives \( 1 \), \( 2t \), and \( 3t^2 \) respectively. This process involves applying basic derivative rules:
  • The derivative of \( t^n \) is \( nt^{n-1} \).
This technique is not only powerful for physics-related problems but is also widely applicable across various disciplines.
3D Space
Operating in **3D space** introduces a whole new layer of complexity and excitement. Unlike the 2D plane, every point in 3D is defined by three coordinates: \( x \), \( y \), and \( z \). This is depicted by our position vector in the form \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \).
  • \( \mathbf{i} \) represents the x-axis.
  • \( \mathbf{j} \) denotes the y-axis.
  • \( \mathbf{k} \) indicates the z-axis.
Understanding 3D space is pivotal for visualizing movements and rotations, allowing calculations of real-world trajectories. It aids in comprehending the interaction between different dimensions and seeing how vectors play a part in modeling our universe.

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Most popular questions from this chapter

The surface of a large cup is formed by revolving the graph of the function \(y=0.25 x^{1.6}\) from \(x=0\) to \(x=5\) about the \(y\) -axis (measured in centimeters). Use technology to graph the surface.

Find the domains of the vector-valued functions. \(\quad \mathbf{r}(t)=\left\langle e^{t}, \frac{1}{\sqrt{4-t}}, \sec (t)\right\rangle\)

Find the tangential and normal components of acceleration. \(\mathbf{r}(t)=\left\langle 2 t, t^{2}, \frac{t^{3}}{3}\right\rangle\)

The following problems consider launching a cannonball out of a cannon. The cannonball is shot out of the cannon with an angle \(\theta\) and initial velocity \(\mathbf{v}_{0}\). The only force acting on the cannonball is gravity, so we begin with a constant acceleration \(\mathbf{a}(t)=-g \mathbf{j}\). . Find the velocity vector function \(\mathbf{v}(t)\).

A projectile is shot in the air from ground level with an initial velocity of \(500 \mathrm{~m} / \mathrm{sec}\) at an angle of \(60^{\circ}\) with the horizontal. The graph is shown here: . At what time does the projectile reach maximum height?
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