Note

We use the term “particle” when we treat an object as a moving point, ignoring its internal structure.

EXAMPLE 1

Sketch the curve with parametric equations \[ x = 2t-4,\qquad y = 3+t^2\tag{2} \]

Solution First compute the \(x\)- and \(y\)-coordinates for several values of \(t\) as in Table 11.1, and plot the corresponding points \((x,y)\) as in Figure 11.2. Then join the points by a smooth curve, indicating the direction of motion with an arrow.

Figure 11.2: The parametric curve \(x = 2t-4\), \(y = 3+t^2\).

CONCEPTUAL INSIGHT

The graph of a function \(y=f(x)\) can always be parametrized in a simple way as \[ c(t)=(t,f(t)) \]

For example, the parabola \(y=x^2\) is parametrized by \(c(t) = (t,t^2)\) and the curve \(y = e^t\) by \(c(t) = (t,e^t)\). An advantage of parametric equations is that they enable us to describe curves that are not graphs of functions. For example, the curve in Figure 11.3 is not of the form \(y=f(x)\) but it can be expressed parametrically.

EXAMPLE 2 Eliminating the Parameter

Describe the parametric curve \[ c(t) = (2t- 4,3+t^2) \] of the previous example in the form \(y=f(x)\).

Solution We “eliminate the parameter” by solving for \(y\) as a function of \(x\). First, express \(t\) in terms of \(x\): Since \(x = 2t-4\), we have \(t = \frac12x+ 2\). Then substitute \[ y = 3+t^2 = 3+\left(\frac12x+ 2\right)^2 = 7+2x +\frac14x^2 \]

Thus, \(c(t)\) traces out the graph of \(f(x) =7+2x +\frac14x^2\) shown in Figure 11.2.

EXAMPLE 3

A bullet follows the trajectory \[ c(t) = (80t, 200t-4.9t^2) \] until it hits the ground, with \(t\) in seconds and distance in meters (Figure 11.4). Find:

Figure 11.4: Trajectory of bullet.

• (a) The bullet’s height at \(t=5\) s.
• (b) Its maximum height.

Solution The height of the bullet at time \(t\) is \(y(t) = 200t - 4.9t^2\).

• (a) The height at \(t=5\) s is \[ y(5) = 200(5)-4.9(5^2) = 877.5 \mathrm{m} \]
• (b) The maximum height occurs at the critical point of \(y(t)\): \[ y'(t) =\frac{d}{dx}{t}(200t - 4.9t^2) = 200 - 9.8t = 0\quad\Rightarrow\quad t = \frac{200}{9.8}\approx 20.4 \,\,\mathrm{s} \]

The bullet’s maximum height is \(y(20.4) = 200(20.4) - 4.9(20.4)^2 \approx 2041 \mathrm{m}\).

THEOREM 1 Parametrization of a Line

(a) The line through \(P=(a,b)\) of slope \(m\) is parametrized by \[ \boxed{\bbox[#fef7e5,5pt]{ x = a+rt,\qquad y = b+st \qquad -\infty < t < \infty}}\tag{3} \] for any \(r\) and \(s\) (with \(r\ne 0\)) such that \(m=s/r\).

(b) The line through \(P=(a,b)\) and \(Q= (c,d)\) has parametrization \[ \boxed{\bbox[#fef7e5,5pt]{ x = a +t(c-a),\qquad y = b + t(d-b)\qquad -\infty < t < \infty}}\tag{4} \] The segment from \(P\) to \(Q\) corresponds to \(0\le 1\le t\).

Solution (a) Solve \(x = a+rt\) for \(t\) in terms of \(x\) to obtain \(t = (x-a)/r\). Then \[ y = b+st = b +s\left(\frac{x-a}r\right) = b+m(x-a) \qquad\textrm{or}\qquad y-b=m(x-a) \] This is the equation of the line through \(P=(a,b)\) of slope \(m\). The choice \(r=1\) and \(s=m\) yields the parametrization in Figure 11.5.

Figure 11.5: The line \[ y-a=m(x-b) \] has parametrization \[ c(t) = (a+t, b+mt) \] This corresponds to \(r=1\), \(s=m\) in Eq 3.

The parametrization in (b) defines a line that satisfies \((x(0),y(0))=(a,b)\) and \((x(1),y(1))=(c,d)\). Thus, it parametrizes the line through \(P\) and \(Q\) and traces the segment from \(P\) to \(Q\) as \(t\) varies from \(0\) to \(1\).

EXAMPLE 4 Parametrization of a Line

Find parametric equations for the line through \(P=(3,-1)\) of slope \(m=4\).

Solution We can parametrize the line by taking \(r=1\) and \(s=4\) in Eq. (3): \[ x = 3+t, \qquad y = -1+4t \]

This is also written as \(c(t) = (3+t, -1+4t)\). Another parametrization of the line is \(c(t) = (3+5t, -1+20t)\), corresponding to \(r=5\) and \(s=20\) in Eq. (3).

EXAMPLE 5 Parametrization of an Ellipse

Verify that the ellipse with equation \(\big(\tfrac{x}a\big)^2+\big(\tfrac{y}b\big)^2=1\) is parametrized by \[ \boxed{\bbox[#fef7e5,5pt]{ c(t) = (a\cos t,\,b\sin t) \qquad\quad \textrm{(for \(-\pi\le t <\pi\))}}} \] Plot the case \(a= 4\), \(b=2\).

Solution To verify that \(c(t)\) parametrizes the ellipse, show that the equation of the ellipse is satisfied with \(x = a\cos t\), \(y = b\sin t\): \[ \left(\frac{x}a\right)^2 + \left(\frac{y}b\right)^2 = \left(\frac{a\cos t}a\right)^2 + \left(\frac{b\sin t}b\right)^2 = \cos^2t+\sin^2t = 1 \]

To plot the case \(a= 4\), \(b=2\), we connect the points corresponding to the \(t\)-values in Table 11.2 (see Figure 11.7). This gives us the top half of the ellipse corresponding to \(0\le t\le \pi\). Then we observe that \(x(t)=4\cos t\) is even and \(y(t) = 2\sin t\) is odd. As noted above, this tells us that the bottom half of the ellipse is obtained by symmetry with respect to the \(x\)-axis.

Figure 11.7: Ellipse with parametric equations \(x = 4\cos t\), \(y = 2\sin t\).

CONCEPTUAL INSIGHT

The parametric equations for the ellipse in Example 5 illustrate a key difference between the path \(c(t)\) and its underlying curve \(C\). The curve \(C\) is an ellipse in the plane, whereas \(c(t)\) describes a particular, counterclockwise motion of a particle along the ellipse. If we let \(t\) vary from \(0\) to \(4\pi\), then the particle goes around the ellipse twice.

A key feature of parametrizations is that they are not unique. In fact, every curve can be parametrized in infinitely many different ways. For instance, the parabola \(y=x^2\) is parametrized not only by \((t,t^2)\) but also by \((t^3,t^6)\), or \((t^5,t^{10})\), and so on.

EXAMPLE 6 Different Parametrizations of the Same Curve

Describe the motion of a particle moving along each of the following paths.

• (a) \(c_1(t) = (t^3, t^6)\)
• (b) \(c_2(t) = (t^2, t^4)\)
• (c) \(c_3(t) = (\cos t, \cos^2t)\)

Figure 11.8: Three parametrizations of portions of the parabola.

Solution Each of these parametrizations satisfies \(y=x^2\), so all three parametrize portions of the parabola \(y=x^2\).

• (a) As \(t\) varies from \(-\infty\) to \(\infty\), the function \(t^3\) also varies from \(-\infty\) to \(\infty\). Therefore, \(c_1(t) = (t^3, t^6)\) traces the entire parabola \(y=x^2\), moving from left to right and passing through each point once [Figure 11.8].
• (b) Since \(x = t^2 \ge 0\), the path \(c_2(t) = (t^2, t^4)\) traces only the right half of the parabola. The particle comes in toward the origin as \(t\) varies from \(-\infty\) to \(0\), and it goes back out to the right as \(t\) varies from \(0\) to \(\infty\) [Figure 11.8].
• (c) As \(t\) varies from \(-\infty\) and \(\infty\), \(\cos t\) oscillates between 1 and \(-1\). Thus a particle following the path \(c_3(t) = (\cos t, \cos^2t)\) oscillates back and forth between the points \((1,1)\) and \((-1,1)\) on the parabola. [Figure 11.8].

EXAMPLE 7 Using Symmetry to Sketch a Loop

Sketch the curve \[ c(t) = (t^2+1, t^3-4t) \] Label the points corresponding to \(t = 0, \pm 1, \pm 2, \pm 2.5\).

Solution

Step 1. Use symmetry.

Observe that \(x(t) = t^2+1\) is an even function and that \(y(t)=t^3-4t\) is an odd function. As noted before Example 5, this tells us that \(c(t)\) is symmetric with respect to the \(x\)-axis. Therefore, we will plot the curve for \(t\ge 0\) and reflect across the \(x\)-axis to obtain the part for \(t\le 0\).

Step 2. Analyze \(x(t)\), \(y(t)\) as functions of \(t\)

We have \(x(t)= t^2 +1\) and \(y(t) = t^3 -4t\). The \(x\)-coordinate \(x(t) = t^2+1\) increases to \(\infty\) as \(t\to \infty\). To analyze the \(y\)-coordinate, we graph \(y(t)=t^3-4t= t(t-2)(t+2)\) as a function of \(t\) (not as a function of \(x\)). Since \(y(t)\) is the height above the \(x\)-axis, Figure 11.9 shows that \begin{alignat*}{4} y(t)&<0& \qquad&\textrm{for}&\qquad &0 < t < 2,&\quad&\Rightarrow\quad\textrm{curve below \(x\)-axis}\\ y(t)&>0& \qquad &\textrm{for}&\qquad &t > 2,&\quad&\Rightarrow\quad\textrm{curve above \(x\)-axis} \end{alignat*}

So the curve starts at \(c(0)=(1,0)\), dips below the \(x\)-axis and returns to the \(x\)-axis at \(t=2\). Both \(x(t)\) and \(y(t)\) tend to \(\infty\) as \(t \to \infty\). The curve is concave up because \(y(t)\) increases more rapidly than \(x(t)\).

Step 3. Plot points and join by an arc.

The points \(c(0)\), \(c(1)\), \(c(2)\), \(c(2.5)\) tabulated in Table 11.3 are plotted and joined by an arc to create the sketch for \(t\ge 0\) as in Figure 11.9. The sketch is completed by reflecting across the \(x\)-axis as in Figure 11.9.

Figure 11.9: The curve \(c(t)=(t^2+1,t^3-4t)\).

EXAMPLE 8 Parametrizing the Cycloid

Find parametric equations for the cycloid generated by a point \(P\) on the unit circle.

Solution The point \(P\) is located at the origin at \(t=0\). At time \(t\), the circle has rolled \(t\) radians along the \(x\) axis and the center \(C\) of the circle then has coordinates \((t,1)\) as in Figure 11.11. Figure 11.11 shows that we get from \(C\) to \(P\) by moving down \(\cos t\) units and to the left \(\sin t\) units, giving us the parametric equations \[ \boxed{\bbox[#fef7e5,5pt]{ x(t) = t - \sin t,\qquad y(t) = 1 - \cos t}}\tag{5} \]

Notation

In this section, we write \(f'(t), x'(t), y'(t)\), and so on to denote the derivative with respect to \(t\).

Caution

Do not confuse \(dy/dx\) with the derivatives \(dx/dt\) and \(dy/dt\), which are derivatives with respect to the parameter \(t\). Only \(dy/dx\) is the slope of the tangent line.

THEOREM 2 Slope of the Tangent Line

Let \(c(t) = (x(t),y(t))\), where \(x(t)\) and \(y(t)\) are differentiable. Assume that \(x'(t)\) is continuous and \(x'(t)\ne 0\). Then \[ \boxed{\bbox[#fef7e5,5pt]{\frac{dy}{dx} = \frac{dy/dt}{dx/dt} =\frac{y'(t)}{x'(t)}}}\tag{7} \]

EXAMPLE 9

Let \(c(t) = (t^2+1, t^3-4t)\). Find:

• (a) An equation of the tangent line at \(t =3\)
• (b) The points where the tangent is horizontal (Figure 11.12).

Figure 11.12: Horizontal tangent lines on \(c(t) = (t^2+1, t^3-4t)\).

Solution We have \[ \frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{(t^3-4t)'}{(t^2+1)'} = \frac{3t^2-4}{2t} \]

(a) The slope at \(t = 3\) is \[ \frac{dy}{dx} =\frac{3t^2-4}{2t}\bigg|_{t=3} = \frac{3(3)^2-4}{2(3)}= \frac{23}6 \]

Since \(c(3) = (10,15)\), the equation of the tangent line in point-slope form is \[ y - 15 = \frac{23}{6}(x-10) \]

(b) The slope \(dy/dx\) is zero if \(y'(t)=0\) and \(x'(t)\ne 0\). We have \(y'(t)=3t^2-4=0\) for \(t = \pm 2/\sqrt{3}\) (and \(x'(t)=2t \neq 0\) for these values of \(t\)). Therefore, the tangent line is horizontal at the points \[ c\left(-\frac{2}{\sqrt 3}\right) = \left(\frac73, \frac{16}{3\sqrt 3}\right),\qquad c\left(\frac{2}{\sqrt 3}\right) = \left(\frac73, -\frac{16}{3\sqrt 3}\right) \]

Note

Bézier curves were invented in the 1960s by the French engineer Pierre Bézier (1910–1999), who worked for the Renault car company. They are based on the properties of Bernstein polynomials, introduced 50 years earlier by the Russian mathematician Sergei Bernstein to study the approximation of continuous functions by polynomials. Today, Béezier curves are used in standard graphics programs, such as Adobe Illustrator\(^{\scriptscriptstyle\mathsf{TM}}\) and Corel Draw\(^{\scriptscriptstyle\mathsf{TM}}\), and in the construction and storage of computer fonts such as TrueType\(^{\scriptscriptstyle\mathsf{TM}}\) and PostScript\(^{\scriptscriptstyle\mathsf{TM}}\) fonts.

EXAMPLE 10

Show that the Bézier curve is tangent to segment \(\overline{P_0P_1}\) at \(P_0\).

Hand sketch made in 1964 by Pierre Bézier for the French automobile manufacturer Renault.

Solution The Bézier curve passes through \(P_0\) at \(t = 0\), so we must show that the slope of the tangent line at \(t=0\) is equal to the slope of \(\overline{P_0P_1}\). To find the slope, we compute the derivatives: \begin{align*} x'(t) &= -3a_0(1-t)^2+ 3a_1(1-4t+3t^2)+a_2(2t-3t^2) +3a_3t^2\\ y'(t) &= -3b_0(1-t)^2+3b_1(1-4t+3t^2)+b_2(2t-3t^2) +3b_3t^2 \end{align*}

Evaluating at \(t=0\), we obtain \(x'(0)=3(a_1-a_0)\), \(y'(0)=3(b_1-b_0)\), and \[ \frac{dy}{dx}\bigg|_{t=0} =\frac{y'(0)}{x'(0)} = \frac{3(b_1-b_0)}{3(a_1-a_0)}= \frac{b_1-b_0}{a_1-a_0} \]

This is equal to the slope of the line through \(P_0=(a_0,b_0)\) and \(P_1=(a_1,b_1)\) as claimed (provided that \(a_1 \neq a_0\)).