1. We start by computing \(\frac{dy}{dx}\text{.}\) With \(\fp(\theta) = 2\cos(\theta)\text{,}\) we have \begin{align*} \frac{dy}{dx} \amp = \frac{2\cos(\theta) \sin(\theta) + \cos(\theta) (1+2\sin(\theta) )}{2\cos^2(\theta) -\sin(\theta) (1+2\sin(\theta) )}\\ \amp = \frac{\cos(\theta) (4\sin(\theta) +1)}{2(\cos^2(\theta) -\sin^2(\theta) )-\sin(\theta) }. \end{align*} When \(\theta=\pi/4\text{,}\) \(\frac{dy}{dx}=-2\sqrt{2}-1\) (this requires a bit of simplification). In rectangular coordinates, the point on the graph at \(\theta=\pi/4\) is \((1+\sqrt{2}/2,1+\sqrt{2}/2)\text{.}\) Thus the rectangular equation of the line tangent to the limaçon at \(\theta=\pi/4\) is \begin{equation*} y=(-2\sqrt{2}-1)\big(x-(1+\sqrt{2}/2)\big)+1+\sqrt{2}/2 \approx -3.83 x+8.24. \end{equation*} The limaçon and the tangent line are graphed in Figure 9.5.3. The normal line has the opposite–reciprocal slope as the tangent line, so its equation is \begin{equation*} y \approx \frac{1}{3.83}x+1.26. \end{equation*}

2. To find the horizontal lines of tangency, we find where \(\frac{dy}{dx}=0\text{;}\) thus we find where the numerator of our equation for \(\frac{dy}{dx}\) is 0. \begin{equation*} \cos(\theta) (4\sin(\theta) +1)=0 \Rightarrow \cos(\theta) =0 \text{ or } 4\sin(\theta) +1=0. \end{equation*} On \([0,2\pi]\text{,}\) \(\cos(\theta) =0\) when \(\theta=\pi/2,\ 3\pi/2\text{.}\) Setting \(4\sin(\theta) +1=0\) gives \(\theta=\sin^{-1}(-1/4)\approx -0.2527 = -14.48^\circ\text{.}\) We want the results in \([0,2\pi]\text{;}\) we also recognize there are two solutions, one in the 3\(^\text{ rd }\) quadrant and one in the 4\(^\text{ th }\text{.}\) Using reference angles, we have our two solutions as \(\theta =3.39\) and \(6.03\) radians. The four points we obtained where the limaçon has a horizontal tangent line are given in Figure 9.5.3 with black–filled dots. To find the vertical lines of tangency, we set the denominator of \(\frac{dy}{dx}=0\text{.}\) \begin{align*} 2(\cos^2(\theta) -\sin^2(\theta) )-\sin(\theta) \amp = 0 .\\ \end{align*} Convert the \(\cos^2(\theta)\) term to \(1-\sin^2(\theta)\text{:}\) \begin{align*} 2(1-\sin^2(\theta) -\sin^2(\theta) )-\sin(\theta) \amp = 0\\ 4\sin^2(\theta) + \sin(\theta) -1 \amp = 0.\\ \end{align*} Recognize this as a quadratic in the variable \(\sin(\theta)\text{.}\) Using the quadratic formula, we have \begin{align*} \sin(\theta) \amp = \frac{-1\pm\sqrt{33}}{8}. \end{align*} We solve \(\sin(\theta) = \frac{-1+\sqrt{33}}8\) and \(\sin(\theta) = \frac{-1-\sqrt{33}}8\text{:}\) \begin{align*} \sin(\theta) \amp =\frac{-1+\sqrt{33}}8 \amp \sin(\theta) \amp = \frac{-1-\sqrt{33}}{8}\\ \theta \amp = \sin^{-1}\left(\frac{-1+\sqrt{33}}8\right) \amp \theta \amp = \sin^{-1}\left(\frac{-1-\sqrt{33}}8\right)\\ \theta \amp = 0.6399 \amp \theta \amp = -1.0030 \end{align*} In each of the solutions above, we only get one of the possible two solutions as \(\sin^{-1}(x)\) only returns solutions in \([-\pi/2,\pi/2]\text{,}\) the 4\(^\text{ th }\) and \(1^\text{ st }\) quadrants. Again using reference angles, we have: \begin{equation*} \sin(\theta) = \frac{-1+\sqrt{33}}8 \Rightarrow \theta = 0.6399,\ 3.7815 \text{ radians } \end{equation*} and \begin{equation*} \sin(\theta) = \frac{-1-\sqrt{33}}8 \Rightarrow \theta = 4.1446,\ 5.2802 \text{ radians. } \end{equation*} These points are also shown in Figure 9.5.3 with white–filled dots.

We need to know when \(r=0\text{.}\) \begin{align*} 1+2\sin(\theta) \amp = 0\\ \sin(\theta) \amp = -1/2\\ \theta \amp = \frac{7\pi}{6},\ \frac{11\pi}6. \end{align*}

Thus the equations of the tangent lines, in polar, are \(\theta = 7\pi/6\) and \(\theta = 11\pi/6\text{.}\) In rectangular form, the tangent lines are \(y=\tan(7\pi/6)x\) and \(y=\tan(11\pi/6)x\text{.}\) The full limaçon can be seen in Figure 9.5.3; we zoom in on the tangent lines in Figure 9.5.5.

This is a direct application of Theorem 9.5.9. The circle is traced out on \([0,\pi]\text{,}\) leading to the integral \begin{align*} \text{ Area } \amp = \frac12\int_0^\pi \cos^2(\theta) \ d \theta\\ \amp = \frac12\int_0^\pi \frac{1+\cos(2\theta)}{2}\ d\theta\\ \amp = \frac14\big(\theta +\frac12\sin(2\theta)\big)\Bigg|_0^\pi\\ \amp = \frac14\pi. \end{align*}

Of course, we already knew the area of a circle with radius \(1/2\text{.}\) We did this example to demonstrate that the area formula is correct.

This is again a direct application of Theorem 9.5.9. \begin{align*} \text{ Area } \amp = \frac12\int_{\pi/6}^{\pi/3} (1+\cos(\theta) )^2\ d\theta\\ \amp = \frac12\int_{\pi/6}^{\pi/3} (1+2\cos(\theta) +\cos^2(\theta) )\ d\theta\\ \amp = \frac12\left(\theta+2\sin(\theta) +\frac12\theta+\frac14\sin(2\theta)\right)\Bigg|_{\pi/6}^{\pi/3}\\ \amp = \frac18\big(\pi+4\sqrt{3}-4\big) \approx 0.7587. \end{align*}

We need to find the points of intersection between these two functions. Setting them equal to each other, we find: \begin{align*} 1+\cos(\theta) \amp = 3\cos(\theta) \\ \cos(\theta) \amp =1/2\\ \theta \amp = \pm \pi/3 \end{align*}

Thus we integrate \(\frac12\big((3\cos(\theta) )^2-(1+\cos(\theta) )^2\big)\) on \([-\pi/3,\pi/3]\text{.}\) \begin{align*} \text{ Area } \amp = \frac12\int_{-\pi/3}^{\pi/3} \big((3\cos(\theta) )^2-(1+\cos(\theta) )^2\big)\ d\theta\\ \amp = \frac12\int_{-\pi/3}^{\pi/3} \big( 8\cos^2(\theta) -2\cos(\theta) -1\big)\ d\theta\\ \amp = \big(2\sin(2\theta) - 2\sin(\theta) +3\theta\big)\Bigg|_{-\pi/3}^{\pi/3}\\ \amp = 2\pi. \end{align*}

Amazingly enough, the area between these curves has a “nice” value.

We need to find the point of intersection between the two curves. Setting the two functions equal to each other, we have \begin{equation*} 2\cos(2\theta) = 1 \Rightarrow \cos(2\theta) = \frac12 \Rightarrow 2\theta = \pi/3 \Rightarrow \theta=\pi/6. \end{equation*}

In part (b) of the figure, we zoom in on the region and note that it is not really bounded between two polar curves, but rather by two polar curves, along with \(\theta=0\text{.}\) The dashed line breaks the region into its component parts. Below the dashed line, the region is defined by \(r=1\text{,}\) \(\theta=0\) and \(\theta = \pi/6\text{.}\) (Note: the dashed line lies on the line \(\theta=\pi/6\text{.}\)) Above the dashed line the region is bounded by \(r=2\cos(2\theta)\) and \(\theta =\pi/6\text{.}\) Since we have two separate regions, we find the area using two separate integrals.

Call the area below the dashed line \(A_1\) and the area above the dashed line \(A_2\text{.}\) They are determined by the following integrals: \begin{equation*} A_1 = \frac12\int_0^{\pi/6} (1)^2\ d\theta\qquad A_2 = \frac12\int_{\pi/6}^{\pi/4} \big(2\cos(2\theta)\big)^2\ d\theta. \end{equation*}

(The upper bound of the integral computing \(A_2\) is \(\pi/4\) as \(r=2\cos(2\theta)\) is at the pole when \(\theta=\pi/4\text{.}\))

We omit the integration details and let the reader verify that \(A_1 = \pi/12\) and \(A_2 = \pi/12-\sqrt{3}/8\text{;}\) the total area is \(A = \pi/6-\sqrt{3}/8\text{.}\)

With \(r=1+2\sin(t)\text{,}\) we have \(r\,' = 2\cos(t)\text{.}\) The limaçon is traced out once on \([0,2\pi]\text{,}\) giving us our bounds of integration. Applying Key Idea 9.5.19, we have \begin{align*} L \amp = \int_0^{2\pi} \sqrt{(2\cos(\theta) )^2+(1+2\sin(\theta) )^2}\ d\theta\\ \amp = \int_0^{2\pi} \sqrt{4\cos^2(\theta) +4\sin^2(\theta) +4\sin(\theta) +1}\ d\theta\\ \amp = \int_0^{2\pi} \sqrt{4\sin(\theta) +5}\ d\theta\\ \amp \approx 13.3649. \end{align*}

The final integral cannot be solved in terms of elementary functions, so we resorted to a numerical approximation. (Simpson's Rule, with \(n=4\text{,}\) approximates the value with \(13.0608\text{.}\) Using \(n=22\) gives the value above, which is accurate to 4 places after the decimal.)

We choose, as implied by the figure, to revolve the portion of the curve that lies on \([0,\pi/4]\) about the initial ray. Using Key Idea 9.5.22 and the fact that \(\fp(\theta) = -2\sin(2\theta)\text{,}\) we have \begin{align*} \text{ Surface Area } \amp = 2\pi\int_0^{\pi/4} \cos(2\theta)\sin(\theta)\sqrt{\big(-2\sin(2\theta)\big)^2+\big(\cos(2\theta)\big)^2}\ d\theta\\ \amp \approx 1.36707. \end{align*}

The integral is another that cannot be evaluated in terms of elementary functions. Simpson's Rule, with \(n=4\text{,}\) approximates the value at \(1.36751\text{.}\)