4. Parametric Equations and Polar Coordinates

Partition the interval \(I=[-3,3]\) into \(t_0=-3, t_1=-2, t_2=-1, \ldots, t_7=3\) and evaluate \((x(t_i), y(t_i))\) and plot points. The resulting curve is given below. The orientation is clockwise.

The initial point is \((9,-2)\) and final point is \((9,4)\). It seems as though the curve is a parabola. To find a Cartesian equation, start with \begin{align*} x&= t^2\\ y&= t+1 \end{align*} and from the \(y\)-equation we get \(t=y-1\) and thus \(x = (y-1)^2\).

Use the estimate \(\sqrt{2}\approx 1.4\). Evaluate the parametric equations along convenient \(\theta\) values:

The curve seems to be a portion of an ellipse. Recall that an ellipse centered at \((x_0, y_0)\) with horizontal radius \(a\) and vertical radius \(b\) has Cartesian equation \[ \frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1 \] Thus, we suspect that the Cartesian equation of the curve is \[ \frac{x^2}{4} + \frac{y^2}{9} = 1 \] To eliminate the parameter start with \begin{align*} x &= 2\cos(\theta)\\ y &= 3\sin(\theta) \end{align*} and then \begin{align*} \frac{x}{2} &= \cos(\theta)\\ \frac{y}{3} &= \sin(\theta) \end{align*} Now \(\cos^2(\theta) + \sin^2(\theta)= 1\) and thus \[ \left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = \cos^2(\theta) + \sin^2(\theta) \] and thus \[ \frac{x^2}{4} + \frac{y^2}{9} = 1. \]

First determine a CCW orientation and then change the signs accordingly. The ellipse is:

The general equations of the parametrization of an ellipse in this case is \begin{align*} x(t) &= -1 + 6\cos(t)\\ y(t) &= 3 + 1\sin(t) \end{align*} We need the interval to be \(I=[-\tfrac{\pi}{2}, \tfrac{5\pi}{2}]\). This gives a CCW orientation. To change the orientation, we can change the sign in front of the \(\cos(t)\) term: \begin{align*} x(t) &= -1 - 6\cos(t)\\ y(t) &= 3 + 1\sin(t) \end{align*}

A parametrization is \begin{align*} x &= x(t) = t\\ y &= y(t) = 2t^3 \end{align*} with \(-1\leq t\leq 2\). We can find a right-to-left parametrization by changing all \(t\)'s to \(-t\) and changing the interval to \(-b\leq t\leq -a\). In this case, a right-to-left parametrization is then \begin{align*} x &= x(t) = -t\\ y &= y(t) = 2(-t)^3 = -2t^3 \end{align*} with interval \(-2\leq t\leq 1\).

The slope of the line is \[ m = \frac{11-2}{\tfrac{2}{7} + 1} = \frac{9}{9/7} = 7 \] The equation of the line is then \(y = 7x + 9\). A parametrization of the entire line is \begin{align*} x &= t\\ y &= 7t+9 \end{align*} and since we only want the line segment where \(-1\leq x\leq \tfrac{2}{7}\) then the interval is \(I=[-1,\tfrac{2}{7}]\).

We could solve for \(y\) in terms of \(x\): \[ y = -\sqrt{9\left(1-\frac{x^2}{4}\right)} \] However, in some cases we may only have a parametric equation for a curve and even if we had a Cartesian equation we may not be able to solve for \(y\) (could use implicit differentiation). Use instead a parametrization \begin{align*} x(t) &= 2\cos(t) \\ y(t) &= 3\sin(t)\\ 0 &\leq t \leq 2\pi \end{align*} We now need a way to find the slope of the tangent line in terms of the parametric equations. We do know that \(y=f(x)\) near \(P\) and thus \(y(t) = f(x(t))\). Therefore, by the chain rule \[ y'(t) = f'(x(t)) \cdot x'(t) \] and therefore \[ f'(x(t)) = \frac{y'(t)}{x'(t)} \] Because \(f'(x) = \frac{dy}{dx}\) this is sometimes written as \[ \frac{dy}{dx} = \frac{y'(t)}{x'(t)} \] Hence, in this case \[ \frac{dy}{dx} = \frac{3\cos(t)}{-2\sin(t)} \] At what \(t\) value do we evaluate? It has to correspond to the \(t\) value where the parametrization passes through the point \(P\). The value \(t^*\) where the parametrization passes through the point \(P\) occurs when \[ 2\cos(t^*) = 1,\qquad 3\sin(t^*) = -3\frac{\sqrt{3}}{2}. \] From \(\cos(t^*) = 1/2\) either \(t^* = \pi/3\) or \(t^* = 5\pi/3\). In this case, need to take \(t^* = 5\pi/3\). Hence, we obtain \[ m = \frac{3\cos(t)}{-2\sin(t)} \Big|_{t=t^*} = \frac{3 (1/2)}{-2 (-\sqrt{3}/2)} = \sqrt{3}/2 \] Therefore, equation of line is \[ y = y_0 + m(x-x_0) = -\frac{3\sqrt{3}}{2} + \sqrt{3}/2 ( x - 1) \] which simplifies to \[ y = \sqrt{3}/2 x - 2\sqrt{3}. \]

The curve is shown in Figure 4.7.

1. We compute \[ m = \frac{y'(t)}{x'(t)} = \frac{3t^2 - 3}{2t} \] The curve passes through the point \(P=(4,1)\) when \(t^2 = 4\), so \(t=\pm 2\) but need \(y(t) = t^3-3t=1\). At \(t=2\) get \(y(2) = 8-6+3=5\) but at \(y(-2) = -2+3=1\). Hence, \(t\) value is \(t^* = -2\). Hence, \[ m = \frac{3(-2)^2-3}{2(-2)} = -\frac{9}{4}. \] Equation of line is \[ y = -\frac{9}{4}(x-4) + 1 = -\frac{9}{4}x + 10 \]
2. Tangent line is horizontal when \(y'(t) = 0\) which occurs at \(t\) values \(t=\pm 1\). The points are therefore \((x(1),y(1)) = (1,1)\) and \((x(-1),y(-1)) = (1,5)\).
3. Tangent line is vertical when \(x'(t) = 0\) which occurs at \(t=0\). The point is \((x(0),y(0)) = (0,0)\).
4. To find a Cartesian equation notice that \[ (y-3)^2 = t^6 -6t^4 + 9t^2 = (t^2)^3 - 6(t^2)^2 + 9t^2 \] and thus \((y-3)^2 = x^3 - 6x^2 + x\) is a Cartesian equation.

1. \(r\sin(\theta) = r^2\cos^2(\theta)\) which can be factored as \[ r (r\cos^2(\theta) - \sin(\theta)) = 0 \] Now \(r=0\) represents only the origin. The other points on \(y=x^2\) therefore satisfy \(r\cos^2(\theta) - \sin(\theta)=0\) which can be written as \[ r = \tan(\theta)\sec(\theta),  \theta\neq \frac{\pi}{2} \]
2. Since \(x^2+y^2=r^2\) then \(r^2 = 7\) or \(r=\sqrt{7}\).
3. Expanding gives \(x^2+y^2-6y+9=9\) and then \(r^2 - 6r\sin(\theta) = 0\) or \(r(r-6\sin(\theta))=0\). Use \(r=6\sin(\theta)\) because when \(\theta=\pi\) get also \(r=0\).

1. \(x-4=0\) or \(x=4\) (vertical line)
2. \(x^2+y^2 = 4x\) and completing square gives \((x-2)^2+y^2=4\) (circle at \((2,0)\) of radius \(4\))
3. Write as \(2r\cos(\theta)-r\sin(\theta)=4\) and then \(2x-y=4\) or \(y=2x-4\) (line)
4. Get \(r=2\frac{x}{r} - \frac{y}{r}\) and thus \(r^2 = 2x - y\) or \(x^2-2x + y^2 + y = 0\) and complete square to get \((x-1)^2+(y+\tfrac{1}{2})^2 = \frac{5}{4}\).

Create a table of \((r,\theta)\) coordinates by varying \(\theta\) at step-size \(\tfrac{\pi}{4}\):

This curve is called a cardioid. Since \(r(\theta) = 1+\sin(\theta)\), we can obtain a parametrization for this curve as follows: \begin{align*} x(\theta) &= r(\theta)\cos(\theta) = (1+\sin(\theta))\cos(\theta)\\ y(\theta) &= r(\theta)\sin(\theta) = (1+\sin(\theta))\sin(\theta)\\ 0 &\leq\theta\leq 2\pi \end{align*} In general, if a curve is expressed in polar coordinates as \(r=r(\theta)\) for \(a\leq \theta\leq b\) then a parametrization for the curve is: \begin{align*} x(\theta) &= r(\theta)\cos(\theta)\\ y(\theta) &= r(\theta)\sin(\theta)\\ a &\leq\theta\leq b \end{align*}

Recall that \(m = \frac{dy}{dx} = \frac{y'(t)}{x'(t)}\) if given a parametrization \(x=x(t)\) and \(y=y(t)\). Here we have \begin{align*} x(\theta) &= (1+\sin(\theta))\cos(\theta)\\ y(\theta) &= (1+\sin(\theta))\sin(\theta) \end{align*} We get \[ m = \frac{\cos(\theta)\sin(\theta)+(1+\sin(\theta))\cos(\theta)}{\cos^2(\theta) - (1+\sin(\theta))\sin(\theta)} \] and evaluating at \(\theta=\pi/4\) we obtain \(m=-(1+\sqrt{2})\). The point on the curve at \(\theta=\pi/4\) is \((x_0,y_0) = (1+\sqrt{2}/2)\sqrt{2}/{2}, (1+\sqrt{2}/2)\sqrt{2}/{2})\) and thus the equation of the line is \(y=y_0 + m(x-x_0)\).

A complete revolution of the cardioid requires \(0\leq \theta\leq 2\pi\). Now \(r(\theta) = 1+\sin(\theta)\) and then \begin{align*} x(\theta) &= (1+\sin(\theta))\cos(\theta)\\ y(\theta) &= (1+\sin(\theta))\sin(\theta) \end{align*} Now \(r'(\theta) = \cos(\theta)\) and thus \begin{align*} L &= \int_{0}^{2\pi} \sqrt{ (r(\theta))^2 + (r'(\theta))^2}\,d\theta\\ & = \int_0^{2\pi} \sqrt{2+2\sin(\theta)}\,d\theta \end{align*}

Compute \begin{align*} L = \int_0^{\sqrt{5}} \sqrt{ r^2 + (r')^2}\,d\theta &=\int_0^{\sqrt{5}} \sqrt{ \theta^4 + 4\theta^2}\,d\theta\\ &= \int_0^{\sqrt{5}} \theta\sqrt{ \theta^2 + 4}\,d\theta\\ &= \frac{19}{3} \end{align*}

Apply the formula: \begin{align*} A &= \int_0^{\pi} \frac{1}{2}(1+\sin(\theta))^2\,d\theta\\ &= \int_0^{\pi} \frac{1}{2}(1+2\sin(\theta)+\sin^2(\theta))\,d\theta\\ &= \frac{\pi}{2} + 2 + \frac{1}{2}\int_0^\pi \frac{1-\cos(2\theta)}{2}\,d\theta\\ &= \frac{\pi}{2} + 2 + \frac{\pi}{4} = 2+3\frac{\pi}{4} \end{align*}

1. Evaluating \(x(\theta)=r(\theta)\cos(\theta)\) and \(y(\theta)=r(\theta)\sin(\theta)\) from \(\theta=0\) to \(\theta=2\pi\) at step-size of \(\pi/4\), we obtain the following graph:

   

2. \begin{align*} A &= \int_0^{2\pi} \frac{1}{2}r(\theta)^2\,d\theta\\[2ex] &= \int_0^{2\pi} \frac{1}{2}(2-\cos(\theta))^2\, d\theta\\[2ex] &= \frac{1}{2}\left(\int_0^{2\pi}4\,d\theta - 4\int_0^{2\pi} \cos(\theta)\,d\theta +\int_0^{2\pi}\cos^2(\theta)\,d\theta \right)\\[2ex] &= \frac{9\pi}{2} \end{align*}
3. \begin{align*} L &= \int_0^{2\pi} \sqrt{r^2 + (r')^2}\,d\theta\\[2ex] &= \int_0^{2\pi} \sqrt{(2-\cos(\theta))^2 + \sin^2(\theta)}\,d\theta\\[2ex] &= \int_0^{2\pi} \sqrt{5-4\cos(\theta)}\,d\theta \end{align*}
4. Recall that the Trapezoidal rule is for estimating an integral \[ \int_a^b f(x)\,dx. \] Partitioning \([a,b]\) into \(n\) equal subintervals, each subinterval has length \(\Delta x = \frac{b-a}{n}\). The Trapezoidal rule is then \[ T = \frac{\Delta x}{2}\left[ f(x_0) + 2f(x_1) + 2f(x_2)+\cdots+2f(x_{n-1}) + f(x_n)\right] \] where \(x_0,x_1,\ldots,x_n\) are the points obtained after subdividing the interval \([a,b]\). We wish to apply the Trapezoidal rule to \[ L = \int_0^{2\pi} \sqrt{5-4\cos(\theta)}\,d\theta \] Hence, here \(f(\theta) = \sqrt{5-4\cos(\theta)}\), \(n=4\), \(\Delta\theta=\frac{2\pi}{n} = \frac{\pi}{2}\), and the points to evaluate \(f\) are \(\theta_0=0\), \(\theta_1=\frac{\pi}{2}\), \(\theta_2=\pi\), \(\theta_3=3\pi/2\), and \(\theta_4=2\pi\). Hence, \begin{align*} T &= \frac{\Delta\theta}{2}\left[f(\theta_0) + 2f(\theta_1) + 2f(\theta_2)+2f(\theta_{3}) + f(\theta_4) \right]\\[2ex] &= \frac{\pi}{4}\left[\sqrt{1}+2\sqrt{5}+2\sqrt{9}+2\sqrt{5}+1\right]\\[2ex] &= \frac{\pi}{4}(8+4\sqrt{5})\approx L \end{align*}
5. Recall that given a parametrized curve \(x=x(t)\) and \(y=y(t)\), the slope of the line tangent to the curve at \(t\) is \[ m= \frac{y'(t)}{x'(t)} \] The tangent line is vertical when \(x'(t)=0\). Here, \begin{align*} x(\theta)&= (2-\cos(\theta))\cos(\theta)=2\cos(\theta)-\cos^2(\theta)\\[2ex] y(\theta)&=(2-\cos(\theta))\sin(\theta)=2\sin(\theta)-\cos(\theta)\sin(\theta) \end{align*} Then \begin{align*} x'(\theta) &= -2\sin(\theta)+2\cos(\theta)\sin(\theta)\\ &= 2\sin(\theta)(\cos(\theta)-1) \end{align*} In the interval \(0\leq\theta\leq 2\pi\), \(x'(\theta) = 0\) when \(\theta=0\) and \(\theta=\pi\). Hence, the points are \((x(0),y(0))=(1,0)\) and \((x(\pi),y(\pi)) = (-3,0)\). This agrees with the sketch of the curve.

1. Evaluating \(x(\theta)=r(\theta)\cos(\theta)\) and \(y(\theta)=r(\theta)\sin(\theta)\) from \(\theta=0\) to \(\theta=2\pi\) at step-size of \(\pi/16\), we obtain the curve shown in Figure 4.14.

   

2. Since \(r'(\theta) = 2\cos(\theta)\), the arc length of the curve along for \(0\leq\theta\leq \pi/2\) is \begin{align*} L &= \int_0^{\pi/2} \sqrt{r(\theta) + (r'(\theta))^2}\, d\theta \\ &= \int_0^{\pi/2} \sqrt{\sin^2(2\theta) + 4\cos^2(\theta)}\, d\theta \end{align*}
3. Let \(f(\theta) = \sqrt{\sin^2(2\theta) + 4\cos^2(2\theta)}\). Applying Simpson's rule, we obtain that \(\Delta \theta = \frac{\pi/2}{4} = \pi/8\) and then the grid points are \(\theta_0 = 0\), \(\theta_1 = \pi/8\), \(\theta_2 = \pi/4\), \(\theta_3 = 3\pi/8\), and \(\theta_4 = \pi/2\). The symbolic form of Simpson's rule is then \[ S = \frac{\Delta\theta}{3}\big[f(\theta_0) + 4f(\theta_1)+2f(\theta_2)+4f(\theta_3) + f(\theta_4)\big] \]
4. The area enclosed by the curve is \begin{align*} A &= \int_0^{2\pi} \frac{1}{2} r^2(\theta)\, d\theta\\ &= \frac{1}{2}\int_0^{2\pi} \sin^2(2\theta)\, d\theta\\ &= \frac{1}{4}\int_0^{2\pi} (1-\sin(4\theta))\, d\theta\\ &= \frac{1}{4}(\theta + \frac{1}{4}\cos(4\theta))\Big|_{0}^{2\pi}\\ &= \frac{1}{4}(2\pi + \frac{1}{4}\cos(8\pi) - \frac{1}{4}\cos(0))\\ &= \pi/2 \end{align*}