Solution. We follow the same procedure here as we have time and time again when asked to graph anything new – choose values for \(t\), then plot and connect the corresponding points.

Since we are told \(t \geq -2\), we start there and as we plot successive points, we draw an arrow to indicate the direction of the path for increasing values of \(t\).

□

Solution.

1. To get a feel for the curve described by the system \(\left \{ x = t^3, \, y = 2t^2 \right .\) we first sketch the graphs of \(x = t^3\) and \(y = 2t^2\) over the interval \([-1,1]\) below on the left in the middle, respectively.

   We note that as \(t\) takes on values in the interval \([-1,1]\), \(x = t^3\) ranges between \(-1\) and \(1\), and \(y = 2t^2\) ranges between \(0\) and \(2\). This means that all of the action is happening on a portion of the plane, namely \(\left \{ (x,y) \, | \, -1 \leq x \leq 1, \; 0 \leq y \leq 2 \right \}\).

   Next, we plot a few points to get a sense of the position and orientation of the curve. Certainly, \(t=-1\) and \(t=1\) are good values to pick since these are the extreme values of \(t\). We also choose \(t=0\), since that corresponds to a (local) minimum ^[4] (You should review Definitions incdeccnstdefn and localmaxmindefn if you’ve forgotten what ‘increasing’, ‘decreasing’ and ‘local minimum’ mean.) on the graph of \(y = 2t^2\). Plugging in \(t = -1\) gives the point \((-1,2)\), \(t = 0\) gives \((0,0)\) and \(t=1\) gives \((1,2)\).

   More generally, we see that \(x = t^3\) is increasing over the entire interval \([-1,1]\) whereas \(y = 2t^2\) is decreasing over the interval \([-1,0]\) and then increasing over \([0,1]\).

   Geometrically, this means that in order to trace out the path described by the parametric equations, we start at \((-1,2)\) (where \(t=-1\)), then move to the right (since \(x\) is increasing) and down (since \(y\) is decreasing) to \((0,0)\) (where \(t = 0\)).

   We continue to move to the right (since \(x\) is still increasing) but now move upwards (since \(y\) is now increasing) until we reach \((1,2)\) (where \(t=1\)).

   Finally, to get a good sense of the shape of the curve, we eliminate the parameter. Solving \(x = t^3\) for \(t\), we get \(t = \sqrt [3]{x}\). Substituting this into \(y = 2t^2\) gives \(y = 2(\sqrt [3]{x})^2 = 2x^{2/3}\). Our experience in Section PowerFunctions yields the graph of our final answer below on the right.

   \(x = t^3\), \(-1 \leq t \leq 1\)	\(y = 2t^2\), \(-1 \leq t \leq 1\)	\(\left \{ x = t^3, \, y = 2t^2 \right .\), \(-1 \leq t \leq 1\)

2. For the system \(\left \{ x = 2e^{-t}, \, y=e^{-2t} \right .\) for \(t \geq 0\), we proceed as in the previous example and graph \(x = 2e^{-t}\) and \(y = e^{-2t}\) over the interval \([0, \infty )\) below on the left and in the middle, respectively.

   We find that the range of \(x\) in this case is \((0,2]\) and the range of \(y\) is \((0,1]\), so our graph will reside in a portion of Quadrant I: \(\left \{ (x,y) \, | \, 0 < x \leq 2, \; 0 < y \leq 1 \right \}\).

   Next, we plug in some friendly values of \(t\) to get a sense of the orientation of the curve. Since \(t\) lies in the exponent here, ‘friendly’ values of \(t\) involve natural logarithms. Starting with \(t=\ln (1) = 0\) we get ^[5] (The reader is encouraged to review Sections LogarithmicFunctions and PropertiesofLogarithms as needed.) \((2,1)\), for \(t = \ln (2)\) we get \(\left (1,\frac {1}{4}\right )\) and for \(t = \ln (3)\) we get \(\left (\frac {2}{3}, \frac {1}{9}\right )\).

   Since \(t\) is ranging over the unbounded interval \([0, \infty )\), we take the time to analyze the end behavior of both \(x\) and \(y\). We find \(\ds {\lim _{t \rightarrow \infty } x(t) = \lim _{t \rightarrow \infty } 2e^{-t} = 0}\) and \(\ds {\lim _{t \rightarrow \infty } y(t) = \lim _{t \rightarrow \infty } e^{-2t} = 0}\) This means the graph of \(\left \{ x = 2e^{-t}, \, y=e^{-2t} \right .\) approaches the point \((0,0)\).

   Since both \(x = 2e^{-t}\) and \(y = e^{-2t}\) are always decreasing for \(t \geq 0\), we know that our final graph will start at \((2,1)\) (where \(t=0\)), and move consistently to the left (since \(x\) is decreasing) and down (since \(y\) is decreasing) to approach the origin.

   To eliminate the parameter, one way to proceed is to solve \(x = 2e^{-t}\) for \(t\) to get \(t = -\ln \left (\frac {x}{2}\right )\). Substituting this for \(t\) in \(y = e^{-2t}\) gives \(y = e^{-2(-\ln (x/2))} = e^{2\ln (x/2)} = e^{\ln (x/2)^2} = \left (\frac {x}{2}\right )^2 = \frac {x^2}{4}\).

   Alternatively, we could recognize that \(y = e^{-2t} = \left (e^{-t}\right )^2\), and since \(x = 2e^{-t}\) means \(e^{-t} = \frac {x}{2}\), we get \(y = \left (\frac {x}{2}\right )^2 = \frac {x^2}{4}\) this way as well.

   Either way, the graph of \(\left \{ x = 2e^{-t}, \, y=e^{-2t} \right .\) for \(t \geq 0\) is a portion of the parabola \(y = \frac {x^2}{4}\) which starts at the point \((2,1)\) and heads towards, but never reaches, ^[6] (Note the open circle at the origin. See our discussion about holes in graphs in Example volumeex1 in Section FunctionsandtheirRepresentations.) \((0,0)\) as seen below on the right.

   \(x =2e^{-t}\), \(t \geq 0\)	\(y = e^{-2t}\), \(t \geq 0\)	\(\left \{ x = 2e^{-t}, \, y=e^{-2t} \right .\), \(t \geq 0\)

3. For the system \(\left \{ x = \sin (t), \, y = \csc (t) \right .\) for \(0 < t < \pi \), we start by graphing \(x = \sin (t)\) and \(y = \csc (t)\) over the interval \((0,\pi )\) below on the left and in the middle, respectively.

   We find that the range of \(x\) is \((0,1]\) while the range of \(y\) is \([1,\infty )\) which means our graph will lie in the first quadrant.

   Plotting a few friendly points, we see that \(t = \frac {\pi }{6}\) gives the point \(\left (\frac {1}{2}, 2\right )\), \(t = \frac {\pi }{2}\) gives \((1,1)\) and \(t = \frac {5\pi }{6}\) returns us to \(\left ( \frac {1}{2}, 2\right )\).

   Since \(t=0\) and \(t=\pi \) aren’t included in the domain for \(t\), (because \(y = \csc (t)\) is undefined at these \(t\)-values), we analyze the behavior of the system as \(t\) approaches \(0\) and \(\pi \).

   We have \(\ds {\lim _{t \rightarrow 0^{+}} x(t) = \lim _{t \rightarrow 0^{+}} \sin (t) = 0}\) and \(\ds {\lim _{t \rightarrow \pi ^{-}} x(t) = \lim _{t \rightarrow \pi ^{-}} \sin (t) = 0}\). Also, \(\ds {\lim _{t \rightarrow 0^{+}} y(t) = \lim _{t \rightarrow 0^{+}} \csc (t) = \infty }\) and \(\ds {\lim _{t \rightarrow \pi ^{-}} y(t) = \lim _{t \rightarrow \pi ^{-}} \csc (t) = \infty }\). Piecing all of this information together, we get that for \(t\) near \(0\), we have points with very small positive \(x\)-values, but very large positive \(y\)-values.

   As \(t\) ranges through the interval \(\left (0, \frac {\pi }{2}\right ]\), \(x = \sin (t)\) is increasing and \(y = \csc (t)\) is decreasing. This means that we are moving to the right and downwards, through \(\left ( \frac {1}{2}, 2\right )\) when \(t = \frac {\pi }{6}\) to \((1,1)\) when \(t = \frac {\pi }{2}\). Once \(t = \frac {\pi }{2}\), the orientation reverses, and we start to head to the left, since \(x = \sin (t)\) is now decreasing, and up, since \(y = \csc (t)\) is now increasing. We pass back through \(\left ( \frac {1}{2}, 2\right )\) when \(t = \frac {5\pi }{6}\) back to the points with small positive \(x\)-coordinates and large positive \(y\)-coordinates.

   \(x = \sin (t)\), \(0 < t < \pi \)	\(y = \csc (t)\), \(0 < t < \pi \)	\(\left \{ x = \sin (t), \, y = \csc (t) \right .\), \(0 < t < \pi \)

   To better explain this behavior, we eliminate the parameter. Using a reciprocal identity, we write \(y = \csc (t) = \frac {1}{\sin (t)}\). Since \(x =\sin (t)\), the curve traced out by this parametrization is a portion of the graph of \(y = \frac {1}{x}\). We now can explain the unusual behavior as \(t \rightarrow 0^{+}\) and \(t \rightarrow \pi ^{-}\) – for these values of \(t\), we are hugging the vertical asymptote \(x=0\) of the graph of \(y = \frac {1}{x}\).

   We see that the parametrization given above traces out the portion of \(y = \frac {1}{x}\) for \(0< x \leq 1\) twice as \(t\) runs through the interval \((0,\pi )\) as indicated above on the right.

4. Proceeding as above, we set about graphing \(\left \{ x = 1 + 3\cos (t), \, y =2\sin (t) \right .\) for \(0 \leq t \leq \frac {3\pi }{2}\) by first graphing \(x = 1 + 3\cos (t)\) and \(y = 2\sin (t)\) on the interval \(\left [0, \frac {3\pi }{2}\right ]\) below on the left and middle, respectivelt.

   We see that \(x\) ranges from \(-2\) to \(4\) and \(y\) ranges from \(-2\) to \(2\). Hence our graph will reside in the region \(\left \{ (x,y) \, | \, -2 \leq x \leq 4, \; -2 \leq y \leq 2 \right \}\).

   Plugging in \(t = 0\), \(\frac {\pi }{2}\), \(\pi \) and \(\frac {3\pi }{2}\) gives the points \((4,0)\), \((1,2)\), \((-2,0)\) and \((1,-2)\), respectively.

   As \(t\) ranges from \(0\) to \(\frac {\pi }{2}\), \(x = 1 + 3\cos (t)\) is decreasing, while \(y = 2\sin (t)\) is increasing. This means that we start tracing out our answer at \((4,0)\) and continue moving to the left and upwards towards \((1,2)\). For \(\frac {\pi }{2} \leq t \leq \pi \), \(x\) is decreasing, as is \(y\), so the motion is still right to left, but now is downwards from \((1,2)\) to \((-2,0)\). On the interval \(\left [\pi , \frac {3\pi }{2}\right ]\), \(x\) begins to increase, while \(y\) continues to decrease. Hence, the motion becomes left to right but continues downwards, connecting \((-2,0)\) to \((1,-2)\).

   To eliminate the parameter here, we note that the trigonometric functions involved, namely \(\cos (t)\) and \(\sin (t)\), are related by the Pythagorean Identity \(\cos ^{2}(t) + \sin ^{2}(t) = 1\). Hence, we solve \(x = 1+3\cos (t)\) for \(\cos (t)\) to get \(\cos (t) = \frac {x-1}{3}\), and we solve \(y = 2\sin (t)\) for \(\sin (t)\) to get \(\sin (t) = \frac {y}{2}\).

   Substituting these expressions into \(\cos ^{2}(t) + \sin ^{2}(t) = 1\) gives \(\left (\frac {x-1}{3}\right )^2 + \left (\frac {y}{2}\right )^2 = 1\), or \(\frac {(x-1)^2}{9} + \frac {y^2}{4} = 1\).

   From Section Ellipses, we know that the graph of this equation is an ellipse centered at \((1,0)\) with vertices at \((-2,0)\) and \((4,0)\) with a minor axis of length \(4\). Our parametric equations here are tracing out three-quarters of this ellipse, in a counter-clockwise direction.

   \(x =1+3\cos (t)\), \(0 \leq t \leq \frac {3\pi }{2}\)	\(y = 2\sin (t)\), \(0 \leq t \leq \frac {3\pi }{2}\)	\(\left \{ x = 1 + 3\cos (t), \, y = 2\sin (t) \right .\), \(0 \leq t \leq \frac {3\pi }{2}\)

   □

1. \(y = x^2\) from \(x = -3\) to \(x = 2\)
2. \(y = f^{-1}(x)\) where \(f(x) = x^5 + 2x + 1\)
3. The line segment which starts at \((2,-3)\) and ends at \((1,5)\)
4. The circle \(x^2 + 2x + y^2 - 4y = 4\)
5. The left half of the ellipse \(\frac {x^2}{4} + \frac {y^2}{9} = 1\)

Solution.

1. Since \(y = x^2\) is written in the form \(y = f(x)\), we let \(x = t\) and \(y = f(t) = t^2\). Since \(x=t\), the bounds on \(t\) match precisely the bounds on \(x\) we get \(\left \{ x = t, \, y = t^2 \right .\) for \(-3 \leq t \leq 2\).

   The check is almost trivial; with \(x=t\) we have \(y = t^2 = x^2\) as \(t = x\) runs from \(-3\) to \(2\).

2. We are told to parametrize \(y = f^{-1}(x)\) for \(f(x) = x^5 + 2x + 1\) so it is safe to assume that \(f\) is one-to-one. (Otherwise, \(f^{-1}\) would not exist.) To find a formula \(y = f^{-1}(x)\), we follow the procedure outlined on page ?? – we start with the equation \(y = f(x)\), interchange \(x\) and \(y\) and solve for \(y\).

   Doing so gives us the equation \(x = y^5+2y+1\). While we could attempt to solve this equation for \(y\) to get an explicit formula for \(f^{-1}(x)\), we don’t need to. We can parametrize the implicit function ^[7] (See the discussion preceding Example implicitfcnex in Section Relations for a review of this concept.) \(x = f(y) = y^5+2y+1\) by setting \(y = t\) so that \(x = t^5 + 2t + 1\).

   We know from Section GraphsofPolynomials that since \(f(x) = x^5 + 2x + 1\) is an odd-degree polynomial, the range of \(y = f(x) = x^5 + 2x + 1\) is \((-\infty , \infty )\). Hence, in order to trace out the entire graph of \(x = f(y) = y^5+2y+1\), we need to let \(y\) run through all real numbers.

   Hence, our final answer to this problem is \(\left \{ x = t^5+2t+1, \, y = t \right .\) for \(-\infty < t < \infty \). As in the previous problem, our solution is trivial to check. ^[8] (Provided you followed the inverse function theory, of course.)

3. To parametrize line segment which starts at \((2,-3)\) and ends at \((1,5)\), we make use of the formulas \(x = x_{\text {\tiny $0$}} + (x_{\text {\tiny $1$}} - x_{\text {\tiny $0$}}) t\) and \(y = y_{\text {\tiny $0$}} + (y_{\text {\tiny $1$}} - y_{\text {\tiny $0$}}) t\) for \(0 \leq t \leq 1\). While these equations at first glance are quite a handful, ^[9] (Compare and contrast this with Exercise 2dvectorsgiveuslines in Section Vectors.) they can be summarized as ‘\(\text {starting point} + (\text {displacement})t\)’.

   To find the equation for \(x\), we have that the line segment starts at \(x= 2\) and ends at \(x = 1\). This means the displacement in the \(x\)-direction is \(\Delta x = (1-2) = -1\). Hence, the equation for \(x\) is \(x = 2 + (-1)t = 2-t\).

   Similarly for \(y\), we note that the line segment starts at \(y=-3\) and ends at \(y=5\). Hence, the displacement in the \(y\)-direction is \(\Delta y = (5-(-3)) = 8\), so we get \(y = -3+8t\).

   Putting together our answers for \(x\) and \(y\), we get \(\left \{ x = 2-t, \, y = -3+8t \right .\) for \(0 \leq t \leq 1\).

   To check, we can solve \(x = 2-t\) for \(t\) to get \(t = 2-x\). Substituting this into \(y = -3+8t\) gives \(y = -3+8t = -3+8(2-x)\), or \(y = -8x+13\). We know this is the graph of a line, so all we need to check is that it starts and stops at the correct points.

   When \(t=0\), \(x= 2-t = 2\), and when \(t=1\), \(x = 2-t = 1\). Plugging in \(x=2\) gives \(y = -8(2)+13 = -3\), for an initial point of \((2,-3)\). When \(x = 1\), \(y = -8(1)+13 = 5\) for an ending point of \((1,5)\), as required.

4. In order to use the formulas above to parametrize the circle \(x^2 + 2x + y^2 - 4y = 4\), we first need to put the equation into the correct form.

   After completing the squares, we get \((x+1)^2+(y-2)^2 = 9\), or \(\frac {(x+1)^2}{9} + \frac {(y-2)^2}{9} = 1\).

   Once again, the formulas \(x = h+a\cos (t)\) and \(y=k+b\sin (t)\) can be a challenge to memorize, but they come from the Pythagorean Identity \(\cos ^{2}(t) + \sin ^{2}(t) = 1\), so we can always use the identity to get our parametrization instead of relying on memorizing a formula.

   In the equation \(\frac {(x+1)^2}{9} + \frac {(y-2)^2}{9} = 1\), we identify \(\cos (t) = \frac {x+1}{3}\) and \(\sin (t) = \frac {y-2}{3}\). Rearranging these last two equations, we get \(x = -1+3 \cos (t)\) and \(y = 2 +3 \sin (t)\).

   In order to complete one revolution around the circle, we let \(t\) range through the interval \([0,2\pi )\), so our final answer \(\left \{ x = -1+3 \cos (t), \, y = 2 +3 \sin (t) \right .\) for \(0 \leq t < 2\pi \).

   To check our answer, we could eliminate the parameter by solving \(x = -1+3\cos (t)\) for \(\cos (t)\) and \(y = 2+3\sin (t)\) for \(\sin (t)\), invoking a Pythagorean Identity, and then manipulating the resulting equation in \(x\) and \(y\) into the original equation \(x^2+2x+y^2-4y = 4\).

   Instead, we opt for a more direct approach. We substitute \(x = -1+3\cos (t)\) and \(y = 2+3\sin (t)\) into the equation \(x^2 + 2x + y^2 - 4y = 4\) and show that the latter is satisfied for all \(t\) such that \(0 \leq t < 2\pi \).

   \[ \begin{array}{rcl} x^2 + 2x + y^2 - 4y & = & 4 \\ (-1+3\cos (t))^2 + 2(-1+3\cos (t)) + (2+3\sin (t))^2 - 4(2+3\sin (t)) & \stackrel {\text {?}}{=} & 4 \\ 1 - 6\cos (t) + 9\cos ^{2}(t) - 2 + 6\cos (t) + 4 + 12\sin (t) + 9\sin ^{2}(t) - 8 - 12\sin (t) & \stackrel {\text {?}}{=} & 4 \\ 9\cos ^{2}(t) + 9\sin ^{2}(t) -5 & \stackrel {\text {?}}{=} & 4 \\ 9\left (\cos ^{2}(t) + \sin ^{2}(t)\right ) -5 & \stackrel {\text {?}}{=} & 4 \\ 9\left (1\right ) -5 & \stackrel {\text {?}}{=} & 4 \\ 4 & = & 4 \, \checkmark \\ \end{array} \]

   Now that we know the parametric equations give us points on the circle, we can go through the usual analysis as demonstrated in Example parametrictorect to show that the entire circle is covered as \(t\) ranges through the interval \([0,2\pi )\).

5. In the equation \(\frac {x^2}{4} + \frac {y^2}{9} = 1\), we can either use the formulas above or think back to the Pythagorean Identity to get \(x = 2\cos (t)\) and \(y = 3\sin (t)\).

   The normal range on the parameter in this case is \(0 \leq t < 2\pi \), but since we are interested in only the left half of the ellipse, we restrict \(t\) to the values which correspond to Quadrant II and Quadrant III angles, namely \(\frac {\pi }{2} \leq t \leq \frac {3\pi }{2}\). Hence, our final answer is \(\left \{ x = 2\cos (t), \, y = 3\sin (t) \right .\) for \(\frac {\pi }{2} \leq t \leq \frac {3\pi }{2}\).

   Substituting \(x = 2\cos (t)\) and \(y = 3\sin (t)\) into \(\frac {x^2}{4} + \frac {y^2}{9} = 1\) gives \(\frac {4\cos ^{2}(t)}{4} + \frac {9 \sin ^{2}(t)}{9} = 1\), which reduces to the Pythagorean Identity \(\cos ^{2}(t) + \sin ^{2}(t) = 1\). This proves the points generated by the parametric equations \(\left \{ x = 2\cos (t), \, y = 3\sin (t) \right .\) lie on the ellipse \(\frac {x^2}{4} + \frac {y^2}{9} = 1\).

   Employing the techniques demonstrated in Example parametrictorect, we find that the restriction \(\frac {\pi }{2} \leq t \leq \frac {3\pi }{2}\) generates the left half of the ellipse, as required. □

1. The curve which starts at \((2,4)\) and follows the parabola \(y = x^2\) to end at \((-1,1)\). Shift the parameter so that the path starts at \(t=0\).
2. The two part path which starts at \((0,0)\), travels along a line to \((3,4)\), then travels along a line to \((5,0)\).
3. The Unit Circle, oriented clockwise, with \(t=0\) corresponding to \((0,-1)\).

Solution. \(~\)

1. We can parametrize \(y = x^2\) from \(x=-1\) to \(x=2\) using the formula given on Page 7 as \(\left \{x = t, \, y = t^2 \right .\) for \(-1 \leq t \leq 2\). This parametrization, however, starts at \((-1,1)\) and ends at \((2,4)\). Hence, we need to reverse the orientation.

   To this end, we replace every occurrence of \(t\) with \(-t\): \(\left \{x = -t, \, y = (-t)^2 \right .\) for \(-1 \leq -t \leq 2\). After simplifying, we get \(\left \{x = -t, \, y = t^2 \right .\) for \(-2 \leq t \leq 1\).

   We would like \(t\) to begin at \(t=0\) instead of \(t=-2\). The problem here is that the parametrization we have starts \(2\) units ‘too soon’, so we need to introduce a ‘time delay’ of \(2\).

   Replacing every occurrence of \(t\) with \((t-2)\) gives \(\left \{x = -(t-2), \, y =(t-2) ^2 \right .\) for \(-2 \leq t -2 \leq 1\). Simplifying yields \(\left \{x = 2-t, \, y =t^2-4t+4\right .\) for \(0 \leq t \leq 3\).

   We leave it to the reader to verify this system traces \(y = x^2\) starting with \((2,4)\) and ending at \((-1,1)\).

2. Again, when parameterizing line segments, we think: ‘\(\text {starting point} + (\text {displacement})t\)’. For the first part of the path, we get \(\left \{ x = 3t, \, y = 4t \right .\) for \(0 \leq t \leq 1\), and for the second part we get \(\left \{ x = 3 + 2t, \, y = 4 - 4t \right .\) for \(0 \leq t \leq 1\).

   Since the first parametrization leaves off at \(t=1\), we shift the parameter in the second part so it starts at \(t=1\). Our current description of the second part starts at \(t=0\), so we need to introduce a ‘time delay’ of \(1\) unit to the second set of parametric equations.

   Replacing \(t\) with \((t-1)\) in the second set of equations gives \(\left \{ x = 3 + 2(t-1), \, y = 4 - 4(t-1) \right .\) for \(0 \leq t-1 \leq 1\). Simplifying yields \(\left \{ x = 1+2t, \, y = 8 -4t \right .\) for \(1 \leq t \leq 2\). Hence, we may parametrize the path as \(\left \{ x = f(t), \, y = g(t) \right .\) for \(0 \leq t \leq 2\) where

   \[ \begin{array}{ccc} f(t) = \left \{ \begin{array}{rl} 3t, & \text {for $0 \leq t \leq 1$} \\ 1+2t, & \text {for $1 \leq t \leq 2$} \end{array} \right . & \text {and} & g(t) = \left \{ \begin{array}{rl} 4t, & \text {for $0 \leq t \leq 1$} \\ 8-4t, & \text {for $1 \leq t \leq 2$} \end{array} \right . \\ \end{array}\]

   Again, we encourage the reader to check our solution.

3. We know that \(\left \{ x = \cos (t), \, y = \sin (t) \right .\) for \(0 \leq t < 2\pi \) gives a counter-clockwise parametrization of the Unit Circle with \(t = 0\) corresponding to \((1,0)\), so our first task is to reverse orientation.

   Replacing \(t\) with \(-t\) gives \(\left \{ x = \cos (-t), \, y = \sin (-t) \right .\) for \(0 \leq -t < 2\pi \). Using the Even/Odd Identities, we simplify: \(\left \{ x = \cos (t), \, y = -\sin (t) \right .\) for \(-2\pi < t \leq 0\). This parametrization gives a clockwise orientation, but \(t=0\) still corresponds to the point \((1,0)\); the point \((0, -1)\) is reached when \(t = -\frac {3\pi }{2}\).

   Our strategy is to first get the parametrization to ‘start’ at the point \((0,-1)\) and then shift the parameter accordingly so the ‘start’ coincides with \(t = 0\).

   We know that any interval of length \(2\pi \) will parametrize the entire circle, so we keep the equations \(\left \{ x = \cos (t), \, y = -\sin (t) \right .\), but start the parameter \(t\) at \(-\frac {3\pi }{2}\), and find the upper bound by adding \(2\pi \) so \(-\frac {3\pi }{2} \leq t < \frac {\pi }{2}\). We leave it to the reader to verify that \(\left \{ x = \cos (t), \, y = -\sin (t) \right .\) for \(-\frac {3\pi }{2} \leq t < \frac {\pi }{2}\) traces out the Unit Circle clockwise starting at the point \((0, -1)\).

   We now shift the parameter by introducing a ‘time delay’ of \(\frac {3\pi }{2}\) units by replacing every occurrence of \(t\) with \(\left (t - \frac {3\pi }{2}\right )\). We get \(\left \{ x = \cos \left (t - \frac {3\pi }{2}\right ), \, y = -\sin \left (t - \frac {3\pi }{2}\right ) \right .\) for \(-\frac {3\pi }{2} \leq t - \frac {3\pi }{2} < \frac {\pi }{2}\). This simplifies courtesy of the Sum/Difference Formulas to \(\left \{ x = -\sin (t), \, y = -\cos (t) \right .\) for \( 0 \leq t < 2\pi \).

   We leave the check of our solution to the reader. □

Solution. We have \(r = 3\) which gives the equations \(\left \{ x = 3(t -\sin (t)), \, y = 3(1-\cos (t)) \right .\) for \(t \geq 0\). (Here we have returned to the convention of using \(t\) as the parameter.)

Sketching the cycloid by hand is a wonderful exercise in Calculus, but for the purposes of this book, we use a graphing utility. Below is the graph of this cycloid created by desmos.

We see the equations create a series of ‘arches’ and can (partially) verify the reasonableness the graph by finding the \(x\)-intercepts. To do this, we set \(y = 3(1-\cos (t)) = 0\), which amounts to solving \(\cos (t) = 1\).

We get \(t = 2 \pi k\) and since \(t \geq 0\), \(k\) can be any nonnegative integer. Substituting a few of these values for \(t\), \(t = 0\), \(t = 2\pi \), \(t = 4\pi \), and \(t=6\pi \) into the equations \(x = 3(t -\sin (t))\) and \(y = 3(1-\cos (t))\) we obtain the points \((0,0)\), \((6\pi , 0) \approx (18.85, 0)\), \((12 \pi ,0) \approx (37.70, 0)\) and \((18\pi , 0) \approx ( 56.55 , 0)\), which match the graph. In general, the \(x\)-intercepts are \((6\pi k, 0)\) for nonnegative integers \(k\). We leave the details to the reader.

We note it is also possible to analytically determine the (local) maximums of the graph using the techniques demonstrated in Example parametrictorect by analyzing \(y = 3(1-\cos (t))\) . The maximums occur when \(t = (2k +1) \pi k\) where \(k\) is a nonnegative integer, which isn’t too surprising just looking at the problem from a symmetry perspective. Substituting these values for \(t\) into our equations for \(x\) and \(y\) produce points of the form \((3(2k+1) \pi , 6)\). We leave the details to the reader. □