/*! 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 62 Finding the Arc Length of a Pola... [FREE SOLUTION] | 91Ó°ÊÓ

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Finding the Arc Length of a Polar Curve In Exercises \(59-64\) , use a graphing utility to graph the polar equation over the given interval. Use the integration capabilities of the graphing utility to approximate the length of the curve. $$r=e^{\theta}, \quad[0, \pi]$$

Short Answer

Expert verified
The approximate length of the curve for the polar equation \(r = e^\theta\) over the interval [0, \(\pi\)] is \(\sqrt{2}(e^\pi - 1)\).

Step by step solution

01

Derive the Equation

\(\frac{dr}{d\theta}\) for \(r = e^\theta\) can be derived as: \(dr/d\theta = e^\theta\).
02

Find the Expression under Square Root

Next, we compute the value under the square root in the Arc Length formula, which is \(r^2 + (dr/d\theta)^2\) for \(r = e^\theta\) and \(dr/d\theta = e^\theta\). This equals \((e^\theta)^2 + (e^\theta)^2 = 2e^{2\theta}\).
03

Integrate the Expression

Now we need to evaluate the integral \(L = \int_a^b \sqrt{r^2 + (dr/d\theta)^2} d\theta = \int_0^\pi \sqrt{2e^{2\theta}} d\theta = \sqrt{2} \int_0^\pi e^\theta d\theta\). This evaluates to \(\sqrt{2}(e^\pi - 1)\), using the fact that \(\int e^\theta d\theta = e^\theta\).

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Key Concepts

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

Arc Length
Arc length is the distance along a curve between two points. In polar coordinates, it can require a special approach due to the nature of the equations involved. For a polar curve described by the function \( r(\theta) \), the arc length from \( \theta = a \) to \( \theta = b \) is calculated using the integral:
  • \( L = \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta \)
This formula accounts for the function's curve and its rate of change, ensuring an accurate representation of the path's length. Utilizing calculus and polar curves, the formula is derived using principles similar to Cartesian coordinates, but adjusted for the polar format. Understanding this integral setup is essential for tackling any problems involving finding arc lengths of curves described in polar coordinates.
Graphing Utility
In mathematics and related fields, a graphing utility is a tool that helps visualize equations or functions. These utilities come as standalone graphing calculators or software within computers and online platforms. They are essential in exploring polar curves, as these curves can be complex to draw manually.
  • For a given polar equation \( r = e^\theta \), a graphing utility helps quickly visualize the spiral shape it forms.
  • These tools also support computations like integration, providing approximations of values such as arc length more efficiently than manual calculations.
Using a graphing utility streamlines the process, making it not only quicker but also more accurate. It's especially helpful when dealing with intricate curves and extensive calculations.
Integration
Integration is a calculus concept used to find accumulated quantities like areas under curves or, as in our context, arc lengths of polar curves. When integrating functions related to polar coordinates, it effectively adds up infinitely small segments along the curve to get the full measure.
In the given exercise, we integrate the function derived \( \sqrt{2} \int_0^\pi e^\theta \, d\theta \). This can be interpreted as summing up the tiny lengths of straight lines joining points on the polar curve \( r = e^\theta \) from \( \theta = 0 \) to \( \theta = \pi \).
  • This results in calculating \( \sqrt{2} (e^\pi - 1) \), which is the arc length of the curve.
Understanding how to correctly apply and solve integrals is crucial for handling such polar curve problems.
Polar Coordinates
Polar coordinates offer a different way to express points on a plane, using a radius \( r \) and an angle \( \theta \). These coordinates are especially useful for curves with rotational symmetry or those that cannot be described easily with Cartesian coordinates.
For example, the curve described by the polar equation \( r = e^\theta \) illustrates a logarithmic spiral. In this system:
  • \( \theta \) represents the angle from the positive x-axis.
  • \( r \) describes the distance from the origin, creating a dynamic representation of the position on the curve.
Understanding polar coordinates is essential for analyzing complex functions and their properties, such as the arc length in this case. Transitioning smoothly between Cartesian and polar systems can enhance problem-solving skills across different math and physics contexts.

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Most popular questions from this chapter

Finding an Angle In Exercises \(107-112,\) use the result of Exercise 106 to find the angle \(\psi\) between the radial and tangent lines to the graph for the indicated value of \(\theta\) . Use a graphing utility to graph the polar equation, the radial line, and the tangent line for the indicated value of \(\theta .\) Identify the angle \(\psi\) . $$r=2 \cos 3 \theta \quad \theta=\frac{\pi}{4}$$

Finding the Arc Length of a Polar Curve In Exercises \(53-58\) , find the length of the curve over the given interval. $$r=4 \sin \theta, \quad[0, \pi]$$

Integration by Substitution Use integration by substitution to show that if \(y\) is a continuous function of \(x\) on the interval \(a \leq x \leq b,\) where \(x=f(t)\) and \(y=g(t),\) then $$\begin{array}{l}{\int_{a}^{b} y d x=\int_{t_{1}}^{t_{2}} g(t) f^{\prime}(t) d t} \\ {\text { where } f\left(t_{1}\right)=a, f\left(t_{2}\right)=b, \text { and both } g \text { and } f^{\prime} \text { are continuous }} \\ {\text { on }\left[t_{1}, t_{2}\right].}\end{array}$$

Rotated Parabola Write the equation for the parabola rotated \(\pi / 4\) radian counterclockwise from the parabola \(r=\frac{9}{1+\sin \theta}\)

Finding a Polar Equation In Exercises \(33-38\) , find a polar equation for the conic with its focus at the pole and the given eccentricity and directrix. (For convenience, the equation for the directrix is given in rectangular form.) $$\begin{array}{ll}{\text { Conic }} & {\text { Eccentricity }} \\ {\text { Hyperbola}} & {\quad e=\frac{4}{3} }\end{array} \begin{array}{l}{\text { Directrix }} \\ {x=-2}\end{array}$$

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