Section 13.2: Derivatives and Integrals of Vector Functions

Later in this chapter we are going to use vector functions to describe the motion of planets and other objects through space. Here we prepare the way by developing the calculus of vector functions.

Derivatives

The derivative \(\mathbf{r}'\) of a vector function \(\mathbf{r}\) is defined in much the same way as for real-valued functions:

\[ \frac{d\mathbf{r}}{dt} = \mathbf{r}'(t) = \lim_{h \to 0} \frac{\mathbf{r}(t+h) - \mathbf{r}(t)}{h} \tag{1} \]

if this limit exists. The geometric significance of this definition is shown in Figure 1. If the points P and Q have position vectors \(\mathbf{r}(t)\) and \(\mathbf{r}(t+h)\), then \(\vec{PQ}\) represents the vector \(\mathbf{r}(t+h) - \mathbf{r}(t)\), which can therefore be regarded as a secant vector. If \(h > 0\), the scalar multiple \((1/h)(\mathbf{r}(t+h) - \mathbf{r}(t))\) has the same direction as \(\mathbf{r}(t+h) - \mathbf{r}(t)\). As \(h \to 0\), it appears that this vector approaches a vector that lies on the tangent line. For this reason, the vector \(\mathbf{r}'(t)\) is called the tangent vector to the curve defined by \(\mathbf{r}\) at the point P, provided that \(\mathbf{r}'(t)\) exists and \(\mathbf{r}'(t) \neq \mathbf{0}\). The tangent line to C at P is defined to be the line through P parallel to the tangent vector \(\mathbf{r}'(t)\). We will also have occasion to consider the unit tangent vector, which is

\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \]

The following theorem gives us a convenient method for computing the derivative of a vector function \(\mathbf{r}\): just differentiate each component of \(\mathbf{r}\).

Theorem 2 If \(\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}\), where f, g, and h are differentiable functions, then \[ \mathbf{r}'(t) = \langle f'(t), g'(t), h'(t) \rangle = f'(t)\mathbf{i} + g'(t)\mathbf{j} + h'(t)\mathbf{k} \]

PROOF \[ \mathbf{r}'(t) = \lim_{\Delta t \to 0} \frac{1}{\Delta t}[\mathbf{r}(t+\Delta t) - \mathbf{r}(t)] \] \[ = \lim_{\Delta t \to 0} \frac{1}{\Delta t}[\langle f(t+\Delta t), g(t+\Delta t), h(t+\Delta t) \rangle - \langle f(t), g(t), h(t) \rangle] \] \[ = \lim_{\Delta t \to 0} \left\langle \frac{f(t+\Delta t) - f(t)}{\Delta t}, \frac{g(t+\Delta t) - g(t)}{\Delta t}, \frac{h(t+\Delta t) - h(t)}{\Delta t} \right\rangle \] \[ = \left\langle \lim_{\Delta t \to 0} \frac{f(t+\Delta t) - f(t)}{\Delta t}, \lim_{\Delta t \to 0} \frac{g(t+\Delta t) - g(t)}{\Delta t}, \lim_{\Delta t \to 0} \frac{h(t+\Delta t) - h(t)}{\Delta t} \right\rangle \] \[ = \langle f'(t), g'(t), h'(t) \rangle \]

EXAMPLE 1 (a) Find the derivative of \(\mathbf{r}(t) = (1+t^3)\mathbf{i} + te^{-t}\mathbf{j} + \sin 2t \mathbf{k}\). (b) Find the unit tangent vector at the point where \(t=0\).

EXAMPLE 2 For the curve \(\mathbf{r}(t) = \sqrt{t}\mathbf{i} + (2-t)\mathbf{j}\), find \(\mathbf{r}'(t)\) and sketch the position vector \(\mathbf{r}(1)\) and the tangent vector \(\mathbf{r}'(1)\).

EXAMPLE 3 Find parametric equations for the tangent line to the helix with parametric equations \[ x = 2\cos t \quad y = \sin t \quad z=t \] at the point \((0, 1, \pi/2)\).

Just as for real-valued functions, the second derivative of a vector function \(\mathbf{r}\) is the derivative of \(\mathbf{r}'\), that is, \(\mathbf{r}'' = (\mathbf{r}')'\). For instance, the second derivative of the function in Example 3 is \[ \mathbf{r}''(t) = \langle -2\cos t, -\sin t, 0 \rangle \]

The next theorem shows that the differentiation formulas for real-valued functions have their counterparts for vector-valued functions.

Theorem 3 Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are differentiable vector functions, c is a scalar, and f is a real-valued function. Then 1. \(\frac{d}{dt}[\mathbf{u}(t) + \mathbf{v}(t)] = \mathbf{u}'(t) + \mathbf{v}'(t)\) 2. \(\frac{d}{dt}[c\mathbf{u}(t)] = c\mathbf{u}'(t)\) 3. \(\frac{d}{dt}[f(t)\mathbf{u}(t)] = f'(t)\mathbf{u}(t) + f(t)\mathbf{u}'(t)\) 4. \(\frac{d}{dt}[\mathbf{u}(t) \cdot \mathbf{v}(t)] = \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t)\) 5. \(\frac{d}{dt}[\mathbf{u}(t) \times \mathbf{v}(t)] = \mathbf{u}'(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}'(t)\) 6. \(\frac{d}{dt}[\mathbf{u}(f(t))] = f'(t)\mathbf{u}'(f(t))\) (Chain Rule)

This theorem can be proved either directly from Definition 1 or by using Theorem 2 and the corresponding differentiation formulas for real-valued functions. The proof of Formula 4 follows; the remaining formulas are left as exercises.

PROOF OF FORMULA 4 Let \[ \mathbf{u}(t) = \langle f_1(t), f_2(t), f_3(t) \rangle \quad \mathbf{v}(t) = \langle g_1(t), g_2(t), g_3(t) \rangle \] Then \[ \mathbf{u}(t) \cdot \mathbf{v}(t) = f_1(t)g_1(t) + f_2(t)g_2(t) + f_3(t)g_3(t) = \sum_{i=1}^3 f_i(t)g_i(t) \] so the ordinary Product Rule gives \[ \frac{d}{dt}[\mathbf{u}(t) \cdot \mathbf{v}(t)] = \frac{d}{dt} \sum_{i=1}^3 f_i(t)g_i(t) = \sum_{i=1}^3 \frac{d}{dt}[f_i(t)g_i(t)] \] \[ = \sum_{i=1}^3 [f_i'(t)g_i(t) + f_i(t)g_i'(t)] \] \[ = \sum_{i=1}^3 f_i'(t)g_i(t) + \sum_{i=1}^3 f_i(t)g_i'(t) \] \[ = \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t) \]

EXAMPLE 4 Show that if \(|\mathbf{r}(t)| = c\) (a constant), then \(\mathbf{r}'(t)\) is orthogonal to \(\mathbf{r}(t)\) for all t.

The definite integral of a continuous vector function \(\mathbf{r}(t)\) can be defined in much the same way as for real-valued functions except that the integral is a vector. But then we can express the integral of \(\mathbf{r}\) in terms of the integrals of its component functions f, g, and h as follows.

\[ \int_a^b \mathbf{r}(t) dt = \lim_{n \to \infty} \sum_{i=1}^n \mathbf{r}(t_i^*) \Delta t \] \[ = \lim_{n \to \infty} \left[ \left( \sum_{i=1}^n f(t_i^*) \Delta t \right)\mathbf{i} + \left( \sum_{i=1}^n g(t_i^*) \Delta t \right)\mathbf{j} + \left( \sum_{i=1}^n h(t_i^*) \Delta t \right)\mathbf{k} \right] \] and so \[ \int_a^b \mathbf{r}(t) dt = \left( \int_a^b f(t) dt \right)\mathbf{i} + \left( \int_a^b g(t) dt \right)\mathbf{j} + \left( \int_a^b h(t) dt \right)\mathbf{k} \] This means that we can evaluate an integral of a vector function by integrating each component function. We can extend the Fundamental Theorem of Calculus to continuous vector functions as follows: \[ \int_a^b \mathbf{r}(t) dt = \mathbf{R}(t) \Big]_a^b = \mathbf{R}(b) - \mathbf{R}(a) \] where \(\mathbf{R}\) is an antiderivative of \(\mathbf{r}\), that is, \(\mathbf{R}'(t) = \mathbf{r}(t)\). We use the notation \(\int \mathbf{r}(t) dt\) for indefinite integrals (antiderivatives).

EXAMPLE 5 If \(\mathbf{r}(t) = 2\cos t \mathbf{i} + \sin t \mathbf{j} + 2t \mathbf{k}\), then \[ \int \mathbf{r}(t) dt = \left( \int 2\cos t dt \right)\mathbf{i} + \left( \int \sin t dt \right)\mathbf{j} + \left( \int 2t dt \right)\mathbf{k} \] \[ = 2\sin t \mathbf{i} - \cos t \mathbf{j} + t^2 \mathbf{k} + \mathbf{C} \] where \(\mathbf{C}\) is a vector constant of integration, and \[ \int_0^{\pi/2} \mathbf{r}(t) dt = [2\sin t \mathbf{i} - \cos t \mathbf{j} + t^2 \mathbf{k}]_0^{\pi/2} = 2\mathbf{i} + \mathbf{j} + \frac{\pi^2}{4}\mathbf{k} \]

The figure shows a curve C given by a vector function \(\mathbf{r}(t)\). (a) Draw the vectors \(\mathbf{r}(4.5) - \mathbf{r}(4)\) and \(\mathbf{r}(4.2) - \mathbf{r}(4)\). (b) Draw the vectors \[ \frac{\mathbf{r}(4.5) - \mathbf{r}(4)}{0.5} \quad \text{and} \quad \frac{\mathbf{r}(4.2) - \mathbf{r}(4)}{0.2} \] (c) Write expressions for \(\mathbf{r}'(4)\) and the unit tangent vector \(\mathbf{T}(4)\). (d) Draw the vector \(\mathbf{T}(4)\).

1. Make a large sketch of the curve described by the vector function \(\mathbf{r}(t) = \langle t^2, t \rangle\), \(0 \le t \le 2\), and draw the vectors \(\mathbf{r}(1)\), \(\mathbf{r}(1.1)\), and \(\mathbf{r}(1.1) - \mathbf{r}(1)\).
2. Draw the vector \(\mathbf{r}'(1)\) starting at (1, 1), and compare it with the vector \[ \frac{\mathbf{r}(1.1) - \mathbf{r}(1)}{0.1} \] Explain why these vectors are so close to each other in length and direction.

1. Sketch the plane curve with the given vector equation.
2. Find \(\mathbf{r}'(t)\).
3. Sketch the position vector \(\mathbf{r}(t)\) and the tangent vector \(\mathbf{r}'(t)\) for the given value of t. \(\mathbf{r}(t) = \langle t-2, t^2+1 \rangle, t=-1\)

1. Sketch the plane curve with the given vector equation.
2. Find \(\mathbf{r}'(t)\).
3. Sketch the position vector \(\mathbf{r}(t)\) and the tangent vector \(\mathbf{r}'(t)\) for the given value of t. \(\mathbf{r}(t) = \langle t^2, t^3 \rangle, t=1\)

1. Sketch the plane curve with the given vector equation.
2. Find \(\mathbf{r}'(t)\).
3. Sketch the position vector \(\mathbf{r}(t)\) and the tangent vector \(\mathbf{r}'(t)\) for the given value of t. \(\mathbf{r}(t) = e^{2t}\mathbf{i} + e^t\mathbf{j}, t=0\)

1. Sketch the plane curve with the given vector equation.
2. Find \(\mathbf{r}'(t)\).
3. Sketch the position vector \(\mathbf{r}(t)\) and the tangent vector \(\mathbf{r}'(t)\) for the given value of t. \(\mathbf{r}(t) = e^t\mathbf{i} + 2t\mathbf{j}, t=0\)

1. Sketch the plane curve with the given vector equation.
2. Find \(\mathbf{r}'(t)\).
3. Sketch the position vector \(\mathbf{r}(t)\) and the tangent vector \(\mathbf{r}'(t)\) for the given value of t. \(\mathbf{r}(t) = 4\sin t \mathbf{i} - 2\cos t \mathbf{j}, t=3\pi/4\)

1. Sketch the plane curve with the given vector equation.
2. Find \(\mathbf{r}'(t)\).
3. Sketch the position vector \(\mathbf{r}(t)\) and the tangent vector \(\mathbf{r}'(t)\) for the given value of t. \(\mathbf{r}(t) = (\cos t + 1)\mathbf{i} + (\sin t - 1)\mathbf{j}, t=-\pi/3\)

Find the derivative of the vector function. \(\mathbf{r}(t) = \langle \sqrt{t}-2, 3, 1/t^2 \rangle\)

Find the derivative of the vector function. \(\mathbf{r}(t) = \langle e^t, t-t^3, \ln t \rangle\)

Find the derivative of the vector function. \(\mathbf{r}(t) = t^2\mathbf{i} + \cos(t^2)\mathbf{j} + \sin^2 t \mathbf{k}\)

Find the derivative of the vector function. \(\mathbf{r}(t) = \frac{1}{1+t}\mathbf{i} + \frac{t}{1+t}\mathbf{j} + \frac{t^2}{1+t}\mathbf{k}\)

Find the derivative of the vector function. \(\mathbf{r}(t) = t\sin t \mathbf{i} + e^t\cos t \mathbf{j} + \sin t \cos t \mathbf{k}\)

Find the derivative of the vector function. \(\mathbf{r}(t) = \sin^2(at)\mathbf{i} + te^{bt}\mathbf{j} + \cos^2(ct)\mathbf{k}\)

Find the derivative of the vector function. \(\mathbf{r}(t) = \mathbf{a} + t\mathbf{b} + t^2\mathbf{c}\)

Find the derivative of the vector function. \(\mathbf{r}(t) = t\mathbf{a} \times (\mathbf{b} + t\mathbf{c})\)

Find the unit tangent vector \(\mathbf{T}(t)\) at the point with the given value of the parameter t. \(\mathbf{r}(t) = \langle t^2-2t, 1+3t, \frac{1}{3}t^3 + \frac{1}{2}t^2 \rangle, t=2\)

Find the unit tangent vector \(\mathbf{T}(t)\) at the point with the given value of the parameter t. \(\mathbf{r}(t) = \langle \tan^{-1} t, 2e^{2t}, 8te^t \rangle, t=0\)

Find the unit tangent vector \(\mathbf{T}(t)\) at the point with the given value of the parameter t. \(\mathbf{r}(t) = \cos t \mathbf{i} + 3t \mathbf{j} + 2\sin 2t \mathbf{k}, t=0\)

Find the unit tangent vector \(\mathbf{T}(t)\) at the point with the given value of the parameter t. \(\mathbf{r}(t) = \sin^2 t \mathbf{i} + \cos^2 t \mathbf{j} + \tan^2 t \mathbf{k}, t=\pi/4\)

If \(\mathbf{r}(t) = \langle t, t^2, t^3 \rangle\), find \(\mathbf{r}'(t)\), \(\mathbf{T}(1)\), \(\mathbf{r}''(t)\), and \(\mathbf{r}'(t) \times \mathbf{r}''(t)\).

If \(\mathbf{r}(t) = \langle e^{2t}, e^{-2t}, te^{2t} \rangle\), find \(\mathbf{T}(0)\), \(\mathbf{r}''(0)\), and \(\mathbf{r}'(t) \cdot \mathbf{r}''(t)\).

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. \(x = t^2+1, y=4\sqrt{t}, z=e^{t^2-t}; (2, 4, 1)\)

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. \(x = \ln(t+1), y=t\cos 2t, z=2^t; (0, 0, 1)\)

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. \(x = e^{-t}\cos t, y=e^{-t}\sin t, z=e^{-t}; (1, 0, 1)\)

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. \(x = \sqrt{t^2+3}, y=\ln(t^2+3), z=t; (2, \ln 4, 1)\)

Find a vector equation for the tangent line to the curve of intersection of the cylinders \(x^2+y^2=25\) and \(y^2+z^2=20\) at the point \((3, 4, 2)\).

Find the point on the curve \(\mathbf{r}(t) = \langle 2\cos t, 2\sin t, e^t \rangle\), \(0 \le t \le \pi\), where the tangent line is parallel to the plane \(\sqrt{3}x+y=1\).

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustrate by graphing both the curve and the tangent line on a common screen. \(x=t, y=e^{-t}, z=2t-t^2; (0, 1, 0)\)

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustrate by graphing both the curve and the tangent line on a common screen. \(x=2\cos t, y=2\sin t, z=4\cos 2t; (\sqrt{3}, 1, 2)\)

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustrate by graphing both the curve and the tangent line on a common screen. \(x=t\cos t, y=t, z=t\sin t; (-\pi, \pi, 0)\)

1. Find the point of intersection of the tangent lines to the curve \(\mathbf{r}(t) = \langle \sin \pi t, 2\sin \pi t, \cos \pi t \rangle\) at the points where \(t=0\) and \(t=0.5\).
2. Illustrate by graphing the curve and both tangent lines.

The curves \(\mathbf{r}_1(t) = \langle t, t^2, t^3 \rangle\) and \(\mathbf{r}_2(t) = \langle \sin t, \sin 2t, t \rangle\) intersect at the origin. Find their angle of intersection correct to the nearest degree.

At what point do the curves \(\mathbf{r}_1(t) = \langle t, 1-t, 3+t^2 \rangle\) and \(\mathbf{r}_2(s) = \langle 3-s, s-2, s^2 \rangle\) intersect? Find their angle of intersection correct to the nearest degree.

Evaluate the integral. \(\int_0^2 (t\mathbf{i} - t^3\mathbf{j} + 3t^5\mathbf{k}) dt\)

Evaluate the integral. \(\int_0^1 (t^{3/2}\mathbf{i} + (t+1)\sqrt{t}\mathbf{k}) dt\)

Evaluate the integral. \(\int_0^1 \left( \frac{1}{t+1}\mathbf{i} + \frac{1}{t^2+1}\mathbf{j} + \frac{t}{t^2+1}\mathbf{k} \right) dt\)

Evaluate the integral. \(\int_0^{\pi/4} (\sec t \tan t \mathbf{i} + t\cos 2t \mathbf{j} + \sin^2 2t \cos 2t \mathbf{k}) dt\)

Evaluate the integral. \(\int (\sec^2 t \mathbf{i} + t(t^2+1)^3\mathbf{j} + t^2\ln t \mathbf{k}) dt\)

Evaluate the integral. \(\int \left( te^{2t}\mathbf{i} + \frac{t}{1-t}\mathbf{j} + \frac{1}{\sqrt{1-t^2}}\mathbf{k} \right) dt\)

Find \(\mathbf{r}(t)\) if \(\mathbf{r}'(t) = 2t\mathbf{i} + 3t^2\mathbf{j} + \sqrt{t}\mathbf{k}\) and \(\mathbf{r}(1) = \mathbf{i} + \mathbf{j}\).

Find \(\mathbf{r}(t)\) if \(\mathbf{r}'(t) = t\mathbf{i} + e^t\mathbf{j} + te^t\mathbf{k}\) and \(\mathbf{r}(0) = \mathbf{i} + \mathbf{j} + \mathbf{k}\).

If \(\mathbf{u}(t) = \langle \sin t, \cos t, t \rangle\) and \(\mathbf{v}(t) = \langle t, \cos t, \sin t \rangle\), use Formula 4 of Theorem 3 to find \(\frac{d}{dt}[\mathbf{u}(t) \cdot \mathbf{v}(t)]\)

If \(\mathbf{u}\) and \(\mathbf{v}\) are the vector functions in Exercise 47, use Formula 5 of Theorem 3 to find \(\frac{d}{dt}[\mathbf{u}(t) \times \mathbf{v}(t)]\)

Find \(f'(2)\), where \(f(t) = \mathbf{u}(t) \cdot \mathbf{v}(t)\), \(\mathbf{u}(2) = \langle 1, 2, -1 \rangle\), \(\mathbf{u}'(2) = \langle 3, 0, 4 \rangle\), and \(\mathbf{v}(t) = \langle t, t^2, t^3 \rangle\).

If \(\mathbf{r}(t) = \mathbf{u}(t) \times \mathbf{v}(t)\), where \(\mathbf{u}\) and \(\mathbf{v}\) are the vector functions in Exercise 49, find \(\mathbf{r}'(2)\).

If \(\mathbf{r}(t) = \mathbf{a}\cos \omega t + \mathbf{b}\sin \omega t\), where \(\mathbf{a}\) and \(\mathbf{b}\) are constant vectors, show that \(\mathbf{r}(t) \times \mathbf{r}'(t) = \omega \mathbf{a} \times \mathbf{b}\).

If \(\mathbf{r}\) is the vector function in Exercise 51, show that \(\mathbf{r}''(t) + \omega^2\mathbf{r}(t) = \mathbf{0}\).

Show that if \(\mathbf{r}\) is a vector function such that \(\mathbf{r}''\) exists, then \(\frac{d}{dt}[\mathbf{r}(t) \times \mathbf{r}'(t)] = \mathbf{r}(t) \times \mathbf{r}''(t)\)

Find an expression for \(\frac{d}{dt}[\mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t))]\).

If \(\mathbf{r}(t) \neq \mathbf{0}\), show that \(\frac{d}{dt}|\mathbf{r}(t)| = \frac{1}{|\mathbf{r}(t)|}\mathbf{r}(t) \cdot \mathbf{r}'(t)\). [Hint: \(|\mathbf{r}(t)|^2 = \mathbf{r}(t) \cdot \mathbf{r}(t)\)]

If a curve has the property that the position vector \(\mathbf{r}(t)\) is always perpendicular to the tangent vector \(\mathbf{r}'(t)\), show that the curve lies on a sphere with center the origin.

If \(\mathbf{u}(t) = \mathbf{r}(t) \cdot [\mathbf{r}'(t) \times \mathbf{r}''(t)]\), show that \(\mathbf{u}'(t) = \mathbf{r}(t) \cdot [\mathbf{r}'(t) \times \mathbf{r}'''(t)]\)

Show that the tangent vector to a curve defined by a vector function \(\mathbf{r}(t)\) points in the direction of increasing t. [Hint: Refer to Figure 1 and consider the cases \(h>0\) and \(h<0\) separately.]