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 12.3). Thus, our notion of curve includes straight lines as special cases.

Figure 12.3: \(L\) 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 12.4). 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 12.4: \({\bf c}(t)=\hbox{(}\cos t,\sin t\,\hbox{)}\) 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 12.5).

Figure 12.5: The image of \({\bf c}\hbox{(}t\hbox{)}=\hbox{(}t,t^2\hbox{)}\) is the parabola \(y=x^2\).

example 4

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

The formula for \({\bf c}(t)\) is given by \[ {\bf c}(t)=\bigg(vt-r\sin\frac{vt}{R}, R-r\cos\frac{vt}{R}\bigg). \]

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 12.6; it is called a cycloid.

Figure 12.6: The curve traced by a point moving on the rim of a rolling circle is called a cycloid.

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.

EXAMPLE 5 Helix

Describe the curve traced by \({\bf c}(t) = ( {-\sin t}, \cos t, t)\) for \(t \geq 0\) in terms of its projections onto the coordinate planes.

Solution The projections are as follows (Figure 12.9):

• \(xy\)-plane (set \(z=0\)): the path \({\bf p}(t)=( {-\sin t}, \cos t,0)\), which describes a point moving counterclockwise around the unit circle starting at \({\bf p}(0) = (0,1,0)\).
• \(xz\)-plane (set \(y=0\)): the path \(( {-\sin t}, 0, t)\), which is a wave in the \(z\)-direction.
• \(yz\)-plane (set \(x=0\)): the path \(( 0, \cos t, t)\), which is a wave in the \(z\)-direction.

The function \({\bf c}(t)\) describes a point moving above the unit circle in the \(xy\)-plane while its height \(z=t\) increases linearly, resulting in the helix of Figure 12.9 .

Figure 12.9: Projections of the helix \({\bf c}(t) = ( -\sin t, \cos t, t)\).

EXAMPLE 6 Parametrizing the Intersection of Surfaces

Parametrize the curve \(C\) obtained as the intersection of the surfaces \(x^2-y^2 = z-1\) and \(x^2+y^2=4\) (Figure 12.10).

Solution We have to express the coordinates \((x,y,z)\) of a point on the curve as functions of a parameter \(t\). Here are two ways of doing this.

First method: Solve the given equations for \(y\) and \(z\) in terms of \(x\). First, solve for \(y\): \[ x^2+y^2=4\quad\Rightarrow\quad y^2 = 4-x^2\quad\Rightarrow\quad y = \pm \sqrt{4-x^2} \]

The equation \(x^2-y^2=z-1\) can be written \(z=x^2-y^2+1\). Thus, we can substitute \(y^2 = 4-x^2\) to solve for \(z\): \[ z = x^2-y^2+1 = x^2 - (4-x^2)+1 = 2x^2-3 \]

Figure 12.10: Intersection of surfaces \(x^2-y^2=z-1\) and \(x^2+y^2=4\).

Now use \(t=x\) as the parameter. Then \(y = \pm\sqrt{4-t^2}\), \(z =2t^2-3\). The two signs of the square root correspond to the two halves of the curve where \(y>0\) and \(y<0\), as shown in Figure 12.11. Therefore, we need two vector-valued functions to parametrize the entire curve: \[ {\bf c}_1(t) = ( t, \sqrt{4-t^2}, 2t^2-3),\quad {\bf c}_2(t) = ( t, -\sqrt{4-t^2}, 2t^2-3),\qquad -2\le t \le 2 \]

Figure 12.11: Two halves of the curve of intersection in Example 3

Second method: Note that \(x^2+y^2=4\) has a trigonometric parametrization: \(x = 2\cos t\), \(y = 2\sin t\) for \(0 \le t < 2\pi\). The equation \(x^2-y^2 = z-1\) gives us \[ z = x^2-y^2+1 = 4\cos^2t-4\sin^2t+1=4\cos 2t +1 \]

Thus, we may parametrize the entire curve by a single vector-valued function: \[ {\bf c}(t) = ( 2\cos t, 2\sin t, 4\cos 2t +1), \qquad \textrm{ \(0 \le t < 2\pi\)} \]

DEFINITION Limit of a Path

A path \({\bf c}(t)\) approaches the limit \({\bf u}\) (a vector) as \(t\) approaches \(t_0\) if \(\displaystyle{\lim_{t\to t_0}\|{{\bf c}(t)-{\bf u}}\|={\bf 0}}\). In this case, we write \[ \lim_{t\to t_0}{\bf c}(t) = {\bf u} \]

THEOREM 1 Limits Are Computed Componentwise

A path \({\bf c}(t)=( x(t), y(t), z(t))\) approaches a limit as \(t\to t_0\) if and only if each component approaches a limit, and in this case, \[ \lim_{t\to t_0}{\bf c}(t) = ( \lim_{t\to t_0}x(t), \lim_{t\to t_0}y(t), \lim_{t\to t_0}z(t))\tag{1} \]

Proof

Let \({\bf u} = ( a,b,c)\) and consider the square of the length \[ \|{{\bf c}(t)-{\bf u}}\|^2 = (x(t)-a)^2+(y(t)-b)^2+(z(t)-c)^2\tag{2} \]

The term on the left approaches zero if and only if each term on the right approaches zero (because these terms are nonnegative). It follows that \(\|{{\bf c}(t)-{\bf u}}\|\) approaches zero if and only if \(|x(t)-a|\), \(|y(t)-b|\), and \(|z(t)-c|\) tend to zero. Therefore, \({\bf c}(t)\) approaches a limit \({\bf u}\) as \(t\to t_0\) if and only if \(x(t)\), \(y(t)\), and \(z(t)\) converge to the components \(a\), \(b\), and \(c\).

Note

The Limit Laws of scalar functions remain valid in the vector-valued case. They are verified by applying the Limit Laws to the components.

EXAMPLE 1

Calculate \(\displaystyle{\lim_{t\to 3}{\bf c}(t)}\), where \(\displaystyle{{\bf c}(t) = ( t^2, 1-t, t^{-1})}\).

Solution By Theorem 1, \[ \lim_{t\to 3}{\bf c}(t) = \lim_{t\to 3}( t^2, 1-t, t^{-1}) = ( \lim_{t\to 3}t^2, \lim_{t\to 3}(1-t), \lim_{t\to 3} t^{-1}) = ( 9, -2, \frac13) \]

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 by \[ {{\bf c}'(t)=\mathop{\rm lim}_{h\to 0} \frac{{\bf c}(t+h)-{\bf c}(t)}{h}.} \]

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 \[ {\bf c}'(t)=(x'(t),y'(t),z'(t))=x'(t){\bf i}+y'(t){\bf j}+z'(t){\bf k}. \]

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 \({\bf c}(t)\).

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 12.14. At \(t=\pi/2,{\bf c}'(\pi/2)=({-}{\rm sin}\,\pi/2,\cos \pi/2,1)=(-1,0,1)=-{\bf i}+{\bf k}\).

Figure 12.14: 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.

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 \({\bf c}(t_0)\).

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 \(x=3+t,y=6-t,z=5\).

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*}