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Chapter 10: Problem 57

Finding the Arc Length of a Polar Curve In Exercises \(53-58\) , find the length of the curve over the given interval. $$r=1+\sin \theta, \quad[0,2 \pi]$$

Short Answer

Expert verified
The length of the polar curve \(r=1 + \sin \theta\) over the interval \([0, 2蟺]\) is 8.

Step by step solution

01

Derivative of r

First, find the derivative \(dr/d\theta\) for the function \(r = 1 + \sin \theta\). Using the basic differentiation rules, the derivative of \(\sin \theta\) is \(\cos \theta\). This implies that \(dr/d\theta = 0 + \cos \theta = \cos \theta\).
02

Applying the arc length formula

Next, use the formula for arc length of a polar curve \(L = \int_a^b \sqrt{r^2 + (dr/d\theta)^2} d\theta\). Substitute \(r = 1 + \sin \theta\) and \(dr/d\theta = \cos \theta\) into the formula. The integral will be \(L = \int_0^{2蟺} \sqrt{(1+\sin \theta)^2 + (\cos \theta)^2} d\theta\).
03

Simplifying the Integral

Simplify the expression under the radical in the integral \(L = \int_0^{2蟺} \sqrt{1 + 2\sin \theta + \sin^2 \theta + \cos^2 \theta} d\theta\). Recognizing that \(\sin^2 \theta + \cos^2 \theta = 1\), this simplifies to \(L = \int_0^{2蟺} \sqrt{2 + 2\sin \theta} d\theta\).
04

Final Solution

Finally, solve this integral. The antiderivative is not something standard and needs to be looked up or calculated using special techniques. The result, calculated with a tool or manually, will be \(L = 8\).

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Key Concepts

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

Polar Coordinates
Polar coordinates offer a unique way to describe points in a plane using distances and angles, rather than the typical x and y coordinates. You specify a point using the distance from the origin, often denoted as \( r \), and the angle \( \theta \) from the positive x-axis.
  • It's useful for systems where rotation plays a key role, like circular or spiral paths.
  • You express curves with equations such as \( r = 1 + \sin \theta \), which can generate interesting patterns when graphed.
  • The interval \([0, 2\pi]\) denotes a full circle, as \( \theta \) varies.
Understanding polar coordinates is the foundation of working with polar curves, as we often calculate factors like arc length, which involves more complex mathematics like calculus.
Calculus Integrals
Calculus integrals are a powerful tool to calculate quantities like areas under curves and, in our case, the arc length of a curve.
  • The definite integral \( \int_a^b f(x) \, dx \) is calculated over an interval \([a, b]\) and provides the accumulated value, such as area, between these bounds.
  • For polar curves, the arc length \( L \) is given by \( L = \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta \), combining both \( r = f(\theta) \) and its derivative.
  • Compute it by substituting \( r \) and \( \frac{dr}{d\theta} \) into the formula, which then often requires simplifying and solving the integral.
Integrals used in this manner help us find lengths and other measurements related to polar functions, illustrating the link between algebraic expressions and geometrical concepts.
Differentiation
Differentiation involves finding the rate at which a function is changing. It's foundational in calculus and crucial for understanding dynamic systems. In the context of polar functions:
  • The derivative \( \frac{dr}{d\theta} \) signifies how \( r \) changes as the angle \( \theta \) changes.
  • For the example function \( r = 1 + \sin \theta \), its derivative \( \cos \theta \) shows the rate of change of \( r \) with respect to \( \theta \).
  • This derivative is essential for the arc length formula, as it affects the structure of the integral you need to solve.
Understanding differentiation allows for the manipulation and solving of complex real-world problems, assessing how changes in one variable might influence another.
Trigonometric Functions
Trigonometric functions like sine and cosine play a key role in mathematics, especially within topics related to waveforms and circles. These functions:
  • Provide the basic # functions \( \sin \theta \) and \( \cos \theta \), used to model periodic phenomena.
  • In polar coordinates, trigonometric expressions describe the relationship between angles and radial distances, as seen in \( r = 1 + \sin \theta \).
  • They also assist in simplifying calculus equations due to properties like the Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1 \).
Trigonometric functions are vital for solving integrals in problems involving circular or oscillatory motion, such as calculating the arc length of polar curves.

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Most popular questions from this chapter

Identifying a Conic In Exercises \(23-26,\) use a graphing utility to graph the polar equation. Identify the graph and find its eccentricity. $$r=\frac{-15}{2+8 \sin \theta}$$

Finding a Polar Equation Find a polar equation for the ellipse with the following characteristics. Focus: \((0,0)\) Eccentricity: \(e=\frac{1}{2}\) Directrix: \(r=4 \sec \theta\)

True or False? In Exercises \(113-116,\) determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(x>0,\) then the point \((x, y)\) on the rectangular coordinate system can be represented by \((r, \theta)\) on the polar coordinate system, where \(r=\sqrt{x^{2}+y^{2}}\) and \(\theta=\arctan (y / x)\)

Eccentricity In Exercises 67 and 68 , let \(r_{0}\) represent the distance from a focus to the nearest vertex, and let \(r_{1}\) represent the distance from the focus to the farthest vertex. Show that the eccentricity of an ellipse can be written as \(e=\frac{r_{1}-r_{0}}{r_{1}+r_{0}}\) Then show that \(\frac{r_{1}}{r_{0}}=\frac{1+e}{1-e}\)

Antenna Radiation The radiation from a transmitting antenna is not uniform in all directions. The intensity from a particular antenna is modeled by\(r=a \cos ^{2} \theta\) (a) Convert the polar equation to rectangular form. (b) Use a graphing utility to graph the model for \(a=4\) and \(a=6 .\) (c) Find the area of the geographical region between the two curves in part (b).
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