the polar curve $r = 2 + 3sin\theta$ is graphed for $0leq\thetaleq2pi$.\na) find the area enclosed by the…

the polar curve $r = 2 + 3sin\theta$ is graphed for $0leq\thetaleq2pi$.\na) find the area enclosed by the inner loop of the curve.\nb)determine the cartesian coordinates of the point on the curve where the tangent line is horizontal, for $\frac{pi}{2}<\theta<pi$.\nc)set up, but do not evaluate, an integral expression for the arc length of the outer loop.

the polar curve $r = 2 + 3sin\theta$ is graphed for $0leq\thetaleq2pi$.\na) find the area enclosed by the inner loop of the curve.\nb)determine the cartesian coordinates of the point on the curve where the tangent line is horizontal, for $\frac{pi}{2}<\theta<pi$.\nc)set up, but do not evaluate, an integral expression for the arc length of the outer loop.

Answer

Explanation:

Step1: Find limits for inner - loop area

First, find when (r = 0). So, (2+3\sin\theta=0), then (\sin\theta=-\frac{2}{3}). The two solutions in ([0,2\pi]) are (\theta_1=\pi+\arcsin(\frac{2}{3})) and (\theta_2 = 2\pi-\arcsin(\frac{2}{3})). The formula for the area (A) in polar coordinates is (A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta). Here, (\alpha=\pi+\arcsin(\frac{2}{3})) and (\beta = 2\pi - \arcsin(\frac{2}{3})), and (r = 2 + 3\sin\theta). So, (A=\frac{1}{2}\int_{\pi+\arcsin(\frac{2}{3})}^{2\pi-\arcsin(\frac{2}{3})}(2 + 3\sin\theta)^{2}d\theta).

Step2: Find horizontal - tangent points

We know that (x=r\cos\theta=(2 + 3\sin\theta)\cos\theta=2\cos\theta+3\sin\theta\cos\theta) and (y=r\sin\theta=(2 + 3\sin\theta)\sin\theta=2\sin\theta+3\sin^{2}\theta). Differentiate (x) and (y) with respect to (\theta): (\frac{dx}{d\theta}=-2\sin\theta + 3\cos^{2}\theta-3\sin^{2}\theta=-2\sin\theta+3\cos(2\theta)) and (\frac{dy}{d\theta}=2\cos\theta + 6\sin\theta\cos\theta=2\cos\theta+3\sin(2\theta)). For a horizontal tangent, (\frac{dy}{d\theta}=0), so (2\cos\theta+3\sin(2\theta)=0), (2\cos\theta+6\sin\theta\cos\theta=0), (\cos\theta(1 + 3\sin\theta)=0). In the interval (\frac{\pi}{2}<\theta<\pi), (\cos\theta\neq0), then (1 + 3\sin\theta=0), (\sin\theta=-\frac{1}{3}), (\cos\theta=-\sqrt{1-\sin^{2}\theta}=-\frac{2\sqrt{2}}{3}). (r=2+3\sin\theta=2 - 1 = 1). The Cartesian coordinates are (x=r\cos\theta=1\times(-\frac{2\sqrt{2}}{3})=-\frac{2\sqrt{2}}{3}) and (y=r\sin\theta=1\times(-\frac{1}{3})=-\frac{1}{3}).

Step3: Set up arc - length integral

The formula for the arc - length (L) of a polar curve (r = f(\theta)) is (L=\int_{\alpha}^{\beta}\sqrt{r^{2}+(\frac{dr}{d\theta})^{2}}d\theta). First, find (\frac{dr}{d\theta}=3\cos\theta). To find the limits for the outer loop, we note that (r) starts and ends at (r = 0). We already know (r = 0) when (\sin\theta=-\frac{2}{3}). The outer - loop is traversed for (\theta) values from (2\pi-\arcsin(\frac{2}{3})) to (\pi+\arcsin(\frac{2}{3})+2\pi). So the arc - length integral is (L=\int_{2\pi-\arcsin(\frac{2}{3})}^{\pi+\arcsin(\frac{2}{3})+2\pi}\sqrt{(2 + 3\sin\theta)^{2}+(3\cos\theta)^{2}}d\theta).

Answer:

a) (\frac{1}{2}\int_{\pi+\arcsin(\frac{2}{3})}^{2\pi-\arcsin(\frac{2}{3})}(2 + 3\sin\theta)^{2}d\theta) b) ((-\frac{2\sqrt{2}}{3},-\frac{1}{3})) c) (\int_{2\pi-\arcsin(\frac{2}{3})}^{\pi+\arcsin(\frac{2}{3})+2\pi}\sqrt{(2 + 3\sin\theta)^{2}+(3\cos\theta)^{2}}d\theta)