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Chapter 13: Problem 18

To illustrate that the length of a smooth space curve does not depend on the parametrization you use to compute it, calculate the length of one turn of the helix in Example 1 with the following parametrizations. a. \(\mathbf{r}(t)=(\cos 4 t) \mathbf{i}+(\sin 4 t) \mathbf{j}+4 t \mathbf{k}, \quad 0 \leq t \leq \pi / 2\) b. \(\mathbf{r}(t)=[\cos (t / 2)] \mathbf{i}+[\sin (t / 2)] \mathbf{j}+(t / 2) \mathbf{k}, \quad 0 \leq t \leq 4 \pi\) c. \(\mathbf{r}(t)=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}-t \mathbf{k}, \quad-2 \pi \leq t \leq 0\)

Short Answer

Expert verified
The lengths from different parametrizations were calculated as 4蟺, 2蟺鈭�3, and 2蟺鈭�2.

Step by step solution

01

Calculate length for parametrization (a)

For the curve \( \mathbf{r}(t)=(\cos 4t) \mathbf{i}+(\sin 4t) \mathbf{j}+4t \mathbf{k} \), find the derivative: \( \mathbf{r}'(t)=(-4\sin 4t) \mathbf{i}+(4\cos 4t) \mathbf{j}+4 \mathbf{k} \). The speed \( \left\| \mathbf{r}'(t) \right\| \) is given by:\[ \sqrt{(-4\sin 4t)^2+(4\cos 4t)^2+4^2} = \sqrt{16 \sin^2 4t + 16 \cos^2 4t + 16} = \sqrt{32}. \]Now, integrate the speed from 0 to \( \frac{\pi}{2} \) to find the length:\[ \int_0^{\frac{\pi}{2}} \sqrt{32} \ dt = \sqrt{32} \times \frac{\pi}{2} = 4\pi. \]
02

Calculate length for parametrization (b)

For the curve \( \mathbf{r}(t) = [\cos (t / 2)] \mathbf{i} + [\sin (t / 2)] \mathbf{j} + (t / 2) \mathbf{k} \), find the derivative: \( \mathbf{r}'(t) = \left( -\frac{1}{2} \sin (t/2) \right) \mathbf{i} + \left( \frac{1}{2} \cos (t/2) \right) \mathbf{j} + \frac{1}{2} \mathbf{k} \). The speed is:\[ \left\| \mathbf{r}'(t) \right\| = \sqrt{\left( -\frac{1}{2} \sin (t/2) \right)^2 + \left(\frac{1}{2} \cos (t/2) \right)^2 + (\frac{1}{2})^2} = \sqrt{ \frac{1}{4} \sin^2 (t/2) + \frac{1}{4} \cos^2 (t/2) + \frac{1}{4} } = \frac{\sqrt{3}}{2}. \]Integrate from 0 to \( 4\pi \) to find the length:\[ \int_0^{4\pi} \frac{\sqrt{3}}{2} \, dt = \frac{\sqrt{3}}{2} \times 4\pi = 2\pi \sqrt{3}. \]
03

Calculate length for parametrization (c)

For the curve \( \mathbf{r}(t) = (\cos t) \mathbf{i} - (\sin t) \mathbf{j} - t \mathbf{k} \), find the derivative: \( \mathbf{r}'(t) = (-\sin t) \mathbf{i} - (\cos t) \mathbf{j} - \mathbf{k} \). The speed is:\[ \left\| \mathbf{r}'(t) \right\| = \sqrt{(-\sin t)^2 + (-\cos t)^2 + (-1)^2} = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}. \]Integrate from -\( 2\pi \) to 0 to find the length:\[ \int_{-2\pi}^0 \sqrt{2} \, dt = \sqrt{2} \times 2\pi = 2\pi \sqrt{2}. \]
04

Compare lengths for different parametrizations

Different parametrizations give different expressions for the curve's length, but they should all correspond to one complete turn of the helix. The answers are calculations of each parametrization separately.

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Key Concepts

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

Parametrization
In mathematics, parametrization plays a crucial role in defining curves, surfaces, and geometrical shapes. A parametrization involves representing a curve as a vector function, usually in terms of a single parameter, typically denoted as \(t\). This provides a means to express the position of points along the curve.
  • Different parametrizations can represent the same curve, allowing for flexibility in analysis and calculations.
  • The key is to transform a curve into a mathematical expression, which can help us compute other properties, like length, more easily.
  • Parametrization simplifies handling curves in multidimensional spaces, such as in vector calculus.
For example, various parametrizations of a helix (a type of three-dimensional curve) allow us to compute its properties from different starting points and with different parameters, all representing the same geometrical structure.
Space Curve
A space curve is a three-dimensional curve represented by a vector function. This function assigns a position vector to each point on the curve. Understanding space curves is essential as they add depth to our study of geometry beyond two dimensions.
  • Space curves are generally expressed as \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \), where \(x(t), y(t),\) and \(z(t)\) describe the position in 3D space.
  • They are important in physics and engineering for describing trajectories and paths of particles or objects.
  • Visualizing space curves involves thinking in terms of three axes 鈥� commonly known as the x, y, and z axes.
In practical terms, a space curve like a helix can be represented in a formula, allowing for precise calculation of properties like length or position at any point in time.
Derivative of Vector Functions
Finding the derivative of vector functions is akin to discovering rates of change of curves. This derivative will tell us how the position vector changes with time or another parameter, providing insights into the curve's dynamics.
  • The derivative \( \mathbf{r}'(t) \) translates to dissecting the curve鈥檚 direction and speed.
  • It results in a new vector function that represents the tangent vector at each point on the curve.
  • The magnitude of this derivative, often referred to as speed, reflects how fast the curve is traversed.
For instance, the derivative can help compute the length of a curve by integrating it over a specified interval, capturing the journey's full extent.
Vector Calculus
Vector calculus is a branch of mathematics focusing on vector fields and operations on vector functions. Foundational in fields such as physics and engineering, this study empowers us to perform calculations involving curves and surfaces.
  • It deals with concepts like gradient, divergence, and curl, fostering a greater understanding of vector fields.
  • Vector calculus helps in calculating arc lengths, surface areas, and ensuring precise modeling of the physical world.
  • Techniques from vector calculus, such as integration and differentiation, allow analyzing how vectors behave as they move through space.
Through vector calculus, it is possible to manipulate and explore more complex systems mathematically, underlaid by the operations in the mathematical analysis of space curves, such as helices.
Integration of Vector Functions
Integration of vector functions is essential to determine specific geometrical properties, like the length of a curve. This process aggregates small changes along the curve to discern the complete nature of the curve's path.
  • In this context, the integral of the tangent vector yields the total length of the trajectory.
  • The integration process considers the magnitude of the curve's derivative vector over a given interval.
  • Accurate integration facilitates modeling real-world systems that involve movement across space.
For example, by integrating the speed function derived from the vector function's derivative over a set interval, we calculate the entire path length of a helix. Each step in parametrization shows that although expressions change, the underlying property 鈥� such as total length 鈥� stays consistent, illustrating the beauty and power of mathematical integration.

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Most popular questions from this chapter

\(\mathbf{r}(t)\) is the position of a particle in space at time \(t\) Find the particle's velocity and acceleration vectors. Then find the particle's speed and direction of motion at the given value of \(t .\) Write the particle's velocity at that time as the product of its speed and direction. $$\mathbf{r}(t)=\left(e^{-t}\right) \mathbf{i}+(2 \cos 3 t) \mathbf{j}+(2 \sin 3 t) \mathbf{k}, \quad t=0$$

\(\mathbf{r}(t)\) is the position of a particle in space at time \(t\) Find the angle between the velocity and acceleration vectors at time \(t=0\) $$\mathbf{r}(t)=(3 t+1) \mathbf{i}+\sqrt{3} t \mathbf{j}+t^{2} \mathbf{k}$$

In Moscow in 1987 , Natalya Lisouskaya set a women's world record by putting an 4 kg shot 22.63 m. Assuming that she launched the shot at a \(40^{\circ}\) angle to the horizontal from \(2 \mathrm{m}\) above the ground, what was the shot's initial speed?

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: a. Plot the plane curve given in parametric or function form over the specified interval to see what it looks like.b. Calculate the curvature \(\kappa\) of the curve at the given value \(t_{0}\) using the appropriate formula from Exercise 5 or \(6 .\) Use the parametrization \(x=t\) and \(y=f(t)\) if the curve is given as a function \(y=f(x).\)c. Find the unit normal vector \(\mathbf{N}\) at \(t_{0} .\) Notice that the signs of the components of \(\mathbf{N}\) depend on whether the unit tangent vector \(\mathbf{T}\) is turning clockwise or counterclockwise at \(\left.t=t_{0} . \text { (See Exercise } 7 .\right)\) d. If \(\mathbf{C}=a \mathbf{i}+b \mathbf{j}\) is the vector from the origin to the center \((a, b)\) of the osculating circle, find the center \(\mathbf{C}\) from the vector equation $$\mathbf{C}=\mathbf{r}\left(t_{0}\right)+\frac{1}{\kappa\left(t_{0}\right)} \mathbf{N}\left(t_{0}\right),$$ The point \(P\left(x_{0}, y_{0}\right)\) on the curve is given by the position vector \(\mathbf{r}\left(t_{0}\right)\) e. Plot implicitly the equation \((x-a)^{2}+(y-b)^{2}=1 / \kappa^{2}\) of the osculating circle. Then plot the curve and osculating circle together. You may need to experiment with the size of the viewing window, but be sure the axes are equally scaled. $$\mathbf{r}(t)=t^{2} \mathbf{i}+\left(t^{3}-3 t\right) \mathbf{j}, \quad-4 \leq t \leq 4, \quad t_{0}=3 / 5.$$

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