2.5 Introduction to Paths and Curves

In this section, we introduce some of the basic geometry and computational methods for paths in the plane and space. This will be an important ingredient for the chain rule treated in the next section. We will return to paths with additional topics in Chapter 4.

Figure 2.38: The map \({\bf c}\) is the path; its image \(C\,\) is the curve we “see.”

Paths and Curves

We often think of a curve as a line drawn on paper, such as a straight line, a circle, or a sine curve. It is useful to think of a curve \(C\) mathematically as the set of values of a function that maps an interval of real numbers into the plane or space. We shall call such a map a path. We usually denote a path by \({\bf c}\). The image \(C\) of the path then corresponds to the curve we see on paper (see Figure 2.38). Often we write t for the independent variable and imagine it to be time, so that \({\bf c}(t)\) is the position at time t of a moving particle, which traces out a curve as \(t\) varies. We also say c parametrizes \(C\). Strictly speaking, we should distinguish between \({\bf c}(t)\) as a point in space and as a vector based at the origin.

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example 1

The straight line \(L\) in \({\mathbb R}^3\) through the point \((x_0,y_0,z_0)\) in the direction of vector \({\bf v}\) is the image of the path \[ {\bf c}(t)=(x_0,y_0,z_0)+t{\bf v} \] for \(t\in {\mathbb R}\) (see Figure 2.39). Thus, our notion of curve includes straight lines as special cases.

Figure 2.39: \(L\) is the straight line in space through \(\hbox{(}x_0,y_0,z_0\hbox{)}\) and in direction \({\bf v}\); its equation is \({\bf c}(t)=\hbox{(}x_0,y_0,z_0\hbox{)}+t{\bf v}\).

example 2

The unit circle \(C\): \(x^2+y^2=1\) in the plane is the image of the path \[ {\bf c}\,\colon\,{\mathbb R}\to{\mathbb R}^2, {\bf c}(t)=(\cos t,\sin t), 0\leq t\leq 2\pi \] (see Figure 2.40). The unit circle is also the image of the path \(\tilde{\bf c} (t)=(\cos 2t,\sin 2t)\), \(0\leq t\leq \pi\). Thus, different paths may parametrize the same curve.

Figure 2.40: \({\bf c}(t)=\hbox{(}\cos t,\sin t\,\hbox{)}\) is a path whose image \(C\) is the unit circle.

Paths and Curves

A path in \({\mathbb R}^n\) is a map \({\bf c}\,{:}\,\,[a,b]\to{\mathbb R}^n\); it is a path in the plane if \(n=2\) and a path in space if \(n=3\). The collection \(C\) of points \({\bf c}(t)\) as \(t\) varies in \([a,b]\) is called a curve, and \({\bf c}(a)\) and \({\bf c}(b)\) are its endpoints. The path c is said to parametrize the curve \(C\). We also say \({\bf c}(t)\) traces out \(C\) as \(t\) varies.

If \({\bf c}\) is a path in \({\mathbb R}^3\), we can write \({\bf c}(t)=(x(t),y(t),z(t))\), and we call \(x(t),y(t)\), and \(z(t)\) the component functions of \({\bf c}\). We form component functions similarly in \({\mathbb R}^2\) or, generally, in \({\mathbb R}^n\). We also consider paths whose domain is the whole real line as in the next example.

example 3

The path \({\bf c}(t)=(t,t^2)\) traces out a parabolic arc. This curve coincides with the graph \(f(x)=x^2\) (see Figure 2.41).

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Figure 2.41: The image of \({\bf c}\hbox{(}t\hbox{)}=\hbox{(}t,t^2\hbox{)}\) is the parabola \(y=x^2\).
Figure 2.42: The vector \({\bf d}(t)\) points from the wheel’s center, \({\bf C}(t)\), to the position \({\bf c}(t)\) of a point on the wheel and rotates in the clockwise direction while the wheel moves to the right.

example 4

A wheel of radius \(R\) rolls to the right along a straight line at speed \(v\). Use vector methods to find the path \({\bf c}(t)\) of the point on the wheel that initially lies at a distance \(r\) below the center.

solution We place the wheel in the \(xy\) plane with its center initially at \((0,R)\), so that the position of the center at time \(t\) is given by the path \({\bf C}(t)=(vt, R)\). (Refer to Figure 2.42.)

The position of the point \({\bf c}(t)\) relative to the center is given by the vector \({\bf d}(t)={\bf c}(t)-{\bf C}(t)\) that has the initial value \(-r{\bf j}\) and rotates in the clockwise direction. The rate of rotation is such that the wheel makes a full rotation after the center has moved a distance \(2\pi\! R\) (equal to the circumference of the wheel). This takes a time \(2\pi\! R/v\), so the angular velocity \(d\theta/dt\) of the wheel is \(v/R\). Because the rotation is clockwise, the vector function \({\bf d}(t)\) is of the form \[ {\bf d}(t)=r\,\bigg(\cos \!\Big[{-}\frac{v}{R}\,t+\theta\Big]{\bf i}+ \sin \Big[{-}\frac{v}{R}\,t+\theta\Big]{\bf j}\bigg) \] for some initial angle \(\theta\). Because \({\bf d}(0)=-r{\bf j}\), we have \(\cos\theta=0\) and \(\sin\,\theta=-1\), so \(\theta=-\pi/2\), and hence \[ {\bf d}(t)=r\,\bigg(\cos \!\Big[{-}\frac{v}{R}\,t-\frac{\pi}{2}\Big]{\bf i}+ \sin\Big[{-}\frac{v}{R}\,t-\frac{\pi}{2}\Big]{\bf j}\bigg). \]

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Figure 2.43: The curve traced by a point moving on the rim of a rolling circle is called a cycloid.

Using \(\cos \,(\varphi-\pi/2)=\sin\,\varphi\) and \(\sin\, (\varphi-\pi/2)=-\cos\varphi\), along with \(\cos\,(-\varphi)=\cos\varphi\) and \(\sin\,(-\varphi)={-}{\sin}\,\varphi\), we get \[ {\bf d}(t)=r\,\bigg({-}{\sin}\frac{vt}{R}\,{\bf i}-\cos\frac{vt}{R}\, {\bf j}\bigg). \]

Finally, the path \({\bf c}(t)\) is given by adding the components of the vector function \({\bf d}(t)\) to the coordinates of the path \({\bf C}(t)\); the result is \[ {\bf c}(t)=\bigg(vt-r\sin\frac{vt}{R}, R-r\cos\frac{vt}{R}\bigg). \]

In the special case \(v=R=r=1\), we get \({\bf c}(t)=(t-\sin t,1-\cos t)\). The image curve \(C\) of this path c is shown in Figure 2.43; it is called a cycloid.

Figure 2.44: An example of a hypocycloid.

The preceding example considered the path of a point not necessarily on the rim of a wheel rolling along a straight line. When the wheel rolls on a circle, the resulting curve is called an epicycle. These are the epicycles discussed in the Ptolemaic theory in the introduction. If the wheel is outside the circle and the point is on the rim, the curve is called an epicycloid, and when the wheel is inside the circle, it is a hypocycloid. An example of the latter is shown in Figure 2.44.

Historical Note

The French mathematician Blaise Pascal studied the cycloid in 1649 as a way of distracting himself at a time when he was suffering from a painful toothache. When the pain disappeared, he took it as a sign that God was not displeased with his thoughts. Pascal’s results stimulated other mathematicians to investigate this curve, and subsequently numerous remarkable properties were found. One of these was discovered by the Dutchman Christian Huygens, who used it in the construction of a “perfect” pendulum clock.

Velocity and Tangents to Paths

If we think of \({\bf c}(t)\) as the curve traced out by a particle and \(t\) as time, it is reasonable to define the velocity vector as follows.

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Definition: Velocity Vector

If \({\bf c}\) is a path and it is differentiable, we say \({\bf c}\) is a differentiable path. The velocity of c at time \(t\) is defined byfootnote # \[ {{\bf c}'(t)=\mathop{\rm lim}_{h\to 0} \frac{{\bf c}(t+h)-{\bf c}(t)}{h}.} \]

We normally draw the vector \({\bf c}'(t)\) with its tail at the point \({\bf c}(t)\). The speed of the path \({\bf c}(t)\) is \(s=\|{\bf c}'(t)\|\), the length of the velocity vector. If \({\bf c}(t)=(x(t),y(t))\) in \({\mathbb R}^2\), then \[ {\bf c}'(t)=(x'(t),y'(t))=x'(t){\bf i}+y'(t){\bf j} \] and if \({\bf c}(t)=(x(t),y(t),z(t))\) in \({\mathbb R}^3\), then \[ {\bf c}'(t)=(x'(t),y'(t),z'(t))=x'(t){\bf i}+y'(t){\bf j}+z'(t){\bf k}. \]

Here, \(x'(t)\) is the one-variable derivative \(dx/dt\). If we accept limits of vectors interpreted componentwise, the formulas for the velocity vector follow from the definition of the derivative. However, the limit can be interpreted in the sense of vectors as well. In Figure 2.45, we see that \([{\bf c}(t+h)-{\bf c}(t)]/h\) approaches the tangent to the path as \(h\to 0\).

Tangent Vector

The velocity \({\bf c}'(t)\) is a vector tangent to the path \({\bf c}(t)\) at time \(t\). If \(C\) is a curve traced out by c and if \({\bf c}'\)(\(t\)) is not equal to 0, then \({\bf c}'(t)\) is a vector tangent to the curve \(C\) at the point \({\bf c}(t)\).

Figure 2.45: The vector \({\bf c}'(t)\) is tangent to the path \({\bf c}(t)\).

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If we think of the derivative \({\bf D}{\bf c}(t)\) as a matrix, it will be a column vector with the entries \(x'(t),y'(t)\), and \(z'(t)\). Thus, the derivative here is consistent with our earlier notion.

example 5

Compute the tangent vector to the path \({\bf c}(t)=(t,t^2,e^t)\) at \(t=0\).

solution Here \({\bf c}'(t)=(1,2t,e^t)\), and so at \(t=0\) we obtain the tangent vector \((1, 0, 1)\).

example 6

Describe the path \({\bf c}(t)=(\cos t,\sin t,t)\). Find the velocity vector at the point on the image curve where \(t=\pi/2\).

solution For a given \(t\), the point \((\cos t, \sin t, 0)\) lies on the circle \(x^2+y^2\,{=}\,1\) in the \(xy\) plane. Therefore, the point \((\cos t,\sin t, t)\) lies \(t\) units above the point \((\cos t, \sin t, 0)\) if \(t\) is positive and \(-t\) units below \((\cos t,\sin t,0)\) if \(t\) is negative. As \(t\) increases, \((\cos t,\sin t, t)\) wraps around the cylinder \(x^2+y^2=1\) with the z coordinate increasing. The curve this traces out is called a helix, which is depicted in Figure 2.46. At \(t=\pi/2,{\bf c}'(\pi/2)=({-}{\rm sin}\,\pi/2,\cos \pi/2,1)=(-1,0,1)=-{\bf i}+{\bf k}\).

Figure 2.46: The helix \({\bf c}\hbox{(}t\hbox{)}=\hbox{(}\cos t,\sin t,t\hbox{)}\) wraps around the cylinder \(x^2+y^{\,2}=1\).

example 7

The cycloidal path of a particle on the edge of a wheel of radius \(R\) with speed \(v\) is given by \({\bf c}(t)=(vt-R\sin\, (vt/R), R-R\cos \,(vt/R))\). (See Example 4.) Find the velocity \({\bf c}'(t)\) of the particle as a function of \(t\). When is the velocity zero? Is the velocity vector ever vertical?

solution To find the velocity, we differentiate: \begin{eqnarray*} {\bf c}'(t) &=& \bigg(\frac{d}{dt}\,\bigg(vt-R\sin\frac{vt}{R}\bigg), \frac{d}{dt}\,\bigg(R-R\cos\frac{vt}{R}\bigg)\bigg)\\[5.3pt] &=& \bigg(v-v\cos\frac{vt}{R},v\sin\frac{vt}{R}\bigg). \end{eqnarray*}

In vector notation, \({\bf c}'(t)=(v-v\cos\, (vt/R)){\bf i}+(v\sin\, (vt/R)){\bf j}\). The component in the direction of \({\bf i}\) is \(v(1-\cos\, (vt/R))\), which is zero whenever \(vt/R\) is an integer multiple of \(2\pi\). For such values of \(t\), \(\sin \,(vt/R)\) is zero as well, so the only times at which the velocity is zero are when \(t=2\pi{n}R/v\) for some integer \(n\). At such times, \({\bf c}(t)=(2\pi{n}R, 0)\), so the moving point is touching the ground. These moments occur at time intervals of \(2\pi\!R/v\) (more frequently for small wheels, as well as for rapidly rolling ones).

The velocity vector is never vertical, because the horizontal component vanishes only when the vertical one does as well.

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Figure 2.47 shows some velocity vectors superimposed on the cycloidal path of Figure 2.43.

Figure 2.47: Velocity vectors for the curve traced out by a point on the rim of a rolling wheel.

Tangent Line

The tangent line to a path at a point is the line through the point in the direction of the tangent vector. Using the point-direction form of the equation of a line, we obtain the parametric equation for the tangent line.

Tangent Line to a Path

If \({\bf c}(t)\) is a path, and if \({\bf c'}(t_0)\ne 0\), the equation of its tangent line at the point \({\bf c}(t_0)\) is \[ \boldsymbol{\mathit{l}}(t)={\bf c}(t_0)+(t-t_0){\bf c}'(t_0). \]

If \(C\) is the curve traced out by \({\bf c}\), then the line traced out by l is the tangent line to the curve \(C\) at \({\bf c}(t_0)\).

Notice that we have written the equation in such a way that l goes through the point \({\bf c}(t_0)\) at \(t=t_0\) (rather than \(t=0\)). See Figure 2.48.

Figure 2.48: The tangent line to a path.

example 8

A path in \({\mathbb R}^3\) goes through the point (3, 6, 5) at \(t=0\) with tangent vector \({\bf i}-{\bf j}\). Find the equation of the tangent line.

solution The equation of the tangent line is \[ \boldsymbol{\mathit{l}}(t)=(3,6,5)+t({\bf i}-{\bf j})=(3,6,5)+t(1,-1,0)=(3+t,6-t,5). \]

In \((x,y,z)\) coordinates, the tangent line is \(x=3+t,y=6-t,z=5\).

Physically, we can interpret motion along the tangent line as the path that a particle on a curve would follow if it were set free at a certain moment.

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example 9

Suppose that a particle follows the path \({\bf c}(t)=(e^t,e^{-t},\cos t)\) until it flies off on a tangent at \(t=1\). Where is it at \(t=3\)?

solution The velocity vector is \((e^t, -e^{-t},{-}{\sin t})\), which at \(t=1\) is the vector \((e,-1/e,{-}{\sin 1})\). The particle is at \((e,1/e,\cos 1)\) at \(t=1\). The equation of the tangent line is \(\boldsymbol{\mathit{l}}(t)=(e,1/e,\cos 1)+(t-1)(e,-1/e,{-}{\sin 1})\). At \(t=3\), the position on this line is \begin{eqnarray*} \boldsymbol{\mathit{l}}(3) &=& \bigg(e,\frac{1}{e},\cos 1\bigg)+2\,\bigg(e,-\frac{1}{e},{-}{\sin 1}\bigg)=\bigg(3e,-\frac{1}{e},\cos 1-2\sin 1\bigg)\\[6pt] &\cong& (8.155,-0.368,-1.143). \\[-36pt] \end{eqnarray*}