/*! 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 57 Spiral arc length Find the lengt... [FREE SOLUTION] | 91Ó°ÊÓ

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Spiral arc length Find the length of the entire spiral \(r=e^{-a \theta}\), for \(\theta \geq 0\) and \(a>0\).

Short Answer

Expert verified
The arc length of the entire spiral is \(L = \frac{\sqrt{1+a^2}}{a}\).

Step by step solution

01

Identify the polar function and its derivative

The polar function is given by \(r = e^{-a\theta}\). To find the arc length, we will need to find the derivative of the function with respect to \(\theta\). The derivative is: $$\frac{dr}{d\theta} = -ae^{-a\theta}$$
02

Set up the arc length formula

Now we will set up the arc length formula, which is given by \(L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}d\theta\). Our integral will be from \(\theta = 0\) to \(\theta = \infty\), since we want to find the length of the entire spiral: $$L = \int_{0}^{\infty} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}d\theta$$
03

Substitute the polar function and its derivative

Now we will substitute the polar function and its derivative into the integral: $$L = \int_{0}^{\infty} \sqrt{(e^{-a\theta})^2 + (-ae^{-a\theta})^2}d\theta$$
04

Simplify the integrand

We can simplify the integrand in the integral by combining the two terms inside the square root: $$L = \int_{0}^{\infty} \sqrt{e^{-2a\theta} + a^2e^{-2a\theta}}d\theta$$ Factor out \(e^{-2a\theta}\) from inside the square root: $$L = \int_{0}^{\infty} \sqrt{e^{-2a\theta}(1 + a^2)}d\theta$$ Take the constant (\(1+a^2\)) outside the square root since it doesn't depend on \(\theta\): $$L = \sqrt{1+a^2}\int_{0}^{\infty} e^{-a\theta}d\theta$$
05

Solve the integral

Since we now have a simple exponential function inside the integral, we can easily integrate it: $$L = \sqrt{1+a^2}\left[-\frac{1}{a}e^{-a\theta}\right]_{0}^{\infty}$$
06

Evaluate the integral

Now we will evaluate the integral at the limits of integration: $$L = \sqrt{1+a^2}\left(-\frac{1}{a}e^{-a\infty} - -\frac{1}{a}e^{-a(0)}\right)$$ Since the exponential function approaches 0 as its argument approaches infinity, \(e^{-a\infty} = 0\): $$L = \sqrt{1+a^2}\left(\frac{1}{a}\right)$$
07

Final Answer

The arc length of the entire spiral \(r = e^{-a\theta}\) is: $$ L = \frac{\sqrt{1+a^2}}{a} $$

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Key Concepts

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

Polar Coordinates
Polar coordinates are a unique way of describing points on a plane using a combination of distance and angle. Instead of using traditional Cartesian coordinates like \((x, y)\), polar coordinates express points as \((r, \theta)\), where:
  • \(r\) is the radial distance from the origin to the point.
  • \(\theta\) is the angular position relative to a reference direction (usually the positive x-axis).
In polar systems, curves can be more elegantly represented using polar equations. For example, a spiral can be defined using equations like \(r = e^{-a\theta}\). This specific type of curve expands or contracts as \(\theta\) changes, which is very different from the linear or parabolic shapes often seen in Cartesian coordinates. In polar coordinates, understanding these relationships is key for calculating properties like arc length.
Integration Techniques
Integration techniques involve finding the area under a curve or solving problems involving continuous sums. To calculate the arc length of a polar curve, we use a specific integration formula: \[ L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}d\theta \]This formula allows us to sum up small segments of the polar curve to find the total arc length.
Here's a step-by-step breakdown:- Identify the polar function and its derivative, \(\frac{dr}{d\theta}\).- Plug these values into the arc length formula.The integral bounds from \(\theta = 0\) to \(\theta = \infty\) describe the entire extent of the spiral, capturing its full length by considering each infinitesimal piece. Integration transforms this idea into a summation that evaluates to the length of the curve. This demonstrates how calculus can be applied to solve complex geometry problems involving curves.
Exponential Functions
Exponential functions are mathematical expressions where the variable is in the exponent. An example in the given problem is \(r = e^{-a\theta}\). These functions are known for their rapid changes:
  • \(e^{-a\theta}\) decreases as \(\theta\) increases if \(a > 0\).
  • They exhibit continuous growth or decay depending on the sign and value of the exponent.
In polar curves like spirals, this exponential nature controls how tightly the spiral wraps as \(\theta\) changes. Integrals involving exponential functions often simplify due to their predictable behavior. This makes them easier to handle compared to polynomials or other complex equations. Understanding these properties helps when evaluating integrals or solving for values over infinite intervals, as seen in the arc length of the spiral problem, where the function simplifies and integrates elegantly.

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Most popular questions from this chapter

Prove the following identities. Assume that \(\mathbf{u}, \mathbf{v}, \mathbf{w}\) and \(\mathbf{x}\) are nonzero vectors in \(\mathbb{R}^{3}\). $$(\mathbf{u} \times \mathbf{v}) \cdot(\mathbf{w} \times \mathbf{x})=(\mathbf{u} \cdot \mathbf{w})(\mathbf{v} \cdot \mathbf{x})-(\mathbf{u} \cdot \mathbf{x})(\mathbf{v} \cdot \mathbf{w})$$

Graph the curve \(\mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle\) and prove that it lies on the surface of a sphere centered at the origin.

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Write the vector \langle 2,-6\rangle in terms of \(\mathbf{I}\) and \(\mathbf{J}\).

Suppose water flows in a thin sheet over the \(x y\) -plane with a uniform velocity given by the vector \(\mathbf{v}=\langle 1,2\rangle ;\) this means that at all points of the plane, the velocity of the water has components \(1 \mathrm{m} / \mathrm{s}\) in the \(x\) -direction and \(2 \mathrm{m} / \mathrm{s}\) in the \(y\) -direction (see figure). Let \(C\) be an imaginary unit circle (that does not interfere with the flow). a. Show that at the point \((x, y)\) on the circle \(C\) the outwardpointing unit vector normal to \(C\) is \(\mathbf{n}=\langle x, y\rangle\) b. Show that at the point \((\cos \theta, \sin \theta)\) on the circle \(C\) the outward-pointing unit vector normal to \(C\) is also $$ \mathbf{n}=\langle\cos \theta, \sin \theta\rangle $$ c. Find all points on \(C\) at which the velocity is normal to \(C\). d. Find all points on \(C\) at which the velocity is tangential to \(C\). e. At each point on \(C\) find the component of \(v\) normal to \(C\) Express the answer as a function of \((x, y)\) and as a function of \(\theta\) f. What is the net flow through the circle? That is, does water accumulate inside the circle?

Consider the ellipse \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle\) for \(0 \leq t \leq 2 \pi,\) where \(a\) and \(b\) are real numbers. Let \(\theta\) be the angle between the position vector and the \(x\) -axis. a. Show that \(\tan \theta=(b / a) \tan t\) b. Find \(\theta^{\prime}(t)\) c. Note that the area bounded by the polar curve \(r=f(\theta)\) on the interval \([0, \theta]\) is \(A(\theta)=\frac{1}{2} \int_{0}^{\theta}(f(u))^{2} d u\) Letting \(f(\theta(t))=|\mathbf{r}(\theta(t))|,\) show that \(A^{\prime}(t)=\frac{1}{2} a b\) d. Conclude that as an object moves around the ellipse, it sweeps out equal areas in equal times.

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