9.7: Derivatives and Integrals of Vector-Valued Functions

The Derivative

In single variable calculus, we define the derivative, $f'\text{,}$ of a given function $f$ by

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h},$$

provided the limit exists. At a given value of $a\text{,}$ $f'(a)$ measures the instantaneous rate of change of $f\text{,}$ and also tells us the slope of the tangent line to the curve $y = f(x)$ at the point $(a, f(a))\text{.}$ The definition of the derivative extends naturally to vector-valued functions and curves in space.

Definition $9.7.1$: Derivative of a Vector-valued Function

The derivative of a vector-valued function $\mathbf{r}$ is defined to be

$$\mathbf{r}'(t) = \lim_{h \to 0} \frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}$$

for those values of $t$ at which the limit exists. We also use the notation $\frac{d\mathbf{r}}{dt}$ and $\frac{d}{dt}[\mathbf{r}(t)]$ for $\mathbf{r}'(t)\text{.}$

Activity $9.7.2$

Let's investigate how we can interpret the derivative $\mathbf{r}'(t)\text{.}$ Let $\mathbf{r}$ be the vector-valued function whose graph is shown in Figure $9.7.2$, and let $h$ be a scalar that represents a small change in time. The vector $\mathbf{r}(t)$ is the blue vector in Figure $9.7.1$ and $\mathbf{r}(t+h)$ is the green vector.

1. Is the quantity $\mathbf{r}(t+h)-\mathbf{r}(t)$ a vector or a scalar? Identify this object in Figure $9.7.1$.
2. Is $\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}$ a vector or a scalar? Sketch a representative vector $\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}$ with $h \lt 1$ in Figure $9.7.2$.
3. Think of $\mathbf{r}(t)$ as providing the position of an object moving along the curve these vectors trace out. What do you think that the vector $\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}$ measures? Why? (Hint: You might think analogously about difference quotients such as $\frac{f(x+h) - f(x)}{h}$ or $\frac{s(t+h) - s(t)}{h}$ from calculus I.)
4. Figure $9.7.2$ presents three snapshots of the vectors $\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}$ as we let $h \to 0\text{.}$ Write 2-3 sentences to describe key attributes of the vector
   $$\lim_{h \to 0} \frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}.$$

   (Hint: Compare to limits such as $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ or $\lim_{h \to 0} \frac{s(t+h) - s(t)}{h}$ from calculus I, keeping in mind that in three dimensions there is no general concept of slope.)

As Activity $9.7.2$ indicates, if $\mathbf{r}(t)$ determines the position of an object at time $t\text{,}$ then $\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}$ represents the average rate of change in the position of the object over the interval $[t,t+h]\text{,}$ which is also the average velocity of the object on this interval. Thus, the derivative

$$\mathbf{r}'(t) = \lim_{h \to 0} \frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}$$

is the instantaneous rate of change of $\mathbf{r}(t)$ at time $t$ (for those values of $t$ for which the limit exists), so $\mathbf{r}'(t) = \mathbf{v}(t)$ is the instantaneous velocity of the object at time $t\text{.}$ Furthermore, we can interpret the derivative $\mathbf{r}'(t)$ as the direction vector of the line tangent to the graph of $\mathbf{r}$ at the value $t\text{.}$

Similarly,

$$\mathbf{v}'(t) = \mathbf{r}''(t) = \lim_{h \to 0} \frac{\mathbf{v}(t+h)-\mathbf{v}(t)}{h}$$

is the instantaneous rate of change of the velocity of the object at time $t\text{,}$ for those values of $t$ for which the limits exists, and thus $\mathbf{v}'(t) = \mathbf{a}(t)$ is the acceleration of the moving object.

Note well: Both the velocity and acceleration are vector quantities: they have magnitude and direction. By contrast, the magnitude of the velocity vector, $| \mathbf{v}(t) |\text{,}$ which is the speed of the object at time $t\text{,}$ is a scalar quantity.

Computing Derivatives

As we learned in single variable calculus, computing derivatives from the definition is often difficult. Fortunately, properties of the limit make it straightforward to calculate the derivative of a vector-valued function similar to how we developed shortcut differentiation rules in calculus I. To see why, recall that the limit of a sum is the sum of the limits, and that we can remove constant factors from limits. Thus, as we observed in a particular example in Preview Activity $9.7.1$, if $\mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k}\text{,}$ it follows that

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\mathbf{j} + [z(t+h)-z(t)] \mathbf{k}}{h}\\[4pt] = \mathstrut & \left(\lim_{h \to 0} \frac{x(t+h)-x(t)}{h} \right) \mathbf{i} + \left( \lim_{h \to 0} \frac{y(t+h)-y(t)}{h} \right) \mathbf{j}\\[4pt] \mathstrut & + \left( \lim_{h \to 0} \frac{z(t+h)-z(t)}{h} \right)\mathbf{k}\\[4pt] \mathstrut & = x&#x27;(t)\mathbf{i} + y&#x27;(t) \mathbf{j} + z&#x27;(t) \mathbf{k}. \end{align*}</annotation></semantics></math></span>

Thus, we can calculate the derivative of a vector-valued function by simply differentiating its components.

The derivative of a vector-valued function

If $\mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k}\text{,}$ then

$$\frac{d}{dt} \mathbf{r}(t) = x'(t) \mathbf{i} + y'(t) \mathbf{j} + z'(t) \mathbf{k}$$

for those values of $t$ at which $x\text{,}$ $y\text{,}$ and $z$ are differentiable.

Activity $9.7.3$

For each of the following vector-valued functions, find $\mathbf{r}'(t)\text{.}$

1. $\mathbf{r}(t) = \langle \cos(t), t\sin(t), \ln(t) \rangle\text{.}$
2. $\mathbf{r}(t) = \langle t^2 + 3t, e^{-2t}, \frac{t}{t^2 + 1} \rangle\text{.}$
3. $\mathbf{r}(t) = \langle \tan(t), \cos(t^2), te^{-t} \rangle\text{.}$
4. $\mathbf{r}(t) = \langle \sqrt{t^4 + 4}, \sin(3t), \cos(4t) \rangle\text{.}$

In first-semester calculus, we developed several important differentiation rules, including the constant multiple, product, quotient, and chain rules. For instance, recall that we formally state the product rule as

$$\frac{d}{dx}[f(x) \cdot g(x)] = f(x)\cdot g'(x) + g(x)\cdot f'(x).$$

There are several analogous rules for vector-valued functions, including a product rule for scalar functions and vector-valued functions. These rules, which are easily verified, are summarized as follows.

Properties of derivatives of vector-valued functions

Let $f$ be a differentiable real-valued function of a real variable $t$ and let $\mathbf{r}$ and $\mathbf{s}$ be differentiable vector-valued functions of the real variable $t\text{.}$ Then

1. $\displaystyle \frac{d}{dt} \left[\mathbf{r}(t) + \mathbf{s}(t) \right] = \mathbf{r}'(t) + \mathbf{s}'(t)$
2. $\displaystyle \frac{d}{dt} [f(t) \mathbf{r}(t)] = f(t) \mathbf{r}'(t) + f'(t) \mathbf{r}(t)$
3. $\displaystyle \frac{d}{dt} \left[\mathbf{r}(t) \cdot \mathbf{s}(t) \right] = \mathbf{r}'(t) \cdot \mathbf{s}(t) + \mathbf{r}(t) \cdot \mathbf{s}'(t)$
4. $\displaystyle \frac{d}{dt} \left[\mathbf{r}(t) \times \mathbf{s}(t) \right] = \mathbf{r}'(t) \times \mathbf{s}(t) + \mathbf{r}(t) \times \mathbf{s}'(t)$
5. $\frac{d}{dt} \left[\mathbf{r}(f(t))\right] = f'(t) \mathbf{r}'(f(t))\text{.}$

Note well. When applying these properties, use care to interpret the quantities involved as either scalars or vectors. For example, $\mathbf{r}(t) \cdot \mathbf{s}(t)$ defines a scalar function because we have taken the dot product of two vector-valued functions. However, $\mathbf{r}(t) \times \mathbf{s}(t)$ defines a vector-valued function since we have taken the cross product of two vector-valued functions.

Activity $9.7.4$

The left side of Figure $9.7.3$ shows the curve described by the vector-valued function $\mathbf{r}$ defined by

$$\mathbf{r}(t) = \left\langle 2t-\frac12 t^2 + 1, t-1\right\rangle.$$

1. Find the object's velocity $\mathbf{v}(t)\text{.}$
2. Find the object's acceleration $\mathbf{a}(t)\text{.}$
3. Indicate on the left of Figure $9.7.4$ the object's position, velocity and acceleration at the times $t=0, 2, 4\text{.}$ Draw the velocity and acceleration vectors with their tails placed at the object's position.
4. Recall that the speed is $|\mathbf{v}| = \sqrt{\mathbf{v}\cdot\mathbf{v}}\text{.}$ Find the object's speed and graph it as a function of time $t$ on the right of Figure $9.7.4$. When is the object's speed the slowest? When is the speed increasing? When is it decreasing?
5. What seems to be true about the angle between $\mathbf{v}$ and $\mathbf{a}$ when the speed is at a minimum? What is the angle between $\mathbf{v}$ and $\mathbf{a}$ when the speed is increasing? when the speed is decreasing?
6. Since the square root is an increasing function, we see that the speed increases precisely when $\mathbf{v}\cdot\mathbf{v}$ is increasing. Use the product rule for the dot product to express $\frac{d}{dt}(\mathbf{v}\cdot\mathbf{v})$ in terms of the velocity $\mathbf{v}$ and acceleration $\mathbf{a}\text{.}$ Use this to explain why the speed is increasing when $\mathbf{v}\cdot\mathbf{a} > 0$ and decreasing when $\mathbf{v}\cdot\mathbf{a} \lt 0\text{.}$ Compare this to part (d).
7. Show that the speed's rate of change is
   $$\frac{d}{dt}|\mathbf{v}(t)| = comp_{\mathbf{v}} \mathbf{a}.$$

Tangent Lines

One of the most important ideas in first-semester calculus is that a differentiable function is locally linear: that is, when viewed up close, the curve generated by a differentiable function looks very much like a line. Indeed, when we zoom in sufficiently far on a particular point, the curve looks indistinguishable from its tangent line.

In the same way, we expect that a smooth curve in 3-space will be locally linear. In the following activity, we investigate how to find the tangent line to such a curve. Recall from our work in Section 9.5 that the vector equation of a line that passes through the point at the tip of the vector $\mathbf{L}_0 = \langle x_0, y_0, z_0 \rangle$ in the direction of the vector $\mathbf{u} = \langle a, b, c \rangle$ can be written as

$$\mathbf{L}(t) = \mathbf{L}_0 + t \mathbf{u}.$$

In parametric form, the line $\mathbf{L}$ is given by

$$x(t) = x_0 + at, \ \ y(t) = y_0 + bt, \ \ z(t) = z_0 + ct.$$

Activity $9.7.5$

Let

$$\mathbf{r}(t) = \cos(t) \mathbf{i} - \sin(t) \mathbf{j} + t \mathbf{k}.$$

Sketch the curve using some appropriate tool.

1. Determine the coordinates of the point on the curve traced out by $\mathbf{r}(t)$ when $t = \pi\text{.}$
2. Find a direction vector for the line tangent to the graph of $\mathbf{r}$ at the point where $t=\pi\text{.}$
3. Find the parametric equations of the line tangent to the graph of $\mathbf{r}$ when $t=\pi\text{.}$
4. Sketch a plot of the curve $\mathbf{r}(t)$ and its tangent line near the point where $t = \pi\text{.}$ In addition, include a sketch of $\mathbf{r}'(\pi)\text{.}$ What is the important role of $\mathbf{r}'(\pi)$ in this activity?

We see that our work in Activity $9.7.5$ can be generalized. Given a differentiable vector-valued function $\mathbf{r}\text{,}$ the tangent line to the curve at the input value $a$ is given by

$$\mathbf{L}(t) = \mathbf{r}(a) + t \mathbf{r}'(a).\tag{$$$9.7.1$}

Here we see that because the tangent line is determined entirely by a given point and direction, the point is provided by the function $\mathbf{r}\text{,}$ evaluated at $t = a\text{,}$ while the direction is provided by the derivative, $\mathbf{r}'\text{,}$ again evaluated at $t = a\text{.}$ Note how analogous the formula for $\mathbf{L}(t)$ is to the tangent line approximation from single-variable calculus: in that context, for a given function $y = f(x)$ at a value $x = a\text{,}$ we found that the tangent line can be expressed by the linear function $y = L(x)$ whose formula is

$$L(x) = f(a) + f'(a)(x-a).$$

Equation ($9.7.1$) for the tangent line $\mathbf{L}(t)$ to the vector-valued function $\mathbf{r}(t)$ is nearly identical. Indeed, because there are multiple parameterizations for a single line, it is even possible to write the parameterization as

$$\mathbf{L}(t) = \mathbf{r}(a) + (t-a) \mathbf{r}'(a).\tag{$$$9.7.2$}

(For example, in Equation ($9.7.1$), $\mathbf{L}(0) = \mathbf{r}(a)\text{,}$ so the line's parameterization “starts” at $t = 0\text{.}$ When we write the parameterization in the form of Equation ($9.7.2$), $\mathbf{L}(a) = \mathbf{r}(a)\text{,}$ so the line's parameterization “starts” at $t = a\text{.}$)

As we will learn more in Chapter 10, a smooth surface in 3-space is also locally linear. That means that the surface will look like a plane, which we call its tangent plane, as we zoom in on the graph. It is possible to use tangent lines to traces of the surface to generate a formula for the tangent plane; see Exercise $9.7.7$.15 at the end of this section for more details.

Projectile Motion

Any time that an object is launched into the air with a given velocity and launch angle, the path the object travels is determined almost exclusively by the force of gravity. Whether in sports such as archery or shotput, in military applications with artillery, or in important fields like firefighting, it is important to be able to know when and where a launched projectile will land. We can use our knowledge of vector-valued functions in order to completely determine the path traveled by an object that is launched from a given position at a given angle from the horizontal with a given initial velocity.

Assume we fire a projectile from a launcher and the only force acting on the fired object is the force of gravity pulling down on the object. That is, we assume no effect due to spin, wind, or air resistance. With these assumptions, the motion of the object will be planar, so we can also assume that the motion occurs in two-dimensional space. Suppose we launch the object from an initial position $(x_0, y_0)$ at an angle $\theta$ with the positive $x$-axis as illustrated in Figure $9.7.5$, and that we fire the object with an initial speed of $v_0 = |\mathbf{v}(0)|\text{,}$ where $\mathbf{v}(t)$ is the velocity vector of the object at time $t\text{.}$ Assume $g$ is the positive constant acceleration force due to gravity, which acts to pull the fired object toward the ground (in the negative $y$ direction). Note particularly that there is no external force acting on the object to move it in the $x$ direction.

We first observe that since gravity only acts in the downward direction and that the acceleration due to gravity is constant, the acceleration vector is $\langle 0, -g \rangle\text{.}$ That is, $\mathbf{a}(t) = \langle 0, -g \rangle\text{.}$ We may use this fact about acceleration, together with the initial position and initial velocity in order to fully determine the position $\mathbf{r}(t)$ of the object at time $t\text{.}$ In Exercise $9.7.5$, you can work through the details to show that the following general formula holds.

The motion of a projectile

If an object is launched from a point $(x_0,y_0)$ with initial velocity $v_0$ at an angle $\theta$ with the horizontal, then the position of the object at time $t$ is given by

$$\mathbf{r}(t) = \left\langle v_0 \cos(\theta)t + x_0, -\frac{g}{2}t^2 + v_0 \sin(\theta)t + y_0 \right\rangle.$$

This assumes that the only force acting on the object is the acceleration $g$ due to gravity.