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Chapter 10: Problem 29

Find the curvature \(K\) of the curve. $$ \mathbf{r}(t)=4 \cos 2 \pi t \mathbf{i}+4 \sin 2 \pi t \mathbf{j} $$

Short Answer

Expert verified
The curvature of the given curve \(\mathbf{r}(t)\) is \(\frac{1}{2 \pi}\).

Step by step solution

01

Compute the first derivative of \(\mathbf{r}(t)\)

The derivative of a vector function is computed component-wise. So, the first derivative \(\mathbf{r}'(t)\) is given by: \(\mathbf{r}'(t) = -8 \pi \sin(2 \pi t) \mathbf{i} + 8 \pi \cos(2 \pi t) \mathbf{j}\).
02

Compute the second derivative of \(\mathbf{r}(t)\)

Following the same procedure as in step 1, the second derivative \(\mathbf{r}''(t)\) is given by: \(\mathbf{r}''(t) = -16 \pi^2 \cos(2 \pi t) \mathbf{i} - 16 \pi^2 \sin(2 \pi t) \mathbf{j}\).
03

Compute the magnitudes of the first and second derivatives

The magnitude of a vector \(\mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j}\) is given by \(|\mathbf{a}| = \sqrt{a_x^2 + a_y^2}\). Therefore, \(|\mathbf{r}'(t)| = \sqrt{(-8 \pi \sin(2 \pi t))^2 + (8 \pi \cos(2 \pi t))^2} = 8 \pi\). Using the same mathematics, we get that \(|\mathbf{r}''(t)| = \sqrt{(-16 \pi^2 \cos(2 \pi t))^2 + (-16 \pi^2 \sin(2 \pi t))^2} = 16 \pi^2\).
04

Compute the cross product

The cross product of 2-dimensional vectors \(\mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j}\) and \(\mathbf{b} = b_x \mathbf{i} + b_y \mathbf{j}\) is given by \(|\mathbf{a} \times \mathbf{b}| = |a_x b_y - a_y b_x|\). Therefore, \(\mathbf{r}'(t) \times \mathbf{r}''(t) = |-8 \pi \sin(2 \pi t)\times -16 \pi^2 \sin(2 \pi t) - 8 \pi \cos(2 \pi t) \times -16 \pi^2 \cos(2 \pi t)| = 256 \pi^3\).
05

Evaluate the curvature

Use the formula for curvature, \(K = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}\). Substituting the magnitudes of \(\mathbf{r}'(t) = 8 \pi\), \(\mathbf{r}''(t) = 16 \pi^2\), and cross product \(256 \pi^3\) gives: \(K = \frac{256 \pi^3}{(8 \pi)^3} = \frac{1}{2 \pi}\).

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Key Concepts

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

Vector Calculus
Vector calculus is a field of mathematics that deals with vector fields and operations on them. In problems involving curves, we often look at vector functions like \( \mathbf{r}(t) \) which represent paths or trajectories in space.
In vector calculus:
  • Vectors can have both magnitude and direction.
  • Operations include addition, scalar multiplication, dot product, and cross product.
Understanding how vectors change and relate to each other is key in analyzing paths of motion, forces, and even fluid flow. Calculus operations extend to vector fields, helping us measure changes over curves and surfaces.
Parametric Equations
Parametric equations define curves by expressing coordinates as functions of a parameter, often \( t \). This allows us to:
  • Represent a wide variety of curves.
  • Describe motion where x and y can vary independently with time.
For example, the parametric curve \( \mathbf{r}(t)=4 \cos(2 \pi t) \mathbf{i}+4 \sin(2 \pi t) \mathbf{j} \) describes a circle of radius 4. As \( t \) varies, the equation traces out the curve by determining each position. Understanding parametric forms is crucial for analyzing dynamic systems and natural phenomena.
Cross Product
The cross product is a way to measure perpendicular interaction between two vectors in three-dimensional space. However, in the 2D plane, we can derive a scalar value from 2D vectors using a modified cross product.
  • This scalar helps find curvature in 2D curves.
  • Calculated as \( |\mathbf{a} \times \mathbf{b}| = |a_x b_y - a_y b_x| \) in 2D.
In our example, finding the cross product of derivatives \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \) helped determine the curvature's value. It's a vital tool in analyzing how curves bend or twist.
Derivative
Derivatives are fundamental in calculus and measure how functions change. For vector functions, derivatives tell us about velocity and acceleration.
  • The first derivative \( \mathbf{r}'(t) \) gives the velocity vector.
  • The second derivative \( \mathbf{r}''(t) \) provides the acceleration vector.
By understanding these changes, we can describe how quickly a curve's position and direction are evolving over time. Derivatives help in computing curvature, effectively measuring how tightly or loosely a curve bends in the plane.
Curves in Plane Geometry
Curves in plane geometry depict shapes confined to a two-dimensional plane. Curvature is a measure of a curve's deviation from being a straight line.
  • Circle arcs, parabolas, and spirals are examples of such curves.
  • Curvature \( K \) quantifies this bending, computed as \( K = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \).
The calculation of curvature involves derivatives and often the cross product, uniting various mathematical concepts. This understanding is essential in mathematics and physics, especially for designing paths and studying motion dynamics.

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Most popular questions from this chapter

Find the open interval(s) on which the curve given by the vector-valued function is smooth. $$ \mathbf{r}(t)=\frac{2 t}{8+t^{3}} \mathbf{i}+\frac{2 t^{2}}{8+t^{3}} \mathbf{j} $$

Use the model for projectile motion, assuming there is no air resistance. Rogers Centre in Toronto, Ontario has a center field fence that is 10 feet high and 400 feet from home plate. A ball is hit 3 feet above the ground and leaves the bat at a speed of 100 miles per hour. (a) The ball leaves the bat at an angle of \(\theta=\theta_{0}\) with the horizontal. Write the vector-valued function for the path of the ball. (b) Use a graphing utility to graph the vector-valued function for \(\theta_{0}=10^{\circ}, \theta_{0}=15^{\circ}, \theta_{0}=20^{\circ},\) and \(\theta_{0}=25^{\circ} .\) Use the graphs to approximate the minimum angle required for the hit to be a home run. (c) Determine analytically the minimum angle required for the hit to be a home run.

Use the given acceleration function to find the velocity and position vectors. Then find the position at time \(t=2\) $$ \begin{array}{l} \mathbf{a}(t)=-\cos t \mathbf{i}-\sin t \mathbf{j} \\ \mathbf{v}(0)=\mathbf{j}+\mathbf{k}, \quad \mathbf{r}(0)=\mathbf{i} \end{array} $$

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