Vector-Valued Functions

A vector-valued function assigns a vector to each value of

t

. In the plane it has two components,

\mathbf{r}(t) = \langle x(t), y(t)\rangle

. Placed with its tail at the origin, the vector

\mathbf{r}(t)

points to

(x(t), y(t))

. As

t

runs over an interval, its tip traces the curve of the parametric equations

x = x(t)

y = y(t)

Derivatives

Every rule for derivatives of real functions applies to each component. The second derivative

\mathbf{r}''(t)

is the derivative of

\mathbf{r}'(t)

, again one component at a time. Where

x'(t) \ne 0

, the slope of the tangent line is the ratio of the components,

\frac{y'(t)}{x'(t)}

, as on the page on parametric derivatives.

Example 1

Let

\mathbf{r}(t) = \left\langle e^{t - 1} + t,\; 4\arctan t\right\rangle

. (a) Find

\mathbf{r}'(t)

and

\mathbf{r}'(1)

. (b) Find the slope of the line tangent to the curve at

t = 1

. (c) Find the vector

\mathbf{r}''(1)

(a) Differentiate each component:

\begin{aligned} \mathbf{r}'(t) &= \left\langle e^{t - 1} + 1,\; \frac{4}{1 + t^2}\right\rangle, \\[4pt] \mathbf{r}'(1) &= \langle 2, 2\rangle. \end{aligned}

(b) The slope is the second component over the first:

\frac{dy}{dx}\bigg|_{t = 1} = \frac{y'(1)}{x'(1)} = \frac{2}{2} = 1.

(c) The chain rule on the second component gives

\begin{aligned} \mathbf{r}''(t) &= \left\langle e^{t - 1},\; \frac{-8t}{\left(1 + t^2\right)^2}\right\rangle, \\[4pt] \mathbf{r}''(1) &= \langle 1, -2\rangle. \end{aligned}

The curve traced by r(t), with r′(1) = ⟨2, 2⟩ drawn at half length from the point r(1) = (2, π). It's tangent to the curve.

An antiderivative of a vector-valued function is also found one component at a time, and the constant of integration is a vector

\langle C_1, C_2\rangle

. One known value of

\mathbf{r}

fixes both constants.

Example 2

Let

\mathbf{r}'(t) = \left\langle 6\sqrt{t},\; \frac{4}{2t + 1}\right\rangle

for

t \ge 0

, with

\mathbf{r}(1) = \langle 3, -1\rangle

. Find

\mathbf{r}(t)

and

\mathbf{r}(4)

Integrate each component:

\begin{aligned} x(t) &= 4t^{3/2} + C_1, \\[4pt] y(t) &= 2\ln(2t + 1) + C_2. \end{aligned}

t = 1

4 + C_1 = 3

gives

C_1 = -1

, and

2\ln 3 + C_2 = -1

gives

C_2 = -1 - 2\ln 3

. So

\begin{aligned} \mathbf{r}(t) &= \Big\langle 4t^{3/2} - 1, \\ &\quad\;\; 2\ln(2t + 1) - 1 - 2\ln 3\Big\rangle, \\[6pt] \mathbf{r}(4) &= \langle 31,\; 2\ln 9 - 1 - 2\ln 3\rangle \\ &= \langle 31,\; 2\ln 3 - 1\rangle \\ &\approx \langle 31, 1.197\rangle. \end{aligned}

The curve traced by r(t) for 0 ≤ t ≤ 4. It passes through (3, −1) at t = 1 and ends near (31, 1.197).

When the components have no antiderivative in closed form, the definite integral gives the change in

\mathbf{r}

\mathbf{r}(b) = \mathbf{r}(a) + \int_{a}^{b} \mathbf{r}'(t)\,dt.