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Chapter 17: Problem 39

Use a scalar line integral to find the length of the following curves. $$\mathbf{r}(t)=\left\langle 20 \sin \frac{t}{4}, 20 \cos \frac{t}{4}, \frac{t}{2}\right\rangle, \text { for } 0 \leq t \leq 2$$

Short Answer

Expert verified
Answer: The length of the curve is \(2\sqrt{\frac{101}{4}}\).

Step by step solution

01

Compute the derivative of the curve

Given the curve \(\mathbf{r}(t)=\left\langle 20 \sin \frac{t}{4}, 20 \cos \frac{t}{4},\frac{t}{2}\right\rangle\), we take the derivative of each component with respect to t: $$\frac{d \mathbf{r}}{dt} = \left\langle \frac{d}{dt}\left(20 \sin \frac{t}{4}\right), \frac{d}{dt}\left(20 \cos \frac{t}{4}\right),\frac{d}{dt}\left(\frac{t}{2}\right) \right\rangle$$ After differentiating, we obtain: $$\frac{d \mathbf{r}}{dt} = \left\langle 5 \cos \frac{t}{4}, -5 \sin \frac{t}{4}, \frac{1}{2} \right\rangle$$
02

Calculate the magnitude of the derivative

Now we find the magnitude of the derivative: $$\left\|\frac{d \mathbf{r}}{dt}\right\| = \sqrt{\left(5 \cos \frac{t}{4}\right)^2 + \left(-5 \sin \frac{t}{4}\right)^2 + \left(\frac{1}{2}\right)^2}$$ After simplification, we have: $$\left\|\frac{d \mathbf{r}}{dt}\right\| = \sqrt{25\left(\cos^2 \frac{t}{4} +\sin^2 \frac{t}{4}\right) + \frac{1}{4}} = \sqrt{25 + \frac{1}{4}} = \sqrt{\frac{101}{4}}$$
03

Integrate the magnitude

Finally, we integrate the magnitude with respect to t over the interval [0, 2]: $$L = \int_0^2 \left\|\frac{d \mathbf{r}}{dt}\right\| dt = \int_0^2 \sqrt{\frac{101}{4}} dt$$ As the integrand is a constant, we can simply multiply it by the length of the interval: $$L = \sqrt{\frac{101}{4}} \cdot (2 - 0) = 2\sqrt{\frac{101}{4}}$$ The length of the curve is \(2\sqrt{\frac{101}{4}}\).

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Key Concepts

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

Curve Length
In mathematical terms, the length of a curve is the total distance along its path from one end to the other. The concept of curve length is significant in various applications, including physics and engineering. To find a curve's length, we consider the integral of the magnitude of its derivative.
The curve given in this context is defined by the vector function \(\mathbf{r}(t)=\left\langle 20 \sin \frac{t}{4}, 20 \cos \frac{t}{4},\frac{t}{2}\right\rangle\) for \(0 \leq t \leq 2\). By using a scalar line integral, the length of the curve can be calculated by summing up, or integrating, little segments of the curve across the parameter \(t\). This process enables us to convert complex curves into calculable formats, thereby estimating their length accurately.
Magnitude of Derivative
The magnitude of the derivative is a critical step in calculating curve length through scalar line integrals. To comprehend this, first, recall that the derivative of a vector function corresponds to the tangent vector. The magnitude of this tangent vector outlines the rate of change in the curve's position as \(t\) varies.
The derivative calculated earlier is \(\frac{d \mathbf{r}}{dt} = \left\langle 5 \cos \frac{t}{4}, -5 \sin \frac{t}{4}, \frac{1}{2} \right\rangle\), where each component is the rate of change of the respective components of \(\mathbf{r}(t)\). Then we compute its magnitude, which is essentially the length of this derivative vector and it is given by:
  • \(5 \cos \frac{t}{4}\)
  • \(-5 \sin \frac{t}{4}\)
  • \(\frac{1}{2}\)
Combining these, the magnitude of the derivative is \(\sqrt{25 + \frac{1}{4}}\) which simplifies our calculations in further steps.
Derivative of Vector Function
Finding the derivative of a vector function is essential in understanding the behavior of a curve. This derivative indicates how each component of a given vector function changes with respect to the parameter \(t\). For \(\mathbf{r}(t)=\left\langle 20 \sin \frac{t}{4}, 20 \cos \frac{t}{4},\frac{t}{2}\right\rangle\), we performed differentiation as follows:
  • \(\frac{d}{dt}\left(20 \sin \frac{t}{4}\right) = 5 \cos \frac{t}{4}\)
  • \(\frac{d}{dt}\left(20 \cos \frac{t}{4}\right) = -5 \sin \frac{t}{4}\)
  • \(\frac{d}{dt}\left(\frac{t}{2}\right) = \frac{1}{2}\)
These derivatives demonstrate how each component vectors behave along the curve's path, providing a clearer picture of their instantaneous rates of change, which is valued in curve analysis.
Integral of a Constant Function
Integrating a constant function is a straightforward step in mathematics. This becomes applicable in our context after identifying that the magnitude of the derivative, which was simplified as \(\sqrt{\frac{101}{4}}\), is indeed a constant.

Integration over an interval \([0, 2]\) involves calculating \(\int_0^2 \sqrt{\frac{101}{4}} dt\).
As the integrand is constant, we can expand our result by simply multiplying this constant with the length of the interval, which is \(2 - 0 = 2\). Thus, \(L = 2\sqrt{\frac{101}{4}}\). This integral simplifies the computation of curve length significantly by reducing it to the product of a constant magnitude and the interval length.

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Most popular questions from this chapter

Consider the rotational velocity field \(\mathbf{v}=\mathbf{a} \times \mathbf{r},\) where \(\mathbf{a}\) is a nonzero constant vector and \(\mathbf{r}=\langle x, y, z\rangle .\) Use the fact that an object moving in a circular path of radius \(R\) with speed \(|\mathbf{v}|\) has an angular speed of \(\omega=|\mathbf{v}| / R\). a. Sketch a position vector \(\mathbf{a},\) which is the axis of rotation for the vector field, and a position vector \(\mathbf{r}\) of a point \(P\) in \(\mathbb{R}^{3}\). Let \(\theta\) be the angle between the two vectors. Show that the perpendicular distance from \(P\) to the axis of rotation is \(R=|\mathbf{r}| \sin \theta\). b. Show that the speed of a particle in the velocity field is \(|\mathbf{a} \times \mathbf{r}|\) and that the angular speed of the object is \(|\mathbf{a}|\). c. Conclude that \(\omega=\frac{1}{2}|\nabla \times \mathbf{v}|\).

Radial fields in \(\mathbb{R}^{3}\) are conservative Prove that the radial field \(\mathbf{F}=\frac{\mathbf{r}}{|\mathbf{r}|^{p}},\) where \(\mathbf{r}=\langle x, y, z\rangle\) and \(p\) is a real number, is conservative on any region not containing the origin. For what values of \(p\) is \(\mathbf{F}\) conservative on a region that contains the origin?

Fourier's Law of heat transfer (or heat conduction ) states that the heat flow vector \(\mathbf{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T,\) which means that heat energy flows from hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of \(J /(m-s-K)\) A temperature function for a region \(D\) is given. Find the net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=-k \iint_{S} \nabla T \cdot \mathbf{n} d S\) across the boundary S of \(D\) In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume \(k=1 .\) $$\begin{aligned} &T(x, y, z)=100+x^{2}+y^{2}+z^{2}\\\ &D=\\{(x, y, z): 0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\\} \end{aligned}$$

Alternative construction of potential functions Use the procedure in Exercise 71 to construct potential functions for the following fields. $$\quad \mathbf{F}=\langle-y,-x\rangle$$

Conditions for Green's Theorem Consider the radial field \(\mathbf{F}=\langle f, g\rangle=\frac{\langle x, y\rangle}{\sqrt{x^{2}+y^{2}}}=\frac{\mathbf{r}}{|\mathbf{r}|}\) a. Explain why the conditions of Green's Theorem do not apply to F on a region that includes the origin. b. Let \(R\) be the unit disk centered at the origin and compute \(\iint_{R}\left(\frac{\partial f}{\partial x}+\frac{\partial g}{\partial y}\right) d A\) c. Evaluate the line integral in the flux form of Green's Theorem on the boundary of \(R\) d. Do the results of parts (b) and (c) agree? Explain.
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