/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Total curvature \(\quad\) We fin... [FREE SOLUTION] | 91Ó°ÊÓ

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Total curvature \(\quad\) We find the total curvature of the portion of a smooth curve that runs from \(s=s_{0}\) to \(s=s_{1}>s_{0}\) by integrating \(\kappa\) from \(s_{0}\) to \(s_{1}\) . If the curve has some other parameter, say \(t,\) then the total curvature is $$ K=\int_{s_{0}}^{s_{1}} \kappa d s=\int_{t_{0}}^{t_{1}} \kappa \frac{d s}{d t} d t=\int_{t_{0}}^{t_{1}} \kappa|\mathbf{v}| d t $$ where \(t_{0}\) and \(t_{1}\) correspond to \(s_{0}\) and \(s_{1} .\) Find the total curvatures of $$ \begin{array}{l}{\text { a. The portion of the helix } \mathbf{r}(t)=(3 \cos t) \mathbf{i}+(3 \sin t) \mathbf{j}+t \mathbf{k}} \\ {0 \leq t \leq 4 \pi .} \\\ {\text { b. The parabola } y=x^{2},-\infty< x < \infty}\end{array} $$

Short Answer

Expert verified
a. \(\frac{6\pi\sqrt{3}}{5}\), b. Undefined (infinite).

Step by step solution

01

Understanding the Problem

We are asked to find the total curvature of two curves: a helix parameterized by \( \mathbf{r}(t) = (3 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + t \mathbf{k} \) over the interval \([0, 4\pi]\), and a parabola \( y = x^2 \) over the interval \((-\infty, \infty)\). We need to integrate the curvature \( \kappa \) with respect to the arc length \( s \) for both curves.
02

Calculating the Curvature of the Helix

The curvature \( \kappa \) for a space curve given by \( \mathbf{r}(t) \) is defined as \( \kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \). For \( \mathbf{r}(t) = (3 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + t \mathbf{k} \), calculate the first derivative \( \mathbf{r}'(t) = (-3 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j} + \mathbf{k} \) and the second derivative \( \mathbf{r}''(t) = (-3 \cos t) \mathbf{i} - (3 \sin t) \mathbf{j} \). The cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) yields \( 9(\mathbf{i}) + 9(\mathbf{j}) - 9 \mathbf{k} \), which has magnitude \( \sqrt{243} = 9\sqrt{3} \). The magnitude of \( \mathbf{r}'(t) \) is \( \sqrt{10} \). Thus, \( \kappa = \frac{9\sqrt{3}}{(\sqrt{10})^3} = \frac{9\sqrt{3}}{30} = \frac{3\sqrt{3}}{10} \).
03

Integrating Curvature for the Helix

To find the total curvature of the helix, integrate \( \kappa \) over the interval \([0, 4\pi]\): \[ K = \int_{0}^{4\pi} \frac{3\sqrt{3}}{10} \, dt = \frac{3\sqrt{3}}{10} \cdot (4\pi - 0) = \frac{3\sqrt{3}(4\pi)}{10} = \frac{12\pi\sqrt{3}}{10} = \frac{6\pi\sqrt{3}}{5} \].
04

Calculating the Curvature of the Parabola

For the parabola \( y = x^2 \), convert it into a parameterized form:\( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \). The curvature \( \kappa \) for a planar curve \( y = f(x) \) is \( \kappa = \frac{|2y'|}{(1+(y')^2)^{3/2}} \) where \( y' = 2x \). Substitute \( y' = 2t \) to get \( \kappa = \frac{|4t|}{(1 + (2t)^2)^{3/2}} \).
05

Integrating Curvature for the Parabola (Conceptual)

Since the parabola has infinite arcs on both directions, theoretically, the total curvature over \((-\infty, \infty)\) is infinite. In practical problems, one often considers segments, but as this is not specified, conceptual evaluation is necessary. Thus, we generally note the total curvature as undefined, as it is conceptually infinite.

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Key Concepts

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

Arc Length
The arc length of a curve is an essential concept that refers to the distance along the curve between two points. To calculate the arc length precisely, one often employs calculus, particularly integration. For a curve represented by a vector function \( \mathbf{r}(t) \), the formula for arc length \( L \) over an interval \([a, b]\) is:
\[ L = \int_{a}^{b} |\mathbf{r}'(t)| \, dt \]
The arc length accumulates the infinitesimal linear distances along the curve from start to end. It's a generalization of straight-line distance for curves, crucially important for both theoretical applications and practical tasks like measuring real-life objects.
Understanding arc length helps in various fields such as physics (for motion along curves), engineering (for designing curves), and computer graphics (for rendering curves accurately).
Space Curve
Space curves are fascinating geometric objects that extend beyond two dimensions into three-dimensional space. Unlike planar curves, which lie on a flat plane, space curves exist in the full extent of 3D space, often described by vector functions. Consider the curve parametrized by:
\( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \)
Each component function \( x(t), y(t), \) and \( z(t) \) describes the curve's projection on the respective coordinate axis, allowing it to twist and turn freely in space.
Space curves are not limited to simple shapes; they can form complex structures like helices, knots, and loops. These curves are crucial in advanced fields such as aerospace design, where trajectories are plotted in 3D, and bioinformatics, where molecular structures are examined.
Curvature Formula
The curvature of a curve provides a measure of how sharply it bends. For a smooth space curve given by \( \mathbf{r}(t) \), the curvature \( \kappa \) is defined as:
\[ \kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \]
It involves the cross product of the first and second derivatives of the curve, reflecting the change in the tangent vector along the curve. For a plane curve, the curvature takes a slightly different form but follows the same principle of measuring bending.
Curvature is key in analyzing the geometric properties of a curve, including identifying points of high curvature which are significant in fields like mechanical engineering (stress points) and computer vision (object recognition).
Helix
A helix is one of the most intriguing examples of a space curve, characterized by its smooth spiraling shape. It resembles a spring or the threads of a screw. Its standard form in three-dimensional space is given by:
\( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} + bt \mathbf{k} \)
Here, \( a \) determines the radius of the helix's base circle, and \( b \) controls the rise per complete revolution around the axis of rotation.
Helices appear in many real-world applications, from the structure of DNA to the design of spiral staircases and screws. Understanding their mathematical form allows engineers and scientists to apply this natural shape in diverse technological and biological contexts.
Parameterized Curve
Parameterized curves allow us to describe a wider variety of curves efficiently by varying a parameter, typically denoted as \( t \). Instead of a single equation, these curves use vector functions like:
\( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \)
By adjusting \( t \), one traces the path of the curve through its entire length, capturing complex shapes that aren't possible to describe with standard functions. In fact, any smooth curve can be parameterized to better study its properties such as curvature or torsion.
Parameterized curves are fundamental in computer graphics for animations, in physics for describing particle trajectories, and in mathematics for a deeper exploration of geometry.

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Most popular questions from this chapter

Evaluate the integrals in Exercises \(1-10\) $$ \int_{1}^{\ln 3}\left[t e^{t} \mathbf{i}+e^{t} \mathbf{j}+\ln t \mathbf{k}\right] d t $$

Use Simpson's Rule with \(n=10\) to approximate the length of arc of \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k}\) from the origin to the point \((2,4,8) .\)

In Exercises \(31-38\) you will use a CAS to explore the osculating circle at a point \(P\) on a plane curve where \(\kappa \neq 0 .\) Use a CAS to perform the following steps: $$ \begin{array}{l}{\text { a. Plot the plane curve given in parametric or function form over }} \\ {\text { the specified interval to see what it looks like. }} \\ {\text { b. Calculate the curvature } \kappa \text { of the curve at the given value } t_{0}} \\ {\text { using the appropriate formula from Exercise } 5 \text { or } 6 . \text { Use the }} \\ {\text { parametrization } x=t \text { and } y=f(t) \text { if the curve is given as a }} \\ {\quad \text { function } y=f(x) \text { . }}\\\\{\text { c. Find the unit normal vector } N \text { at } t_{0} . \text { Notice that the signs of }} \\ {\text { the components of } N \text { depend on whether the unit tangent vector }} \\\ {\text { T is turning clockwise or counterclockwise at } t=t_{0} \text { . (See }} \\ {\text { Exercise } 7 . )}\\\\{\text { d. If } \mathbf{C}=a \mathbf{i}+b \mathbf{j} \text { is the vector from the origin to the center }(a, b)} \\ {\text { of the osculating circle, find the center } \mathbf{C} \text { from the vector }} \\ {\text { equation }} \\\ {\mathbf{C}=\mathbf{r}\left(t_{0}\right)+\frac{1}{\kappa\left(t_{0}\right)} \mathbf{N}\left(t_{0}\right)} \\ {\text { The point } P\left(x_{0}, y_{0}\right) \text { on the curve is given by the position }} \\ {\text { vector } \mathbf{r}\left(t_{0}\right) .}\\\\{\text { e. Plot implicitly the equation }(x-a)^{2}+(y-b)^{2}=1 / \kappa^{2}} \\ {\text { of the osculating circle. Then plot the curve and osculating }} \\ {\text { circle together. You may need to experiment with the size of }} \\ {\text { the viewing window, but be sure the axes are equally scaled. }}\end{array} $$ $$ \begin{array}{l}{\mathbf{r}(t)=(2 t-\sin t) \mathbf{i}+(2-2 \cos t) \mathbf{j}, \quad 0 \leq t \leq 3 \pi} \\ {t_{0}=3 \pi / 2}\end{array} $$

A formula for the curvature of a parametrized plane curve $$ \begin{array}{c}{\text { a. Show that the curvature of a smooth curve } \mathbf{r}(t)=f(t) \mathbf{i}+} \\ {g(t) \mathbf{j} \text { defined by twice- differentiable functions } x=f(t) \text { and }} \\ {y=g(t) \text { is given by the formula }} \\ {\kappa=\frac{|\ddot{x} \ddot{y}-\dot{y} \ddot{x}|}{\left(\dot{x}^{2}+\dot{y}^{2}\right)^{3 / 2}}}\end{array} $$The dots in the formula denote differentiation with respect to \(t,\) one derivative for each dot. Apply the formula to find the curvatures of the following curves. $$ \begin{array}{l}{\text { b. } \mathbf{r}(t)=t \mathbf{i}+(\ln \sin t) \mathbf{j}, \quad 0 < t<\pi} \\ {\text { c. } \mathbf{r}(t)=\left[\tan ^{-1}(\sinh t)\right] \mathbf{i}+(\ln \cosh t) \mathbf{j}}\end{array} $$

Volleyball A volleyball is hit when it is 4 ft above the ground and 12 ft from a 6 -ft-high net. It leaves the point of impact with an initial velocity of 35 \(\mathrm{ft} / \mathrm{sec}\) at an angle of \(27^{\circ}\) and slips by the opposing team untouched. a. Find a vector equation for the path of the volleyball. b. How high does the volleyball go, and when does it reach maximum height? c. Find its range and flight time. d. When is the volleyball 7 ft above the ground? How far e. Suppose that the net is raised to 8 ft. Does this change things? Explain.

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