We start by computing the derivative in rectangular coordinates. Recall that \begin{align*} x(\theta ) &= \left (1+2\sin (\theta )\right )\cdot \cos (\theta ) = \cos (\theta )+2\sin (\theta )\cos (\theta )\\ y(\theta ) &= \left (1+2\sin (\theta )\right )\cdot \sin (\theta ) = \sin (\theta )+2\sin ^2(\theta ) \end{align*}

Consequently, \begin{align*} x'(\theta ) &= \answer [given]{-\sin (\theta ) + 2\cos ^2(\theta )-2\sin ^2(\theta )} \\ y'(\theta ) &= \answer [given]{\cos (\theta ) + 4\sin (\theta )\cos (\theta )} \end{align*}

and therefore

When \(\theta =\pi /4\), \begin{align*} x(\theta ) &= \answer [given]{1+\sqrt {2}/2}\\ y(\theta ) &=\answer [given]{1+\sqrt {2}/2}\\ \frac {dy}{dx} &=\answer [given]{-2\sqrt {2}-1} \end{align*}

Thus the equation of the line (in polarrectangular coordinates) tangent to the limaçon at \(\theta =\pi /4\) is

To find these points where the tangent lines is horizontal, we seek points where \(\frac {dy}{dx} = \answer [given]{0}\). From earlier, we discovered so now we must find where the numerator is \(0\). Since the numerator is equal to and since a product is zero when botheither of the terms is zero, the numerator is zero when

• \(\cos (\theta )=0\) or
• \(1+4\sin (\theta )=0\).

Let’s examine the case of \(\cos (\theta ) = 0\) first, and restrict to the situation where \(\theta \) is between \(0\) and \(2\pi \). If \(\theta \) is in the interval \([0,\pi ]\), then \(\cos (\theta )=0\) when \(\theta =\answer [given]{\pi /2}\). If \(\theta \) is in the interval \([\pi ,2\pi ]\), then \(\cos (\theta )=0\) when \(\theta =\answer [given]{3\pi /2}\).

The corresponding points in rectangular coordinates are \begin{align*} x(\pi /2) &= \left (1+2\sin (\pi /2)\right )\cdot \cos (\pi /2)\\ &= 0,\\ y(\pi /2) &= \left (1+2\sin (\pi /2)\right )\cdot \sin (\pi /2)\\ &= 3, \end{align*}

meaning \((x,y) = (0,3)\), and \begin{align*} x(3\pi /2) &= \left (1+2\sin (3\pi /2)\right )\cdot \cos (3\pi /2)\\ &= 0,\\ y(3\pi /2) &= \left (1+2\sin (3\pi /2)\right )\cdot \sin (3\pi /2)\\ &= 1, \end{align*}

meaning \((x,y) = (0,1)\).

But the numerator was the product of \(\cos (\theta )\) and another term, namely \(1+4\sin (\theta )\) So the numerator is also zero when \begin{align*} 1+4\sin (\theta )&=0\\ 4\sin (\theta )&=-1\\ \sin (\theta )&=\frac {-1}{4}. \end{align*}

At this point we have a choice to make: We can either apply the arcsin function and solve for \(\theta \), or we can work with \(\sin (\theta )\) directly. Perhaps the easier route is to work with \(\sin (\theta )\) directly. To explore \(\sin (\theta )\), we may choose an algebraic method employing the Pythagorean identity, or a geometric method looking at the unit circle with the Pythagorean theorem. Let’s take the geometric route to compute \(\cos (\theta )\).

Hence the derivative is zerois oneis undefined at \begin{align*} x(\theta ) &= \left (1+2\sin (\theta )\right )\cdot \cos (\theta )\\ &= \left (1+2(-1/4)\right )\cdot \sqrt {15}/4\\ &= \sqrt {15}/8,\\ y(\theta ) &= \left (1+2\sin (\theta )\right )\cdot \sin (\theta )\\ &= \left (1+2(-1/4)\right )\cdot (-1/4)\\ &= -1/8, \end{align*}

meaning \((x,y) = (\sqrt {15}/8, -1/8)\), and \begin{align*} x(\theta ) &= \left (1+2\sin (\theta )\right )\cdot \cos (\theta )\\ &= \left (1+2(-1/4)\right )\cdot (-\sqrt {15}/4)\\ &= -\sqrt {15}/8,\\ y(\theta ) &= \left (1+2\sin (\theta )\right )\cdot \sin (\theta )\\ &= \left (1+2(-1/4)\right )\cdot (-1/4)\\ &= -1/8, \end{align*}

meaning \((x,y) = (-\sqrt {15}/8, -1/8)\).

In summary, the graph of \(r(\theta ) = 1+2\sin (\theta )\) on the interval \([0,2\pi ]\) has horizontal tangent lines at the points

• \((x,y) = (0,\answer [given]{3})\),
• \((x,y) = (0,1)\),
• \((x,y) = (\sqrt {15}/8, -1/8)\), and
• \((x,y) = (-\sqrt {15}/8, -1/8)\).

You can confirm this by looking at the graph below:

To find the vertical tangent lines, we seek points where the denominator of is equal to zero. To start, we will convert \(\cos ^2(\theta )\) to \(1-\answer [given]{\sin ^2(\theta )}\), and express the denominator as a quadratic expression in \(\sin (\theta )\): \begin{align*} -\sin (\theta ) + 2\cos ^2(\theta )-2\sin ^2(\theta ) & = -\sin (\theta ) + 2(1-\sin ^2(\theta ))-2\sin ^2(\theta ) \\ &= \answer [given]{-4} \sin ^2(\theta ) + \answer [given]{-1}\sin (\theta ) + \answer [given]{2}. \end{align*}

We then set this equal to zero and note that the equation is quadratic equation in the variable \(\sin (\theta )\). The quadratic formula gives that \begin{align*} \sin (\theta ) &= \frac {-1\pm \sqrt {\answer [given]{33}}}{8}. \end{align*}

Now we can find cosine using either geometrically via the unit circle or algebraically with the Pythagorean identity. This time, let’s take the more algebraic route with the Pythagorean identity. When \(\sin (\theta ) =\frac {-1+\sqrt {33}}{8}\), then \begin{align*} \cos ^2(\theta ) + \sin ^2(\theta ) &= 1\\ \cos ^2(\theta ) + \left (\frac {-1+\sqrt {33}}{8}\right ) &= 1. \end{align*}

So

and when \(\sin (\theta ) =\frac {-1-\sqrt {33}}{8}\), \begin{align*} \cos ^2(\theta ) + \sin ^2(\theta ) &= 1\\ \cos ^2(\theta ) + \left (\frac {-1-\sqrt {33}}{8}\right ) &= 1. \end{align*}

So

So now we know that the graph of \(r(\theta ) =1+2\sin (\theta )\) on \([0,2\pi ]\) has vertical tangent lines at the four points Recalling that \begin{align*} x(\theta ) &= \left (1+2\sin (\theta )\right )\cdot \cos (\theta )\\ y(\theta ) &= \left (1+2\sin (\theta )\right )\cdot \sin (\theta ) \end{align*}

we see that the graph of \(r(\theta ) =1+2\sin (\theta )\) on \([0,2\pi ]\) has vertical tangent lines at

You can confirm this visually by looking at the graph below:

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

Thus the equations of the tangent lines, in polarrectangular coordinates, are \begin{align*} \theta &=7\pi /6 \theta \\ \theta &= \answer [given]{11}\pi /6. \end{align*}

In rectangular form, the tangent lines are \(y=\tan (7\pi /6)x\) and \(y=\answer [given]{\tan (11\pi /6)}x\). You can confirm this by looking at the graph below: