DEFINITION Limit of a Vector-Valued Function

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

THEOREM 1 Vector-Valued Limits Are Computed Componentwise

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

Proof

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

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 \(\|{{\bf r}(t)-{\bf u}}\|\) approaches zero if and only if \(|x(t)-a|\), \(|y(t)-b|\), and \(|z(t)-c|\) tend to zero. Therefore, \({\bf r}(t)\) 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 \(\displaystyle{\lim_{t\to 3}{\bf r}(t)}\), where \(\displaystyle{{\bf r}(t) = \langle t^2, 1-t, t^{-1}\rangle}\).

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

THEOREM 2 Vector-Valued Derivatives Are Computed Componentwise

A vector-valued function \({\bf r}(t)=\langle x(t), y(t), z(t)\rangle\) is differentiable if and only if each component is differentiable. In this case, \[ {\bf r}'(t) = \frac{d}{dt}{\bf r}(t) = \langle x'(t), y'(t), z'(t)\rangle \]

Note

By Theorems 1 and 2, vector-valued limits and derivatives are computed “componentwise,” so they are not more difficult to compute than ordinary limits and derivatives.

EXAMPLE 2

Calculate \({\bf r}''(3)\), where \({\bf r}(t)=\langle \ln t, t, t^2\rangle\).

Solution We perform the differentiation componentwise: \begin{align*} {\bf r}'(t) =\frac{d}{dt}\,\langle \ln t, t, t^2\rangle =\langle t^{-1},1,2t\rangle\\ {\bf r}”(t) =\frac{d}{dt}\,\langle t^{-1},1,2t\rangle =\langle -t^{-2},0,2\rangle \end{align*} Therefore, \({\bf r}''(3) =\langle -\frac19,0,2\rangle\).

Differentiation Rules

Assume that \({\bf r}(t)\), \({\bf r}_1(t)\), and \({\bf r}_2(t)\) are differentiable. Then

• Sum Rule: \(\displaystyle{({\bf r}_1(t)+{\bf r}_2(t))' = {\bf r}_1'(t)+{\bf r}_2'(t)}\)
• Constant Multiple Rule: For any constant \(c\), \(\displaystyle{(c\,{\bf r}(t))' = c\,{\bf r}'(t)}\).
• Product Rule: For any differentiable scalar-valued function \(f(t)\), \[ \frac{d}{dt} \bigl(f(t){\bf r}(t)\bigr) = f(t) {\bf r}'(t)+f'(t){\bf r}(t) \]
• Chain Rule: For any differentiable scalar-valued function \(g(t)\), \[ \frac{d}{dt} \,{\bf r}(g(t)) = g'(t){\bf r}'(g(t)) \]

Proof

Each rule is proved by applying the differentiation rules to the components. For example, to prove the Product Rule (we consider vector-valued functions in the plane, to keep the notation simple), we write \[ f(t){\bf r}(t) = f(t)\langle x(t), y(t)\rangle = \langle f(t)x(t), f(t)y(t)\rangle \] Now apply the Product Rule to each component: \begin{align*} \frac{d}{dt} f(t){\bf r}(t) &= \langle \frac{d}{dt}f(t)x(t), \frac{d}{dt}f(t)y(t)\rangle\\ &= \langle f'(t)x(t)+f(t)x'(t), f'(t)y(t) + f(t)y'(t)\rangle\\ &= \langle f'(t)x(t), f'(t)y(t)\rangle + \langle f(t)x'(t), f(t)y'(t)\rangle \\ &= f'(t)\langle x(t), y(t) \rangle + f(t)\langle x'(t), y'(t)\rangle = f'(t){\bf r}(t)+f(t){\bf r}'(t) \end{align*} The remaining proofs are left as exercises (Exercises 69–70).

EXAMPLE 3

Let \({\bf r}(t)=\langle t^2, 5t, 1\rangle\) and \(f(t)= e^{3t}\). Calculate:

• (a) \(\displaystyle{\frac{d}{dt} f(t){\bf r}(t)}\)
• (b) \(\displaystyle{\frac{d}{dt} {\bf r}(f(t))}\)

Solution We have \({\bf r}'(t)=\langle 2t, 5 , 0\rangle\) and \(f'(t)=3e^{3t}\).

• (a) By the Product Rule, \[\begin{align*} \frac{d}{dt} f(t){\bf r}(t) & = f(t){\bf r}'(t)+f'(t){\bf r}(t) = e^{3t}\langle 2t, 5,0 \rangle + 3e^{3t}\langle t^2 , 5t, 1\rangle\\ &= \langle (3t^2+2t)e^{3t}, (15t+5)e^{3t}, 3e^{3t} \rangle \end{align*}\]
• (b) By the Chain Rule, \begin{align*} \frac{d}{dt} {\bf r}(f(t)) &= f'(t){\bf r}'(f(t)) = 3e^{3t}{\bf r}'(e^{3t}) = 3e^{3t}\langle 2e^{3t}, 5 , 0\rangle = \langle 6e^{6t},15e^{3t} , 0\rangle \end{align*}

THEOREM 3 Product Rule for Dot and Cross Products

Assume that \({\bf r}_1(t)\) and \({\bf r}_2(t)\) are differentiable. Then \[ \begin{eqnarray} &&\textrm{Dot Products:}& \frac{d}{dt} \big({\bf r}_1(t)\cdot{\bf r}_2(t)\big) &=& {\bf r}_1(t)\cdot {\bf r}_2'(t)+{\bf r}_1'(t)\cdot {\bf r}_2(t)\tag{4}\\ &&\textrm{Cross Products:}& \frac{d}{dt} \big({\bf r}_1(t)\times {\bf r}_2(t)\big) &=& \bigl[{\bf r}_1(t)\times {\bf r}_2'(t)\bigr] + \bigl[{\bf r}_1'(t)\times {\bf r}_2(t)\bigr]\tag{5} \end{eqnarray} \]

Proof

We verify Eq. (4) for vector-valued functions in the plane. If \({\bf r}_1(t)=\langle x_1(t), y_1(t)\rangle \) and \({\bf r}_2(t)=\langle x_2(t), y_2(t)\rangle \), then \begin{align*} \frac{d}{dt} \bigl({\bf r}_1(t)\cdot{\bf r}_2(t)\bigr) &= \frac{d}{dt} \big(x_1(t)x_2(t)+y_1(t)y_2(t)\big)\\ &=x_1(t)x_2'(t)+x_1'(t)x_2(t)+y_1(t)y_2'(t)+y_1'(t)y_2(t)\\ &= \big(x_1(t)x_2'(t)+ y_1(t)y_2'(t) \big) + \big(x_1'(t)x_2(t)+y_1'(t)y_2(t) \big)\\ &= {\bf r}_1(t)\cdot{\bf r}_2'(t) + {\bf r}_1'(t)\cdot{\bf r}_2(t) \end{align*}

The proof of Eq. (5) is left as an exercise (Exercise 71).

CAUTION

Order is important in the Product Rule for cross products. The first term in Eq. (5) must be written as \[{\bf r}_1(t)\times {\bf r}_2'(t) \] not \({\bf r}_2'(t)\times {\bf r}_1(t)\). Similarly, the second term is \({\bf r}_1'(t)\times {\bf r}_2(t)\). Why is order not a concern for dot products?

EXAMPLE 4

Prove the formula \(\displaystyle{\frac{d}{dt} \bigl({\bf r}(t)\times{\bf r}'(t)\bigr) = {\bf r}(t)\times{\bf r}”(t)}\).

Solution By the Product Formula for cross products, \[ \frac{d}{dt} \bigl({\bf r}(t)\times{\bf r}'(t)\bigr) = {\bf r}(t)\times{\bf r}”(t) + \underbrace{{\bf r}'(t)\times{\bf r}'(t)}_{\textrm{Equals \({\bf 0}\)}} = {\bf r}(t)\times{\bf r}”(t) \]

Here, \({\bf r}'(t)\times{\bf r}'(t) ={\bf 0}\) because the cross product of a vector with itself is zero.

Note

Although it has been our convention to regard all vectors as based at the origin, the tangent vector \({\bf r}'(t)\) is an exception; we visualize it as a vector based at the terminal point of \({\bf r}(t)\). This makes sense because \({\bf r}'(t)\) then appears as a vector tangent to the curve ( Figure 13.10).

EXAMPLE 5 Plotting Tangent Vectors

Plot \({\bf r}(t) = \langle \cos t, \sin t, 4\cos^2t\rangle\) together with its tangent vectors at \(t = \frac{\pi}4\) and \(\frac{3\pi}2\). Find a parametrization of the tangent line at \(t = \frac{\pi}4\).

Solution  The derivative is \(\displaystyle{ {\bf r}'(t) = \langle -\sin t, \cos t, -8\cos t\sin t\rangle}\), and thus the tangent vectors at \(t = \frac{\pi}4\) and \(\frac{3\pi}2\) are \[ {\bf r}'\left(\frac{\pi}4\right) = \langle -\frac {\sqrt 2}2, \frac{\sqrt 2}2, -4\rangle,\qquad {\bf r}'\left(\frac{3\pi}2\right) = \langle 1, 0, 0\rangle \]

Figure 13.11 shows a plot of \({\bf r}(t)\) with \({\bf r}'\big(\tfrac{\pi}4\big)\) based at \({\bf r}\big(\tfrac{\pi}4\big)\) and \({\bf r}'\big(\tfrac{3\pi}2\big)\) based at \({\bf r}\big(\tfrac{3\pi}2\big)\). At \(t=\frac{\pi}4\), \({\bf r}\big(\frac{\pi}4\big) = \langle \frac{\sqrt 2}2, \frac{\sqrt 2}2, 2 \rangle\) and thus the tangent line is parametrized by \[ {\bf L}(t) = {\bf r}\left(\frac{\pi}4\right) + t\,{\bf r}'\left(\frac{\pi}4\right) = \langle \frac{\sqrt 2}2, \frac{\sqrt 2}2, 2 \rangle + t \langle -\frac {\sqrt 2}2, \frac{\sqrt 2}2, -4\rangle \]

EXAMPLE 6 Horizontal Tangent Vectors on the Cycloid

The function \[ {\bf r}(t) = \langle t - \sin t, 1-\cos t\rangle \] traces a cycloid. Find the points where:

• (a) \({\bf r}'(t)\) is horizontal and nonzero.
• (b) \({\bf r}'(t)\) is the zero vector.

Solution The tangent vector is \({\bf r}'(t) = \langle 1 - \cos t, \sin t\rangle\). The \(y\)-component of \({\bf r}'(t)\) is zero if \(\sin t = 0\)—that is, if \(t = 0, \pi, 2\pi, \ldots.\) We have \begin{eqnarray*} {\bf r}(0) &=& \langle 0,0\rangle,& {\bf r}'(0) &=& \langle 1-\cos 0,\sin 0 \rangle = \langle 0,0\rangle\quad\textrm{(zero vector)}\\ {\bf r}( \pi ) &=& \langle \pi,2\rangle,& {\bf r}'(\pi) &=& \langle 1-\cos \pi,\sin \pi \rangle = \langle 2,0\rangle\quad\textrm{(horizontal)} \end{eqnarray*}

By periodicity, we conclude that \({\bf r}'(t)\) is nonzero and horizontal for \(t =\pi, 3\pi, 5\pi,\dots\) and \({\bf r}'(t)={\bf 0}\) for \(t=0, 2\pi, 4\pi,\dots\) (Figure 13.12).

CONCEPTUAL INSIGHT

The cycloid in Figure 13.12 has sharp points called cusps at points where \(x= 0, 2\pi, 4\pi,\dots\). If we represent the cycloid as the graph of a function \(y=f(x)\), then \(f'(x)\) does not exist at these points. By contrast, the vector derivative \({\bf r}'(t)=\langle 1-\cos t, \sin t\rangle\) exists for all \(t\), but \({\bf r}'(t)={\bf 0}\) at the cusps. In general, \({\bf r}'(t)\) is a direction vector for the tangent line whenever it exists, but we get no information about the tangent line (which may or may not exist) at points where \({\bf r}'(t)={\bf 0}\).

EXAMPLE 7 Orthogonality of \({\bf r}\) and \({\bf r}'\) When \({\bf r}\) Has Constant Length

Prove that if \({\bf r}(t)\) has constant length, then \({\bf r}(t)\) is orthogonal to \({\bf r}'(t)\).

Solution  By the Product Rule for Dot Products, \[ \frac{d}{dt} \|{{\bf r}(t)}\|^2 = \frac{d}{dt} \bigl({\bf r}(t)\cdot{\bf r}(t)\bigr) = {\bf r}(t)\cdot{\bf r}'(t)+{\bf r}'(t)\cdot{\bf r}(t) = 2{\bf r}(t)\cdot{\bf r}'(t) \]

This derivative is zero because \(\displaystyle{\|{{\bf r}(t)}}\|\) is constant. Therefore \displaystyle{{\bf r}(t)\cdot{\bf r}'(t)=0}, and \({\bf r}(t)\) is orthogonal to \({\bf r}'(t)\) [or \({\bf r}'(t)=\mathbf{0}\)].

GRAPHICAL INSIGHT

The result of Example 7 has a geometric explanation. A vector parametrization \({\bf r}(t)\) consisting of vectors of constant length \(R\) traces a curve on the surface of a sphere of radius \(R\) with center at the origin (Figure 13.13). Thus \({\bf r}'(t)\) is tangent to this sphere. But any line that is tangent to a sphere at a point \(P\) is orthogonal to the radial vector through \(P\), and thus \({\bf r}(t)\) is orthogonal to \({\bf r}'(t)\).

THEOREM 4

If \({\bf R}_1(t)\) and \({\bf R}_2(t)\) are differentiable and \({\bf R}_1'(t) = {\bf R}_2'(t)\), then \[ {\bf R}_1(t)={\bf R}_2(t)+c \] for some constant vector \(c\).

Fundamental Theorem of Calculus for Vector-Valued Functions

If \({\bf r}(t)\) is continuous on \([a,b]\), and \({\bf R}(t)\) is an antiderivative of \({\bf r}(t)\), then \[ \int_a^b {\bf r}(t)\, dt = {\bf R}(b)-{\bf R}(a) \]

EXAMPLE 8 Finding Position via Vector-Valued Differential Equations

The path of a particle satisfies \[ \frac{d{\bf r}}{dt} = \langle 1-6\sin 3t, \dfrac15t \rangle \] Find the particle’s location at \(t=4\) if \({\bf r}(0) = \langle 4,1\rangle\).

Figure 13.14: Particle path \[{\bf r}(t) = \langle t+2\cos 3t + 2, \tfrac1{10}t^2+1\rangle \]

Solution  The general solution is obtained by integration: \[ {\bf r}(t) = \int \langle 1-6\sin 3t, \dfrac15t \rangle\, dt = \langle t+2\cos 3t, \dfrac1{10}t^2\rangle + {\bf c} \]

The initial condition \({\bf r}(0)= \langle 4, 1\rangle\) gives us \[ {\bf r}(0) = \langle 2 , 0 \rangle + {\bf c} = \langle 4, 1\rangle \]

Therefore, \(\displaystyle{{\bf c} = \langle 2, 1\rangle}\) and (Figure 13.14) \[ {\bf r}(t) = \langle t+2\cos 3t, \dfrac1{10}t^2 \rangle+ \langle 2, 1\rangle= \langle t+2\cos 3t + 2, \dfrac1{10}t^2+1\rangle \]

The particle’s position at \(t=4\) is \[ {\bf r}(4)=\langle 4+2\cos 12 + 2, \dfrac1{10}(4^2)+1\rangle \approx \langle 7.69, 2.6\rangle \]