Textbook Question
63–74. Arc length of polar curves Find the length of the following polar curves.
The curve r = sin³(θ/3), for 0 ≤ θ ≤ π/2
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16. Parametric Equations & Polar Coordinates
Calculus in Polar Coordinates
3:13 minutesProblem 12.3.78
Textbook Question
Spiral arc length Consider the spiral r=4θ, for θ≥0.
a. Use a trigonometric substitution to find the length of the spiral, for 0≤θ≤√8.
Verified step by step guidance1
Recall the formula for the length of a curve given in polar coordinates: \(L = \int_{\alpha}^{\beta} \sqrt{r(\theta)^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta\).
Identify the given function: \(r(\theta) = 4\theta\), and the interval for \(\theta\) is from \$0$ to \(\sqrt{8}\).
Compute the derivative of \(r(\theta)\) with respect to \(\theta\): \(\frac{dr}{d\theta} = 4\).
Substitute \(r(\theta)\) and \(\frac{dr}{d\theta}\) into the arc length formula to get: \(L = \int_0^{\sqrt{8}} \sqrt{(4\theta)^2 + 4^2} \, d\theta = \int_0^{\sqrt{8}} \sqrt{16\theta^2 + 16} \, d\theta\).
Factor out the constant inside the square root and prepare for a trigonometric substitution: \(L = \int_0^{\sqrt{8}} 4 \sqrt{\theta^2 + 1} \, d\theta\). Use the substitution \(\theta = \sinh u\) or \(\theta = \tan u\) to simplify the integral.
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Video duration:
3mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Polar Coordinates and Parametric Curves
In polar coordinates, a curve is described by r as a function of θ. Understanding how to express the curve in terms of r(θ) and θ is essential, as it allows conversion to parametric form for length calculations using x = r cos θ and y = r sin θ.
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Intro to Polar Coordinates
Arc Length Formula for Polar Curves
The length of a curve defined in polar form r(θ) from θ = a to θ = b is given by the integral ∫ from a to b of √(r(θ)² + (dr/dθ)²) dθ. This formula combines the radius and its rate of change to measure the curve's length accurately.
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Trigonometric Substitution in Integration
Trigonometric substitution is a technique used to simplify integrals involving square roots of quadratic expressions. By substituting variables with trigonometric functions, the integral becomes easier to evaluate, which is useful when finding the arc length of curves like spirals.
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