Question

Math

Posted 2 months ago

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The curve above is drawn in the $x y$-plane and is described by the equation in polar coordinates $r=\theta+\sin (2 \theta)$ for $0 \leq \theta \leq \pi$, where $r$ is measured in meters and $\theta$ is measured in radians. The derivative of $r$ with respect to $\theta$ is given by $\frac{d r}{d \theta}=1+2 \cos (2 \theta)$.
(a) Find the area bounded by the curve and the $x$-axis.
(b) Find the angle $\theta$ that corresponds to the point on the curve with $x$-coordinate -2 .
(c) For $\frac{\pi}{3}<\theta<\frac{2 \pi}{3}, \frac{d r}{d \theta}$ is negative. What does this fact say about $r$ ? What does this fact say about the curve?
(d) Find the value of $\theta$ in the interval $0 \leq \theta \leq \frac{\pi}{2}$ that corresponds to the point on the curve in the first quadrant with greatest distance from the origin. Justify your answer.
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Answer from Sia

Posted 2 months ago

Solution by Steps

step 1

The area bounded by the curve and the $x$-axis in polar coordinates is given by the integral: $\frac{1}{2} \int_{0}^{\pi} r^2 \, d\theta$

step 2

Substitute $r = \theta + \sin(2\theta)$ into the integral: $\frac{1}{2} \int_{0}^{\pi} (\theta + \sin(2\theta))^2 \, d\theta$

step 3

Expand the integrand: $\frac{1}{2} \int_{0}^{\pi} (\theta^2 + 2\theta \sin(2\theta) + \sin^2(2\theta)) \, d\theta$

step 4

Integrate each term separately: $\frac{1}{2} \left[ \int_{0}^{\pi} \theta^2 \, d\theta + 2 \int_{0}^{\pi} \theta \sin(2\theta) \, d\theta + \int_{0}^{\pi} \sin^2(2\theta) \, d\theta \right]$

step 5

Evaluate the integrals: $\frac{1}{2} \left[ \frac{\pi^3}{3} + 0 + \frac{\pi}{2} \right] = \frac{1}{2} \left( \frac{\pi^3}{3} + \frac{\pi}{2} \right) = \frac{\pi^3}{6} + \frac{\pi}{4}$

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(b) Find the angle $\theta$ that corresponds to the point on the curve with $x$-coordinate -2.
```

step 1

The $x$-coordinate in polar coordinates is given by $x = r \cos(\theta)$

step 2

Substitute $r = \theta + \sin(2\theta)$ and $x = -2$: $-2 = (\theta + \sin(2\theta)) \cos(\theta)$

step 3

Solve the equation $-2 = (\theta + \sin(2\theta)) \cos(\theta)$ for $\theta$

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(c) For $\frac{\pi}{3} < \theta < \frac{2\pi}{3}$, $\frac{d r}{d \theta}$ is negative. What does this fact say about $r$? What does this fact say about the curve?
```

step 1

Given $\frac{d r}{d \theta} = 1 + 2 \cos(2\theta)$

step 2

For $\frac{\pi}{3} < \theta < \frac{2\pi}{3}$, $\cos(2\theta)$ is negative, making $1 + 2 \cos(2\theta)$ negative

step 3

Since $\frac{d r}{d \theta}$ is negative, $r$ is decreasing in this interval

step 4

The curve is moving closer to the origin as $\theta$ increases from $\frac{\pi}{3}$ to $\frac{2\pi}{3}$

```
(d) Find the value of $\theta$ in the interval $0 \leq \theta \leq \frac{\pi}{2}$ that corresponds to the point on the curve in the first quadrant with greatest distance from the origin. Justify your answer.
```

step 1

The distance from the origin in polar coordinates is given by $r = \theta + \sin(2\theta)$

step 2

To find the maximum distance, take the derivative and set it to zero: $\frac{d r}{d \theta} = 1 + 2 \cos(2\theta) = 0$

step 3

Solve for $\theta$: $1 + 2 \cos(2\theta) = 0 \Rightarrow \cos(2\theta) = -\frac{1}{2} \Rightarrow 2\theta = \frac{2\pi}{3} \Rightarrow \theta = \frac{\pi}{3}$

step 4

Verify that $\theta = \frac{\pi}{3}$ is within the interval $0 \leq \theta \leq \frac{\pi}{2}$ and corresponds to a maximum by checking the second derivative or evaluating $r$ at endpoints

Answer

(a) $\frac{\pi^3}{6} + \frac{\pi}{4}$

(b) Solve $-2 = (\theta + \sin(2\theta)) \cos(\theta)$ for $\theta$

(c) $r$ is decreasing; the curve moves closer to the origin

(d) $\theta = \frac{\pi}{3}$

Key Concept

Polar coordinates and integrals

Explanation

The problem involves finding areas, solving equations, and analyzing behavior in polar coordinates using calculus techniques.

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