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Chapter 13: Problem 8

Exercises \(5-8\) give the position vectors of particles moving along various curves in the \(x y-\) plane. In each case, find the particle's velocity and acceleration vectors at the stated times and sketch them as vectors on the curve. Motion on the parabola \(y=x^{2}+1\) $$ \mathbf{r}(t)=t \mathbf{i}+\left(t^{2}+1\right) \mathbf{j} ; \quad t=-1,0, \text { and } 1 $$

Short Answer

Expert verified
The velocity vectors at \( t = -1, 0, \) and \( 1 \) are \( \mathbf{i} - 2\mathbf{j} \), \( \mathbf{i} \), and \( \mathbf{i} + 2\mathbf{j} \), respectively. The acceleration is consistently \( 2\mathbf{j} \).

Step by step solution

01

Understand the Problem

We need to find the velocity and acceleration vectors for a particle moving along the parabola given by the position vector \( \mathbf{r}(t) = t\mathbf{i} + (t^2 + 1)\mathbf{j} \) for different values of \( t \), specifically \( t = -1, 0, \) and \( 1 \). The goal is to calculate the velocity \( \mathbf{v}(t) \) and acceleration \( \mathbf{a}(t) \) vectors for these times.
02

Differentiate to Find Velocity

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Calculate:\[\mathbf{v}(t) = \frac{d}{dt}[t\mathbf{i} + (t^2 + 1)\mathbf{j}] = \mathbf{i} + 2t\mathbf{j}\]
03

Differentiate Velocity to Find Acceleration

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). Calculate:\[\mathbf{a}(t) = \frac{d}{dt}[\mathbf{i} + 2t\mathbf{j}] = 2\mathbf{j}\]
04

Evaluate Vectors at Specific Times

Substitute \( t = -1, 0, \) and \( 1 \) into the expressions for \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \):- At \( t = -1 \): \( \mathbf{v}(-1) = \mathbf{i} - 2\mathbf{j} \) and \( \mathbf{a}(-1) = 2\mathbf{j} \).- At \( t = 0 \): \( \mathbf{v}(0) = \mathbf{i} \) and \( \mathbf{a}(0) = 2\mathbf{j} \).- At \( t = 1 \): \( \mathbf{v}(1) = \mathbf{i} + 2\mathbf{j} \) and \( \mathbf{a}(1) = 2\mathbf{j} \).
05

Sketch Vectors on the Curve

Draw the parabola \( y = x^2 + 1 \), and plot the position vectors for \( t = -1, 0, \) and \( 1 \). At each point, sketch the velocity and acceleration vectors:- At \( t = -1 \), plot vector \( \mathbf{v}(-1) \) starting at \( (-1, 2) \) with components \( \mathbf{i} - 2\mathbf{j} \), and \( \mathbf{a}(-1) \) at the same point.- At \( t = 0 \), plot vector \( \mathbf{v}(0) \) from \( (0, 1) \) heading right, and \( \mathbf{a}(0) \) pointing upwards.- At \( t = 1 \), plot vector \( \mathbf{v}(1) \) from \( (1, 2) \), at an angle, and acceleration upwards.

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Key Concepts

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

Position Vector
The position vector is a fundamental element in calculus and physics, representing the position of a point in space relative to a reference point. In our exercise, the position vector describes the path of a particle on a curve in the xy-plane, specifically on a parabola. Given by \( \mathbf{r}(t) = t \mathbf{i} + (t^2 + 1) \mathbf{j} \), each component corresponds to the coordinates in the Cartesian plane. Here:
  • \( t \mathbf{i} \) represents the horizontal component (x-axis).
  • \( (t^2 + 1) \mathbf{j} \) represents the vertical component (y-axis).
This equation describes how the particle moves along the parabola \( y = x^2 + 1 \), illustrating its path over different times \( t \). As \( t \) changes, so do the values of the components, indicating the position at any given time.
Velocity Vector
In calculus, the velocity vector is derived by differentiating the position vector with respect to time. It gives us the instantaneous direction and speed at which a particle is moving. In our example, we find the velocity vector by taking the derivative of the given position vector:
  • \( \mathbf{v}(t) = \frac{d}{dt}[t\mathbf{i} + (t^2 + 1)\mathbf{j}] \)
  • Resulting in \( \mathbf{v}(t) = \mathbf{i} + 2t\mathbf{j} \)
This vector shows a constant horizontal component (\( \mathbf{i} \)) and a variable vertical component (\( 2t\mathbf{j} \)), changing with time. The velocity vector's role is crucial as it reflects how swiftly and in which direction the particle shifts its position along the parabola. Evaluating at different times, like \( t = -1, 0, 1 \), gives specific velocities for those instances.
Acceleration Vector
Acceleration is a vector indicating the rate of change of velocity over time. By differentiating the velocity vector of the particle, we obtain the acceleration vector. This shows how the velocity itself is changing:
  • \( \mathbf{a}(t) = \frac{d}{dt}[\mathbf{i} + 2t\mathbf{j}] \)
  • Becoming \( \mathbf{a}(t) = 2\mathbf{j} \)
Here, we observe that the acceleration is constant over time, meaning that the only change happening to the velocity is in its vertical component. This constant acceleration is typical along parabolic paths and showcases how forces, such as gravity in a more physics-focused problem, can act uniformly.
Differentiation
Differentiation is a core concept in calculus used to find rates of change or slopes of curves. In this exercise, it allows us to transition from a position vector, which describes a point, to a velocity vector, which describes a path's movement, and then to an acceleration vector, which explains changes in that movement.
  • To find the velocity from the position vector, differentiate the position function \( \mathbf{r}(t) \).
  • To find acceleration, differentiate the resultant velocity function \( \mathbf{v}(t) \).
By understanding differentiation, students can parse how key vector quantities reflect dynamics in motion and anticipate the behavior of particles along structures like parabolas. It forms a bridge from static positions to dynamic analysis.
Parabola
A parabola is a symmetric, curved path typically described by a quadratic equation. Its appearance often resembles a U-shape. In our example, the parabola is defined by the equation \( y = x^2 + 1 \). Components of the equation include:
  • \( x^2 \) indicates the quadratic nature, showing that the curve has identical limbs extending upward as \( x \) approaches infinity.
  • The \( +1 \) signifies a vertical shift of the entire parabola, moving it one unit above the x-axis.
Particles traveling along a parabola experience varying velocity and acceleration due to the curve's gradient changes. This exercise encapsulates these principles by providing a practical illustration of how a particle dynamically interacts with a parabolic path during its movement.

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Most popular questions from this chapter

Motion along an ellipse A particle moves around the ellipse \((y / 3)^{2}+(z / 2)^{2}=1\) in the \(y z\) -plane in such a way that its position at time \(t\) is $$ \mathbf{r}(t)=(3 \cos t) \mathbf{j}+(2 \sin t) \mathbf{k} $$ Find the maximum and minimum values of \(|\mathbf{v}|\) and \(|\mathbf{a}| .\) (Hint: Find the extreme values of \(|\mathbf{v}|^{2}\) and \(|\mathbf{a}|^{2}\) first and take square roots later.)

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the space curves. \(\mathbf{r}(t)=(\cosh t) \mathbf{i}-(\sinh t) \mathbf{j}+t \mathbf{k}\)

In Exercises 19 and \(20, \mathbf{r}(t)\) is the position vector of a particle in space at time \(t .\) Find the time or times in the given time interval when the velocity and acceleration vectors are orthogonal. $$ \mathbf{r}(t)=(t-\sin t) \mathbf{i}+(1-\cos t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi $$

Firing from \(\left(x_{0}, y_{0}\right)\) Derive the equations $$ \begin{aligned} x &=x_{0}+\left(v_{0} \cos \alpha\right) t \\ y &=y_{0}+\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2} \end{aligned} $$ (see Equation ( 5\()\) in the text) by solving the following initial value problem for a vector \(\mathbf{r}\) in the plane. $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}} \\ {\text { Initial conditions: }} & {\mathbf{r}(0)=x_{0} \mathbf{i}+y_{0} \mathbf{j}} \\ {} & {\frac{d \mathbf{r}}{d t}(0)=\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(v_{0} \sin \alpha\right) \mathbf{j}}\end{array} $$

Are length Find the length of the curve $$ \mathbf{r}(t)=(\sqrt{2} t) \mathbf{i}+(\sqrt{2} t) \mathbf{j}+\left(1-t^{2}\right) \mathbf{k} $$ from \((0,0,1)\) to \((\sqrt{2}, \sqrt{2}, 0)\)
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