OBJECTIVES

When you finish this section, you should be able to:

1. Graph parametric equations (p. 637)
2. Find a rectangular equation for a curve represented parametrically (p. 638)
3. Use time as the parameter in parametric equations (p. 640)
4. Convert a rectangular equation to parametric equations (p. 641)

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Suppose \(x=x(t)\) and \(y=y(t)\) are two functions of a third variable \(t\), called the parameter, that are defined on the same interval \(I\). Then the equations \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ x=x(t)\qquad y=y(t) }} \]

where \(t\) is in \(I\) are called parametric equations, and the collection of points defined by \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ (x,y) =( x( t) ,y( t) ) }} \]

is called a plane curve.

Graphing a Plane Curve

Graph the plane curve represented by the parametric equations \[ \begin{equation*} x( t) =3t^{2} \qquad y( t) =2t\;-2\leq t\leq 2 \end{equation*} \]

Indicate the orientation of the curve.

Solution Corresponding to each number \(t\), \(-2\leq t\leq 2,\) there are a number \(x\) and a number \(y\) that are the coordinates of a point \(( x,y) \) on the curve. We form a table listing various choices of the parameter \(t\) and the corresponding values for \(x\) and \(y\), as shown in Table 1.

The motion begins when \(t=-2\) at the point \(( 12,-4) \) and ends when \(t=2\) at the point \(( 12,4)\). Figure 2 illustrates the plane curve whose parametric equations are \(x( t) =3t^{2}\) and \(y( t) =2t\). The arrows indicate the orientation of the plane curve for increasing values of the parameter \(t\).

Figure 2 \(x(t)=3t^{2}, y(t)=2t, -2\le t \le 2\)

Finding a Rectangular Equation for a Plane Curve Represented Parametrically

Find a rectangular equation of the curve whose parametric equations are \[ \begin{equation*} x( t) =R\, \cos t \qquad y( t) =R\, \sin t \end{equation*} \]

where \(R>0\) is a constant. Graph the plane curve and indicate its orientation.

Figure 3 The orientation is counterclockwise.

Solution The presence of the sine and cosine functions in the parametric equations suggests using the Pythagorean Identity \(\cos ^{2}t+\sin ^{2}t=1\). Then \[ \begin{array}{rcl@{\quad}crcl} \left( \dfrac{x}{R}\right) ^{2}+\left( \dfrac{y}{R}\right) ^{2}& =&1 &\qquad\color{#0066A7}{\cos \;t = \tfrac{x}{R}\quad \sin t=\dfrac{y}{R}}\\ x^{2}+y^{2}& =& R^{2} \end{array} \]

The graph of the rectangular equation is a circle with center at the origin and radius \(R\). In the parametric equations, as the parameter \(t\) increases, the points \((x,y)\) on the circle are traced out in the counterclockwise direction, as shown in Figure 3.

NOW WORK

Problems 9 and 15.

Finding a Rectangular Equation for a Plane Curve Represented Parametrically

Find rectangular equations for the plane curves represented by each of the parametric equations. Graph each curve and indicate its orientation.

(a) \(x( t) =R\, \cos t\quad y( t) = R\, \sin t\quad 0\leq t\leq \pi \) and \(R> 0\)

(b) \(x( t) =R\, \sin t\quad y( t) = R\, \cos t\quad 0\leq t\leq \pi \) and \(R> 0\)

Solution (a) We eliminate the parameter \(t\) using a Pythagorean Identity. \[ \begin{eqnarray*} \cos^{2}t+\sin ^{2}t &=&1 \\ \left( \dfrac{x}{R}\right) ^{2}+\left( \dfrac{y}{R}\right) ^{2} &=&1 \qquad \color{#0066A7}{\cos t=\dfrac{x}{R},\quad \sin t=\dfrac{y}{R}} \\ x^{2}+y^{2} &=&R^{2} \tag{1} \end{eqnarray*} \]

Figure 4 \(y=\sqrt{R^{2}-x^{2}}\), \(-R\leq x \leq R\).

The rectangular equation represents a circle with radius \(R\) and center at the origin. In the parametric equations, \(0\leq t\leq \pi\), so the curve begins when \(t=0\) at the point \(( R, 0) \), passes through the point \(( 0, R) \) when \(t=\dfrac{\pi }{2}\), and ends when \(t=\pi \) at the point \(( -R,0)\). The curve is an upper semicircle of radius \(R\) with counterclockwise orientation, as shown in Figure 4. If we solve equation (1) for \(y\), we obtain the rectangular equation of the semicircle \[ y=\sqrt{R^{2}-x^{2}}\qquad\hbox{where }-R\leq x\leq R \]

(b) We eliminate the parameter \(t\) as we did in (a), and again we obtain \[ \begin{eqnarray*} x^{2}+y^{2}=R^{2}\tag{2} \end{eqnarray*} \]

Figure 5 \(x=\sqrt{R^{2}-y^{2}}\), \(-R\leq y \leq R\).

The rectangular equation represents a circle with radius \(R\) and center at \(( 0,0) \). But in the parametric equations, \(0\leq t\leq \pi \), so now the curve begins when \(t=0\) at the point \(( 0, R)\), passes through the point \(( R, 0) \) when \(t=\dfrac{\pi }{2}\), and ends at the point \(\left( 0,-R\right) \) when \(t=\pi\). The curve is a right semicircle of radius \(R\) with a clockwise orientation, as shown in Figure 5. If we solve equation (2) for \(x\), we obtain the rectangular equation of the semicircle \[ x=\sqrt{R^{2}-y^{2}}\qquad \hbox{where }-R\leq y\leq R \]

NOTE

In Example 3 (a) we expressed the rectangular equation as a function \(y=f ( x)\), and in (b) we expressed the rectangular equation as a function \(x=g (y)\).

NOW WORK

Problem 17.

Finding a Rectangular Equation for a Plane Curve Represented Parametrically

(a) Find a rectangular equation of the plane curve whose parametric equations are \[ x( t) =\cos ( 2t) \qquad y( t) =\sin t\quad -\dfrac{\pi }{2}\leq t\leq \dfrac{\pi }{2} \]

(b) Graph the rectangular equation.

(c) Determine the restrictions on \(x\) and \(y\) so the graph corresponding to the rectangular equation is identical to the plane curve described by \[ x=x( t)\qquad y=y( t)\quad -\dfrac{\pi }{2} \leq t\leq\dfrac{\pi}{2} \]

(d) Graph the plane curve whose parametric equations are \[ x=\cos ( 2t)\qquad y=\sin t\quad -\dfrac{\pi }{2}\leq t\leq \dfrac{\pi }{2} \]

Figure 6

Solution (a) To eliminate the parameter \(t\), we use a trigonometric identity that involves \(\sin t\) and \(\cos ( 2t)\), namely, \(\sin ^{2}t=\dfrac{1-\cos ( 2t) }{2}\). Then \[ \begin{equation*} y^{2}\underset{\color{#0066A7}{\underset{\hbox{\(y( t) =\sin t\)}}{{\uparrow }}}}{=}\sin ^{2}t=\dfrac{1-\cos (2t) }{2}\underset{\color{#0066A7}{\underset{\hbox{\(x( t)=\cos ( 2t)\)}}{{\uparrow }}}}{=}\dfrac{1-x}{2}\\ \end{equation*} \]

(b) The curve represented by the rectangular equation \(y^{2}=\dfrac{1-x}{2} \) is the parabola shown in Figure 6(a).

(c) The plane curve represented by the parametric equations does not include all the points on the parabola. Since \(x( t) =\cos ( 2t) \) and \(-1\leq \cos ( 2t) \leq 1\), then \(-1\leq x\leq 1\). Also, since \(y( t) =\sin t\), then \(-1\leq y\leq 1\). Finally, the curve is traced out exactly once in the counterclockwise direction from the point \(( -1,-1) \) (when \(t=-\dfrac{\pi }{2}\)) to the point \(( -1,1) \) (when \(t=\dfrac{\pi }{2}\)).

(d) The plane curve represented by the given parametric equations is the part of the parabola shown in Figure 6(b).

NOW WORK

Problem 21.

Using Time as the Parameter in Parametric Equations

Describe the motion of an object that moves along a curve so that at time \(t\) it has coordinates \[ \begin{equation*} x( t) =3\cos t\qquad y( t) =4\sin t \qquad 0\leq t\leq 2\pi \end{equation*} \]

Figure 7 \(\dfrac{x^{2}}{9}+\dfrac{y^{2}}{16}=1\).

Solution We eliminate the parameter \(t\) using the Pythagorean Identity \(\cos ^{2}t+\sin ^{2}t=1\). \[ \begin{equation*} \frac{x^{2}}{9}+\frac{y^{2}}{16}=1\qquad \color{#0066A7}{\cos t= \dfrac{x}{3}, \sin t= \dfrac{y}{4}} \end{equation*} \]

The plane curve is the ellipse shown in Figure 7. When \(t=0\), the object is at the point \((3,0)\). As \(t\) increases, the object moves around the ellipse in a counterclockwise direction, reaching the point \((0,4)\) when \(t=\dfrac{ \pi }{2}\), the point \((-3,0)\) when \(t=\pi\), the point \((0,-4)\) when \(t= \dfrac{3\pi }{2}\), and returning to its starting point \(\left( 3,0\right)\) when \(t=2\pi\).

NOW WORK

Problem 41.

Finding Parametric Equations for a Curve Represented by a Rectangular Equation

Find parametric equations corresponding to the equation \(y=x^{2}-4\).

Solution We let \(x=t\). Then parametric equations are \[ x( t) =t\qquad y( t) =t^{2}-4\qquad -\infty < t < \infty \]

NOW WORK

Problem 47.

Finding Parametric Equations for an Object in Motion

Find parametric equations that trace out the ellipse \[ \begin{equation*} \frac{x^{2}}{4}+y^{2}=1 \end{equation*} \]

where the parameter \(t\) is time (in seconds) and:

(a) The motion around the ellipse is counterclockwise, begins at the point \((2,0)\), and requires \(1\) second for a complete revolution.

(b) The motion around the ellipse is clockwise, begins at the point \((0,1)\) , and requires \(2\) seconds for a complete revolution.

Solution (a) Figure 8 shows the graph of the ellipse \(\dfrac{ x^{2}}{4}+y^{2}=1\).

Figure 8 \(\dfrac{x^{2}}{4}+y^{2}=1\)

Since the motion begins at the point \(( 2,0)\), we want \(x=2\) and \(y=0\) when \(t=0\). We let \[ \begin{equation*} x( t) =2\cos ( \omega t) \qquad\hbox{and}\qquad y( t) =\sin ( \omega t) \end{equation*} \]

for some constant \(\omega\). This choice for \(x=x( t) \) and \(y=y( t) \) satisfies the equation \(\dfrac{x^{2}}{4}+y^{2}=1\) and also satisfies the requirement that when \(t=0\), then \(x=2\) and \(y=0\).

For the motion to be counterclockwise, as \(t\) increases from \(0\), the value of \(x\) must decrease and the value of \(y\) must increase. This requires \(\omega > 0\). [Do you see why? If \(\omega > 0\), then \(x=2\cos ( \omega t) \) is decreasing when \(t > 0\) is near zero, and \(y=\sin ( \omega t) \) is increasing when \(t > 0\) is near zero.]

Finally, since \(1\) revolution takes \(1\) second, the period is \(\dfrac{2\pi }{ \omega }=1\), so \(\omega =2\pi\). Parametric equations that satisfy the conditions given in (a) are \[ \begin{equation*} x( t) =2\cos ( 2\pi t) \qquad y( t) =\sin ( 2\pi t)\;0\leq t\leq 1 \end{equation*} \]

(b) Figure 9 shows the graph of the ellipse \(\dfrac{x^{2}}{4}+y^{2}=1\).

Figure 9 \(\dfrac{x^{2}}{4}+y^{2}=1\)

Since the motion begins at the point \(( 0,1) \), we want \(x=0\) and \(y=1\) when \(t=0\). We let \[ \begin{equation*} x( t) =2\sin ( \omega t) \qquad \hbox{and}\qquad y( t) =\cos ( \omega t) \end{equation*} \]

for some constant \(\omega\). This choice for \(x=x( t) \) and \(y=y( t) \) satisfies the equation \(\dfrac{x^{2}}{4}+y^{2}=1\) and also satisfies the requirement that when \(t=0\), then \(x=2\sin 0=0\) and \(y=\cos 0=1\).

For the motion to be clockwise, as \(t\) increases from \(0\), the value of \(x\) must increase and the value of \(y\) must decrease. This requires that \(\omega > 0\).

Finally, since \(1\) revolution takes \(2\) seconds, the period is \(\dfrac{2\pi }{ \omega }=2\), or \(\omega =\pi\). Parametric equations that satisfy the conditions given in (b) are \[ x( t) =2\sin ( \pi t) \qquad y( t) =\cos ( \pi t)\;0\leq t\leq 2 \]

NOW WORK

Problem 55.

RECALL

For a circle of radius \(r\), a central angle of \(\theta\) radians subtends an arc whose length \(s\) is \(s=r\, \theta\).

THEOREM Parametric Equations of a Cycloid

Parametric equations of a cycloid are \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ x( t) =a(t-\sin t)\qquad y( t) =a(1-\cos t)\quad -\infty < t < \infty }} \]

NOTE

In Greek, brachistochrone means “the shortest time,” and tautochrone “equal time.”