Vector Functions

A function $f: \real \to \real^n$ is called a vector function in $\real^n$ . We'll focus on vector functions in the plane $\real^2$ and space $\real^3$ , but everything goes over to $\real^n$ without any difficulty.

Example. Find $f(0)$ and $f(2)$ for $f: \real \to \real^3$ given by

$$f(t) = (t^2 + 6, \sin \pi t, \ln (t + 2)).$$

$$f(0) = (6, 0, \ln 2) \quad\hbox{and}\quad f(2) = (10, 0, \ln 4).\quad\halmos$$

It is no accident that the function in the last example looked like a parametric curve, because that's what it is. A vector function in $\real^n$ is a curve in $\real^n$ .

Our main concern is the calculus of vector functions, and the basic idea is that we do everything component-by-component, and so many of the things you learned in single-variable calculus carry over with just minor adjustments.

Definition. Suppose $f: \real \to \real^n$ is a vector function, $c \in \real$ , and $L \in \real^n$ . Then $\displaystyle \lim_{t \to c} f(t) = L$ means: For every $\epsilon > 0$ , there is a $\delta$ , such that

$$\delta > |t - c| > 0 \quad\hbox{implies}\quad \epsilon > \|f(t) - L\|.$$

("$\|f(t) - L\|$ " means the length of $f(t) - f(c)$ , regarded as a vector in $\real^n$ .)

Proposition. Suppose $f: \real \to \real^n$ has components

$$f(t) = (f_1(t), f_2(t), \ldots f_n(t)).$$

Then

$$\lim_{t \to c} f(t) = (\lim_{t \to c} f_1(t), \lim_{t \to c} f_2(t), \ldots \lim_{t \to c} f_n(t)).$$

This means that the limit on the left exists if and only if all the component limits on the right exist, and in that case the two sides are equal.

In other words, you take the limit of a vector function by taking the limit of each component. All of the usual rules for computing limits work, one component at a time.

Example. Find the value of $\displaystyle \lim_{t \to 1} \left(\dfrac{t^2 + 3 t - 4}{t - 1}, \cos \dfrac{\pi t}{2}, 4 e^{t - 1}\right)$ if it exists.

The component functions of $f(t) = \left(\dfrac{t^2 + 3 t - 4}{t - 1}, \cos \dfrac{\pi t}{2}, 4 e^{t - 1}\right)$ are $f_1(t) = \dfrac{t^2 + 3 t - 4}{t - 1}$ , $f_2(t) = \cos \dfrac{\pi t}{2}$ , and $f_3(t) = 4 e^{t - 1}$ . You can see that the component functions are ordinary one-variable functions of the kind you see in a first-term calculus course.

Note that

$$\lim_{t \to 1} \dfrac{t^2 + 3 t - 4}{t - 1} = \lim_{t \to 1} \dfrac{(t + 4)(t - 1)}{t - 1} = \lim_{t \to 1} (t + 4) = 5.$$

So

$$\lim_{t \to 1} \left(\dfrac{t^2 + 3 t - 4}{t - 1}, \cos \dfrac{\pi t}{2}, 4 e^{t - 1}\right) = (5, 0, 4).\quad\halmos$$

Definition. Let $f: \real \to \real^n$ be a vector function in $\real^n$ , and let $c \in \real$ . Then f is continuous at c if $\displaystyle \lim_{t \to c} f(t) = f(c)$ .

Proposition. Suppose $f: \real \to \real^n$ has components

$$f(t) = (f_1(t), f_2(t), \ldots f_n(t)).$$

Then f is continuous at c if and only if $f_1$ , $f_2$ , ... $f_n$ are continuous at c, considered as functions $\real \to \real$ .

If that looks a bit technical, don't worry. The meaning is that a vector function is continuous at a point if its component functions are.

Example. Define $f: \real \to \real^2$ by

$$f(t) = \cases{ (t, t^2 + 3) & if $t \ne 0$ \cr (0, 2) & if $t = 0$ \cr}$$

Prove or disprove: f is continuous at $t = 0$ .

$$\lim_{t \to 0} f(t) = \lim_{t \to 0} (t, t^2 + 3) = (0, 3), \quad\hbox{but}\quad f(0) = (0, 2).$$

Since $\displaystyle \lim_{t \to 0} f(t) \ne f(0)$ , the function is not continuous at $t = 0$ .

Definition. Let $f: \real \to \real^n$ be a vector function in $\real^n$ . The derivative of f is the vector function $f': \real \to \real^n$ given by

$$f'(t) = (f_1'(t), f_2'(t), \ldots f_n'(t)).$$

I'll often write $\der {f(t)} t$ or $\der f t$ for $f'(t)$ .

Proposition. Let $f, g: \real \to \real^n$ be vector functions in $\real^n$ , and let $c \in \real$ . Then:

(a) If $c = (c_1, c_2, \ldots c_n)$ is a constant, then $\der c t = \vec{0}$ .

(b) $\der {} t [f(t) + g(t)] = \der f t + \der g t$ .

(c) $\der {} t (c \cdot f(t)) = c \cdot \der f t$ .

(d) ( Dot product) $\der {} t (f(t) \cdot (g(t)) = {\der f t} \cdot g(t) + f(t) \cdot \der g t$ .

Note: In (d), all the products are dot products.

(e) ( Cross product) Suppose $n = 3$ . Then

$$\der {} t (f(t) \times g(t)) = {\der f t} \times g(t) + f(t) \times \der g t.$$

Proof. The proofs amount to proving the results component-wise. For example, consider (d). Suppose

$$f(t) = (f_1(t), f_2(t), \ldots f_n(t)) \quad\hbox{and}\quad g(t) = (g_1(t), g_2(t), \ldots g_n(t)).$$

Then

$$f(t) \cdot g(t) = f_1(t) g_1(t) + f_2(t) g_2(t) + \cdots + f_n(t) g_n(t).$$

Using the Product Rule for functions of one variable, I have

$$\eqalign{ \der {} t f(t) \cdot g(t) & = \der {} t [f_1(t) g_1(t) + f_2(t) g_2(t) + \cdots + f_n(t) g_n(t)] \cr & = [f_1'(t) g_1(t) + f_1(t) g_1'(t)] + [f_2'(t) g_2(t) + f_2(t) g_2'(t)] + \cdots + [f_n'(t) g_n(t) + f_n(t) g_n'(t)] \cr & = [f_1'(t) g_1(t) + f_2'(t) g_2(t) + \cdots + f_n'(t) g_n(t)] + [f_1(t) g_1'(t) + f_2(t) g_2'(t) + \cdots + f_n(t) g_n'(t)] \cr & = (f_1'(t), f_2'(t), \ldots f_n'(t)) \cdot (g_1(t), g_2(t), \ldots g_n(t)) + (f_1(t), f_2(t), \ldots f_n(t)) \cdot (g_1'(t), g_2'(t), \ldots g_n'(t)) \cr & = {\der f t} \cdot g(t) + f(t) \cdot \der g t \cr}$$

The other results are proved in similar fashion.

Example. Let

$$f(t) = (t^2 + 3 t + 1, 17, \sin 4 t).$$

Compute $f'(t)$ and $f'(1)$ .

$$f'(t) = (2 t + 3, 0, 4 \cos 4 t) \quad\hbox{and}\quad f'(1) = (5, 0, 4 \cos 4).\quad\halmos$$

Thinking of $f: \real \to \real^n$ as a curve, $f'(t)$ is a tangent vector to the curve.

$$\hbox{\epsfysize=1.5in \epsffile{vector-functions-1.eps}}$$

Example. Find parametric equations for the tangent line to

$$f(t) = (t^3 + 5, (t + 1)^2, 7 t + 1) \quad\hbox{at}\quad t = 1.$$

The point of tangency is $f(1) = (6, 4, 8)$ . Now

$$f'(t) = (3 t^2, 2(t + 1), 7) \quad\hbox{so}\quad f'(1) = (3, 4, 7).$$

Thus, $(3, 4, 7)$ is a vector tangent to the curve, so it's parallel to the tangent line to the curve. The tangent line is

$$x - 6 = 3 t, \quad y - 4 = 4 t, \quad z - 8 = 7 t.\quad\halmos$$

You can integrate vector functions component-by-component.

Definition. Suppose $f: \real \to \real^n$ has components

$$f(t) = (f_1(t), f_2(t), \ldots f_n(t)).$$

Then

$$\int f(t)\,dt = \left(\int f_1(t)\,dt, \int f_2(t)\,dt, \ldots \int f_n(t)\,dt\right).$$

A similar definition holds for definite integrals.

Proposition. Let $f, g: \real \to \real^n$ be vector functions in $\real^n$ , and let $c \in \real$ . Then:

(a) $\displaystyle \int [f(t) + g(t)]\,dt = \int f(t)\,dt + \int g(t)\,dt$ .

(b) $\displaystyle \int c f(t)\,dt = c \int f(t)\,dt$ .

Example. Compute the integral $\displaystyle \int (4 - (\sec t)^2, e^{6 t}, 12 t^2 - 8 t + 5)\,dt$ .

$$\int (4 - (\sec t)^2, e^{6 t}, 12 t^2 - 8 t + 5)\,dt = \left(4 t - \tan t, \dfrac{1}{6} e^{6 t}, 4 t^3 - 4 t^2 + 5 t\right) + (c_1, c_2, c_3).\quad\halmos$$

Example. Compute the integral $\displaystyle \int_0^1 \left(6 t^2 + 5, \dfrac{3 t + 1}, 6 \cos 3 t\right)\,dt$ .

$$\int_0^1 \left(6 t^2 + 5, \dfrac{3 t + 1}, 6 \cos 3 t\right)\,dt = \left[\left(2 t^3 + 5 t, \dfrac{1}{3} \ln |3 t + 1|, 2 \sin 3 t\right)\right]_0^1 = \left(7, \dfrac{1}{3} \ln 4, 2 \sin 3\right).\quad\halmos$$

Contact information

Bruce Ikenaga's Home Page

Copyright 2018 by Bruce Ikenaga