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Chapter 15: Problem 1

Find general formulas for the tangential and normal components of acceleration and for the curvature of the curve \(C\) determined by \(\mathbf{r}(t)\). $$ \mathbf{r}(t)=t^{2} \mathbf{i}+(3 t+2) \mathbf{j} $$

Short Answer

Expert verified
Tangential: \( a_T = \frac{4t}{\sqrt{4t^2 + 9}} \), Normal: \( a_N = \frac{6}{\sqrt{4t^2 + 9}} \), Curvature: \( \kappa(t) = \frac{6}{(\sqrt{4t^2 + 9})^3} \).

Step by step solution

01

Compute the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Differentiate each component of \( \mathbf{r}(t) \): \( \mathbf{v}(t) = \frac{d}{dt}[t^2] \mathbf{i} + \frac{d}{dt}[3t+2] \mathbf{j} = 2t \mathbf{i} + 3 \mathbf{j} \).
02

Compute the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiate each component of \( \mathbf{v}(t) \): \( \mathbf{a}(t) = \frac{d}{dt}[2t] \mathbf{i} + \frac{d}{dt}[3] \mathbf{j} = 2 \mathbf{i} \).
03

Compute the Speed

The speed \( v(t) \) is the magnitude of the velocity vector \( \mathbf{v}(t) \). Compute the magnitude: \( v(t) = || \mathbf{v}(t) || = \sqrt{(2t)^2 + 3^2} = \sqrt{4t^2 + 9} \).
04

Tangential Component of Acceleration

The tangential component of acceleration is given by \( a_T = \frac{d}{dt}v(t) \). Differentiate \( v(t) \) with respect to \( t \): \( a_T = \frac{d}{dt}\sqrt{4t^2 + 9} = \frac{4t}{\sqrt{4t^2 + 9}} \).
05

Normal Component of Acceleration

The normal component of acceleration is calculated using the formula \( a_N = \sqrt{||\mathbf{a}||^2 - a_T^2} \). Compute \( ||\mathbf{a}|| = 2 \) and \( a_T = \frac{4t}{\sqrt{4t^2 + 9}} \). Thus \( a_N = \sqrt{2^2 - \left(\frac{4t}{\sqrt{4t^2 + 9}}\right)^2} \). Simplify to get \( a_N = \frac{6}{\sqrt{4t^2 + 9}} \).
06

Compute the Curvature

The curvature \( \kappa(t) \) for a plane curve is \( \kappa(t) = \frac{|\mathbf{v}(t) \times \mathbf{a}(t)|}{||\mathbf{v}(t)||^3} \). Compute the cross product \( \mathbf{v}(t) \times \mathbf{a}(t) \), which reduces to \( |2t \cdot 0 - 3 \cdot 2| \)=\( |6| \). Then compute \( \kappa(t) = \frac{6}{(\sqrt{4t^2 + 9})^3} \).

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Key Concepts

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

Curvature of a Curve
Understanding the curvature of a curve is fundamental when exploring the trajectory described by a position vector like \( \mathbf{r}(t) \). Curvature, denoted as \( \kappa(t) \), measures the rate at which a curve deviates from being a straight line. In simpler terms, it tells us how tightly a curve bends at a particular point. The formula to determine the curvature involves both the velocity and acceleration vectors.
To calculate curvature:
  • First compute the cross product of the velocity vector \( \mathbf{v}(t) \) and the acceleration vector \( \mathbf{a}(t) \).
  • The absolute value of this cross product gives us one part of the curvature formula.
  • The magnitude of the velocity vector raised to the third power \( ||\mathbf{v}(t)||^3 \) makes up the denominator.
This is expressed as: \[ \kappa(t) = \frac{|\mathbf{v}(t) \times \mathbf{a}(t)|}{||\mathbf{v}(t)||^3} \] By calculating these components, you gain insight into how sharply a curve turns at each point as a function of time \( t \).
Velocity Vector
The velocity vector \( \mathbf{v}(t) \) is a crucial element in understanding the motion of a particle along its path. It represents both the speed and direction of the particle's movement at a given moment. The velocity vector is derived by differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \).
For example, if \( \mathbf{r}(t) = t^2 \mathbf{i} + (3t + 2) \mathbf{j} \), then finding \( \mathbf{v}(t) \) involves:
  • Computing the derivative of each component of the position vector.
  • This results in \( \mathbf{v}(t) = 2t \mathbf{i} + 3 \mathbf{j} \).
The velocity vector provides a snapshot of how the position is changing over time at any specific instant.
Acceleration Vector
Acceleration vector \( \mathbf{a}(t) \) provides valuable information about how the velocity of a particle changes with time. It indicates whether the particle is speeding up, slowing down, or changing direction. To derive it, simply differentiate the velocity vector \( \mathbf{v}(t) \).
For instance, given \( \mathbf{v}(t) = 2t \mathbf{i} + 3 \mathbf{j} \):
  • Differentiate each component of the velocity vector to get \( \mathbf{a}(t) \).
  • This results in \( \mathbf{a}(t) = 2 \mathbf{i} \).
Understanding the acceleration allows you to predict changes in the velocity along the path, which is essential for determining both the tangential and normal components of the motion.
Differentiation
Differentiation is a mathematical process used to determine the rate of change of a function with respect to a variable. In the context of motion, we use differentiation to derive velocity and acceleration from the position function \( \mathbf{r}(t) \).
Key steps in differentiation include:
  • Identifying the function that needs to be differentiated (e.g., position, velocity).
  • Applying rules of differentiation, such as the power rule, constant rule, and sum rule, to find derivatives.
For position \( \mathbf{r}(t) = t^2 \mathbf{i} + (3t + 2) \mathbf{j} \), differentiation of each term gives:
  • Velocity: \( \mathbf{v}(t) = 2t \mathbf{i} + 3 \mathbf{j} \)
  • Acceleration: \( \mathbf{a}(t) = 2 \mathbf{i} \)
By applying differentiation correctly, you can obtain a deeper comprehension of how motion evolves over time.
Magnitude of a Vector
The magnitude of a vector captures its size or length, without considering its direction. It's a fundamental concept applied when analyzing velocity and acceleration vectors to understand speed and other motion characteristics. The magnitude of a vector \( \mathbf{v}(t) \) expressed as \( x \mathbf{i} + y \mathbf{j} \) is computed using the formula: \[ ||\mathbf{v}(t)|| = \sqrt{x^2 + y^2} \] For example, consider the velocity vector \( \mathbf{v}(t) = 2t \mathbf{i} + 3 \mathbf{j} \):
  • Its magnitude is \( ||\mathbf{v}(t)|| = \sqrt{(2t)^2 + 3^2} = \sqrt{4t^2 + 9} \).
Understanding the magnitude allows you to separate speed (a scalar quantity) from velocity (a vector quantity), making it easier to assess how fast the particle is moving at any point.

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Most popular questions from this chapter

Exer. 1-6: |al Find the unit tangent and normal vectors \(\mathrm{T}(t)\) and \(\mathrm{N}(t)\) for the curve \(C\) determined by \(\mathrm{r}(t)\) (b) Sketch the graph of \(C,\) and show \(\mathrm{T}(t)\) and \(\mathrm{N}(t)\) for the given value of \(t\). $$ \mathbf{r}(t)=2 \sin t \mathbf{i}+3 \mathbf{j}+2 \cos t \mathbf{k} ; \quad t=\pi / 4 $$

Prove that if a point moves along a curve \(C\) with a constant speed, then the acceleration is always normal to \(C\).

If a curve \(C\) has a tangent vector a at a point \(P\), then the normal plane to \(C\) at \(P\) is the plane through \(P\) with normal vector a. Find an equation of the normal plane to the given curve at \(P\). $$ x=t \sin t, \quad y=t \cos t, \quad z=t ; \quad P(\pi / 2,0, \pi / 2) $$

Find \(\mathbf{u}(t)\) if \(\mathbf{u}^{\prime}(t)=e^{-t} \mathbf{i}-4 \sin 2 t \mathbf{j}+3 \sqrt{t} \mathbf{k}\) and \(\mathbf{u}(0)=-\mathbf{i}+2 \mathbf{j}\)

Find general formulas for the tangential and normal components of acceleration and for the curvature of the curve \(C\) determined by \(\mathbf{r}(t)\). $$ \mathbf{r}(t)=t(\cos t \mathbf{i}+\sin t \mathbf{j}) $$
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