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Magazine Chapter 3: Problem 150
A particle moves along the plane curve \(\mathrm{C}\) described by \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} .\) Solve the following problems. Find the curvature of the plane curve at \(t=0,1,2\)
Short Answer
Expert verified
Curvatures are 2 at \(t=0\), \(\frac{2}{3\sqrt{3}}\) at \(t=1\), \(\frac{2}{17^{3/2}}\) at \(t=2\).
Step by step solution
01
Find the Derivative of the Position Vector
Given the position vector \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \), we first find \( \mathbf{r}'(t) \), which is the derivative of the position vector. Taking the derivative, we have \( \mathbf{r}'(t) = \frac{d}{dt}(t) \mathbf{i} + \frac{d}{dt}(t^2) \mathbf{j} = \mathbf{i} + 2t \mathbf{j} \).
02
Find the Second Derivative of the Position Vector
Now, we find the second derivative, \( \mathbf{r}''(t) \), by differentiating \( \mathbf{r}'(t) \). We get \( \mathbf{r}''(t) = \frac{d}{dt}(1) \mathbf{i} + \frac{d}{dt}(2t) \mathbf{j} = 0 \mathbf{i} + 2 \mathbf{j} = 2 \mathbf{j} \).
03
Calculate the Magnitudes of Velocity and Acceleration Vectors
To find the curvature, we need the magnitudes of \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \). The magnitude of \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} \) is \( |\mathbf{r}'(t)| = \sqrt{1^2 + (2t)^2} = \sqrt{1 + 4t^2} \). The magnitude of \( \mathbf{r}''(t) = 2 \mathbf{j} \) is \( |\mathbf{r}''(t)| = \sqrt{0^2 + 2^2} = 2 \).
04
Apply the Curvature Formula
The curvature \( \kappa(t) \) is given by \( \kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \). The cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) is a vector perpendicular to both \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \), with magnitude \( |\mathbf{r}'(t)| \times |\mathbf{r}''(t)| \times \sin(90^\circ) \), which is simply \( |\mathbf{r}'(t)| \times |\mathbf{r}''(t)| \). Thus, \( |\mathbf{r}'(t) \times \mathbf{r}''(t)| = 2\sqrt{1 + 4t^2} \). \( \kappa(t) = \frac{2}{(1 + 4t^2)^{\frac{3}{2}}} \).
05
Find Curvature at Specific Points
Now compute \( \kappa(t) \) at \( t = 0, 1, 2 \): - At \( t = 0 \), \( \kappa(0) = \frac{2}{(1+4\times0^2)^{3/2}} = 2 \).- At \( t = 1 \), \( \kappa(1) = \frac{2}{(1+4\times1^2)^{3/2}} = \frac{2}{9^{1/2}} = \frac{2}{3\sqrt{3}} \).- At \( t = 2 \), \( \kappa(2) = \frac{2}{(1+4\times2^2)^{3/2}} = \frac{2}{17^{3/2}} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Position Vector
In mathematics and physics, a position vector describes the position of a point in space relative to an origin. It's a vector that essentially gives us coordinates indicating where something is located on a plane or in a space.
For a curve defined in a plane, the position vector is crucial. In this exercise, the position of a particle is given by the vector function \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \). This vector function describes a time-dependent path on the plane, with components depending on the variable \( t \).
Understanding the position vector is the first step in analyzing the particle's trajectory. It serves as a foundation for further calculation of derivatives and magnitudes, which help determine characteristics like velocity, acceleration, and curvature.
For a curve defined in a plane, the position vector is crucial. In this exercise, the position of a particle is given by the vector function \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \). This vector function describes a time-dependent path on the plane, with components depending on the variable \( t \).
Understanding the position vector is the first step in analyzing the particle's trajectory. It serves as a foundation for further calculation of derivatives and magnitudes, which help determine characteristics like velocity, acceleration, and curvature.
Derivative
Derivatives are a fundamental concept in calculus, representing the rate of change of a function with respect to its variable. For position vectors, derivatives help us understand how fast and in what direction a particle moves.
The first derivative of a position vector \( \mathbf{r}(t) \), denoted as \( \mathbf{r}'(t) \), represents the particle's velocity. In our problem, the first derivative is \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} \). This indicates that the rate of change in the \( x \)-direction is constant, while it varies linearly with \( t \) in the \( y \)-direction.
The second derivative, \( \mathbf{r}''(t) \), indicates acceleration. In this instance, \( \mathbf{r}''(t) = 2 \mathbf{j} \), showing constant acceleration in the \( y \)-direction, and no acceleration in the \( x \)-direction.
The first derivative of a position vector \( \mathbf{r}(t) \), denoted as \( \mathbf{r}'(t) \), represents the particle's velocity. In our problem, the first derivative is \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} \). This indicates that the rate of change in the \( x \)-direction is constant, while it varies linearly with \( t \) in the \( y \)-direction.
The second derivative, \( \mathbf{r}''(t) \), indicates acceleration. In this instance, \( \mathbf{r}''(t) = 2 \mathbf{j} \), showing constant acceleration in the \( y \)-direction, and no acceleration in the \( x \)-direction.
Magnitude
The magnitude of a vector is a measure of its length or size and is important when analyzing vector quantities like velocity and acceleration.
To find the magnitude of the first derivative \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} \), we use the formula:
\[ |\mathbf{r}'(t)| = \sqrt{1^2 + (2t)^2} = \sqrt{1 + 4t^2} \]
This formula gives us the speed of the particle at any time \( t \).
Similarly, the magnitude of the second derivative \( \mathbf{r}''(t) = 2 \mathbf{j} \) is simply \( |\mathbf{r}''(t)| = 2 \), which is the constant acceleration in the \( y \)-direction. Understanding magnitudes allows us to calculate further properties like the curvature.
To find the magnitude of the first derivative \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} \), we use the formula:
\[ |\mathbf{r}'(t)| = \sqrt{1^2 + (2t)^2} = \sqrt{1 + 4t^2} \]
This formula gives us the speed of the particle at any time \( t \).
Similarly, the magnitude of the second derivative \( \mathbf{r}''(t) = 2 \mathbf{j} \) is simply \( |\mathbf{r}''(t)| = 2 \), which is the constant acceleration in the \( y \)-direction. Understanding magnitudes allows us to calculate further properties like the curvature.
Curvature Formula
Curvature is a measure of how sharply a curve bends. For plane curves, the curvature formula helps us quantify the bending of the path at any given point.
The standard formula for curvature \( \kappa(t) \) of a parametric curve \( \mathbf{r}(t) \) is given by:
\[ \kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \]
In our scenario, the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) results in a vector that is perpendicular to both \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \), specifically capturing how the velocity and acceleration vectors interact. Its magnitude simplifies to \( 2\sqrt{1 + 4t^2} \). Thus, the curvature at any point \( t \) becomes \( \kappa(t) = \frac{2}{(1 + 4t^2)^{\frac{3}{2}}} \). This allows us to compute the curvature at specific times, illustrating the curve's changing nature.
The standard formula for curvature \( \kappa(t) \) of a parametric curve \( \mathbf{r}(t) \) is given by:
\[ \kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \]
In our scenario, the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) results in a vector that is perpendicular to both \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \), specifically capturing how the velocity and acceleration vectors interact. Its magnitude simplifies to \( 2\sqrt{1 + 4t^2} \). Thus, the curvature at any point \( t \) becomes \( \kappa(t) = \frac{2}{(1 + 4t^2)^{\frac{3}{2}}} \). This allows us to compute the curvature at specific times, illustrating the curve's changing nature.
