APEX Calculus

To aid in the drawing, a polar grid is provided below. To place the point \(A\text{,}\) go out 1 unit along the initial ray (putting you on the inner circle shown on the grid), then rotate counter-clockwise \(\pi/4\) radians (or \(45^\circ\)). Alternately, one can consider the rotation first: think about the ray from \(O\) that forms an angle of \(\pi/4\) with the initial ray, then move out 1 unit along this ray (again placing you on the inner circle of the grid).

To plot \(B\text{,}\) go out \(1.5\) units along the initial ray and rotate \(\pi\) radians (\(180^\circ\)).

To plot \(C\text{,}\) go out 2 units along the initial ray then rotate clockwise \(\pi/3\) radians, as the angle given is negative.

To plot \(D\text{,}\) move along the initial ray “\(-1\)” units — in other words, “back up” 1 unit, then rotate counter-clockwise by \(\pi/4\text{.}\) The results are given in Figure 9.4.4.

1. 1. We start with \(P(2,2\pi/3)\text{.}\) Using Key Idea 9.4.6, we have

      \begin{equation*} x= 2\cos(2\pi/3) = -1\qquad y = 2\sin(2\pi/3) = \sqrt{3}\text{.} \end{equation*}

      So the rectangular coordinates are \((-1,\sqrt{3}) \approx (-1,1.732)\text{.}\)

   2. The polar point \(P(-1,5\pi/4)\) is converted to rectangular with:

      \begin{equation*} x=-1\cos(5\pi/4) = \sqrt{2}/2\qquad y= -1\sin(5\pi/4) = \sqrt{2}/2\text{.} \end{equation*}

      So the rectangular coordinates are \((\sqrt{2}/2,\sqrt{2}/2) \approx (0.707,0.707)\text{.}\)

   These points are plotted in Figure 9.4.8.(a). The rectangular coordinate system is drawn lightly under the polar coordinate system so that the relationship between the two can be seen.

   

   

2. 1. To convert the rectangular point \((1,2)\) to polar coordinates, we use the Key Idea to form the following two equations:

      \begin{equation*} 1^2+2^2 = r^2 \qquad \tan(\theta) = \frac{2}{1}\text{.} \end{equation*}

      The first equation tells us that \(r=\sqrt{5}\text{.}\) Using the inverse tangent function, we find

      \begin{equation*} \tan(\theta) = 2 \Rightarrow \theta = \tan^{-1}(2) \approx 1.11\approx 63.43^\circ\text{.} \end{equation*}

      Thus polar coordinates of \((1,2)\) are \(P(\sqrt{5},1.11)\text{.}\)

   2. To convert \((-1,1)\) to polar coordinates, we form the equations

      \begin{equation*} (-1)^2+1^2=r^2 \qquad \tan(\theta) = \frac{1}{-1}\text{.} \end{equation*}

      Thus \(r=\sqrt{2}\text{.}\) We need to be careful in computing \(\theta\text{:}\) using the inverse tangent function, we have

      \begin{equation*} \tan(\theta) = -1 \Rightarrow \theta = \tan^{-1}(-1) = -\pi/4 = -45^\circ\text{.} \end{equation*}

      This is not the angle we desire. The range of \(\tan^{-1}(x)\) is \((-\pi/2,\pi/2)\text{;}\) that is, it returns angles that lie in the \(1\)st and \(4\)th quadrants. To find locations in the \(2\)nd and \(3\)rd quadrants, add \(\pi\) to the result of \(\tan^{-1}(x)\text{.}\) So \(\pi+(-\pi/4)\) puts the angle at \(3\pi/4\text{.}\) Thus the polar point is \(P(\sqrt{2},3\pi/4)\text{.}\) An alternate method is to use the angle \(\theta\) given by arctangent, but change the sign of \(r\text{.}\) Thus we could also refer to \((-1,1)\) as \(P(-\sqrt{2},-\pi/4)\text{.}\)

   These points are plotted in Figure 9.4.8.(b). The polar system is drawn lightly under the rectangular grid with rays to demonstrate the angles used.

1. The equation \(r=1.5\) describes all points that are 1.5 units from the pole; as the angle is not specified, any \(\theta\) is allowable. All points 1.5 units from the pole describes a circle of radius 1.5. We can consider the rectangular equivalent of this equation; using \(r^2=x^2+y^2\text{,}\) we see that \(1.5^2=x^2+y^2\text{,}\) which we recognize as the equation of a circle centered at \((0,0)\) with radius 1.5. This is sketched in Figure 9.4.10.

2. The equation \(\theta = \pi/4\) describes all points such that the line through them and the pole make an angle of \(\pi/4\) with the initial ray. As the radius \(r\) is not specified, it can be any value (even negative). Thus \(\theta = \pi/4\) describes the line through the pole that makes an angle of \(\pi/4 = 45^\circ\) with the initial ray. We can again consider the rectangular equivalent of this equation. Combine \(\tan(\theta) =y/x\) and \(\theta =\pi/4\text{:}\)

   \begin{equation*} \tan(\pi) /4 = y/x \Rightarrow x\tan(\pi) /4 = y \Rightarrow y = x\text{.} \end{equation*}

   This graph is also plotted in Figure 9.4.10.

   

A common question when sketching curves by plotting points is “Which points should I plot?” With rectangular equations, we often choose “easy” values — integers, then add more if needed. When plotting polar equations, start with the “common” angles — multiples of \(\pi/6\) and \(\pi/4\text{.}\) Figure 9.4.12 gives a table of just a few values of \(\theta\) in \([0,\pi]\text{.}\)

Consider the point \(P(2,0)\) determined by the first line of the table. The angle is 0 radians — we do not rotate from the initial ray — then we go out 2 units from the pole. When \(\theta=\pi/6\text{,}\) \(r = 1.866\) (actually, it is \(1+\sqrt{3}/2\)); so rotate by \(\pi/6\) radians and go out 1.866 units.

The graph shown uses more points, connected with straight lines. (The points on the graph that correspond to points in the table are signified with larger dots.) Such a sketch is likely good enough to give one an idea of what the graph looks like.

\(\theta\)	\(r=1+\cos(\theta)\)
\(0\)	\(2\)
\(\pi/6\)	\(1.86603\)
\(\pi/2\)	\(1\)
\(4\pi/3\)	\(0.5\)
\(7 \pi/4\)	\(1.70711\)

We start by making a table of \(\cos(2\theta)\) evaluated at common angles \(\theta\text{,}\) as shown in Figure 9.4.15. These points are then plotted in Figure 9.4.16.(a). This particular graph “moves” around quite a bit and one can easily forget which points should be connected to each other. To help us with this, we numbered each point in the table and on the graph.

Using more points (and the aid of technology) a smoother plot can be made as shown in Figure 9.4.16.(b). This plot is an example of a rose curve.

1. Replace \(y\) with \(r\sin(\theta)\) and replace \(x\) with \(r\cos(\theta)\text{,}\) giving:

   \begin{align*} y \amp =x^2\\ r\sin(\theta) \amp = r^2\cos^2(\theta)\\ \frac{\sin(\theta) }{\cos^2(\theta) } \amp = r \end{align*}

   We have found that \(r=\sin(\theta) /\cos^2(\theta) = \tan(\theta) \sec(\theta)\text{.}\) The domain of this polar function is \((-\pi/2,\pi/2)\text{;}\) plot a few points to see how the familiar parabola is traced out by the polar equation.

2. We again replace \(x\) and \(y\) using the standard identities and work to solve for \(r\text{:}\)

   \begin{align*} xy \amp = 1\\ r\cos(\theta) \cdot r\sin(\theta) \amp = 1\\ r^2 \amp = \frac{1}{\cos(\theta) \sin(\theta) }\\ r \amp = \frac{1}{\sqrt{\cos(\theta) \sin(\theta) }} \end{align*}

   This function is valid only when the product of \(\cos(\theta) \sin(\theta)\) is positive. This occurs in the first and third quadrants, meaning the domain of this polar function is \((0,\pi/2) \cup (\pi,3\pi/2)\text{.}\) We can rewrite the original rectangular equation \(xy=1\) as \(y=1/x\text{.}\) This is graphed in Figure 9.4.18; note how it only exists in the first and third quadrants.

   

3. There is no set way to convert from polar to rectangular; in general, we look to form the products \(r\cos(\theta)\) and \(r\sin(\theta)\text{,}\) and then replace these with \(x\) and \(y\text{,}\) respectively. We start in this problem by multiplying both sides by \(\sin(\theta) -\cos(\theta)\text{:}\)

   \begin{align*} r \amp = \frac{2}{\sin(\theta) -\cos(\theta) }\\ r(\sin(\theta) -\cos(\theta) ) \amp = 2\\ r\sin(\theta) -r\cos(\theta) \amp = 2. \qquad \text{ Now replace with \(y\) and \(x\): }\\ y-x \amp = 2\\ y \amp = x+2\text{.} \end{align*}

   The original polar equation, \(r=2/(\sin(\theta) -\cos(\theta) )\) does not easily reveal that its graph is simply a line. However, our conversion shows that it is. The upcoming gallery of polar curves gives the general equations of lines in polar form.

4. By multiplying both sides by \(r\text{,}\) we obtain both an \(r^2\) term and an \(r\cos(\theta)\) term, which we replace with \(x^2+y^2\) and \(x\text{,}\) respectively.

   \begin{align*} r \amp =2\cos(\theta)\\ r^2 \amp = 2r\cos(\theta)\\ x^2+y^2 \amp = 2x.\\ \end{align*}

   We recognize this as a circle; by completing the square we can find its radius and center.

   \begin{align*} x^2-2x+y^2 \amp = 0\\ (x-1)^2 + y^2 \amp =1\text{.} \end{align*}

   The circle is centered at \((1,0)\) and has radius 1. The upcoming gallery of polar curves gives the equations of some circles in polar form; circles with arbitrary centers have a complicated polar equation that we do not consider here.

As technology is generally readily available, it is usually a good idea to start with a graph. We have graphed the two functions in Figure 9.4.25.(a); to better discern the intersection points, Figure 9.4.25.(b) zooms in around the origin.

We start by setting the two functions equal to each other and solving for \(\theta\text{:}\)

(There are, of course, infinite solutions to the equation \(\cos(\theta) =-1/2\text{;}\) as the limaçon is traced out once on \([0,2\pi]\text{,}\) we restrict our solutions to this interval.)

We need to analyze this solution. When \(\theta = 2\pi/3\) we obtain the point of intersection that lies in the \(4\)th quadrant. When \(\theta = 4\pi/3\text{,}\) we get the point of intersection that lies in the second quadrant. There is more to say about this second intersection point, however. The circle defined by \(r=\cos(\theta)\) is traced out once on \([0,\pi]\text{,}\) meaning that this point of intersection occurs while tracing out the circle a second time. It seems strange to pass by the point once and then recognize it as a point of intersection only when arriving there a “second time.” The first time the circle arrives at this point is when \(\theta = \pi/3\text{.}\) It is key to understand that these two points are the same: \((\cos(\pi/3),\pi/3)\) and \((\cos(4\pi/3),4\pi/3)\text{.}\)

To summarize what we have done so far, we have found two points of intersection: when \(\theta=2\pi/3\) and when \(\theta=4\pi/3\text{.}\) When referencing the circle \(r=\cos(\theta)\text{,}\) the latter point is better referenced as when \(\theta=\pi/3\text{.}\)

There is yet another point of intersection: the pole (or, the origin). We did not recognize this intersection point using our work above as each graph arrives at the pole at a different \(\theta\) value.

A graph intersects the pole when \(r=0\text{.}\) Considering the circle \(r=\cos(\theta)\text{,}\) \(r=0\) when \(\theta = \pi/2\) (and odd multiples thereof, as the circle is repeatedly traced). The limaçon intersects the pole when \(1+3\cos(\theta) =0\text{;}\) this occurs when \(\cos(\theta) = -1/3\text{,}\) or for \(\theta = \cos^{-1}(-1/3)\text{.}\) This is a nonstandard angle, approximately \(\theta = 1.9106 = 109.47^\circ\text{.}\) The limaçon intersects the pole twice in \([0,2\pi]\text{;}\) the other angle at which the limaçon is at the pole is the reflection of the first angle across the \(x\)-axis. That is, \(\theta = 4.3726 = 250.53^\circ\text{.}\)