Derivatives and Integrals of Vector Functions

Calculus of vector functions needs nothing new: differentiate and integrate each component with the rules you learned for antiderivatives and derivatives. What is new is the meaning. The derivative $\mathbf{r}'(t)$ is the velocity of the moving point, and it is tangent to the curve.

Idea clave

$$\mathbf{r}'(t) = \langle f'(t), g'(t), h'(t) \rangle, \qquad \int \mathbf{r}(t)\,dt = \left\langle \int f\,dt, \int g\,dt, \int h\,dt \right\rangle + \mathbf{C}.$$ $\mathbf{r}'(t)$ points along the curve in the direction of motion. The tangent line at $t_0$ is $\mathbf{L}(s) = \mathbf{r}(t_0) + s\,\mathbf{r}'(t_0)$.

Por qué funciona

The derivative is a limit of difference quotients: $$\mathbf{r}'(t) = \lim_{h \to 0} \frac{\mathbf{r}(t + h) - \mathbf{r}(t)}{h}.$$ The vector $\mathbf{r}(t + h) - \mathbf{r}(t)$ is a secant arrow from one point of the curve to a nearby one. Dividing by $h$ rescales it, and as $h \to 0$ the arrow swings into the tangent direction. Because limits go component by component, the result is the vector of ordinary derivatives. The usual rules follow; for example $(\mathbf{u} \cdot \mathbf{v})' = \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}'$, and the same pattern holds for cross products (keep the order).

The red arrow is $\mathbf{r}'(t)$ at the moving point. It always points along the curve, and it is longer where the point moves faster.

Worked examples

Ejemplo resuelto: A derivative

Differentiate $\mathbf{r}(t) = \langle t^3, e^{2t}, \sin 3t \rangle$.

Component by component: $\frac{d}{dt}t^3 = 3t^2$, $\frac{d}{dt}e^{2t} = 2e^{2t}$, $\frac{d}{dt}\sin 3t = 3\cos 3t$.

Respuesta: $\mathbf{r}'(t) = \langle 3t^2, 2e^{2t}, 3\cos 3t \rangle$

Ejemplo resuelto: A tangent line

Find the tangent line to the helix $\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle$ at $t = \frac{\pi}{2}$.

Point: $\mathbf{r}\left(\frac{\pi}{2}\right) = \left\langle 0, 1, \frac{\pi}{2} \right\rangle$.
$\mathbf{r}'(t) = \langle -\sin t, \cos t, 1 \rangle$, so $\mathbf{r}'\left(\frac{\pi}{2}\right) = \langle -1, 0, 1 \rangle$.
$$\mathbf{L}(s) = \left\langle 0, 1, \frac{\pi}{2} \right\rangle + s\langle -1, 0, 1 \rangle = \left\langle -s, 1, \frac{\pi}{2} + s \right\rangle.$$

Respuesta: $\mathbf{L}(s) = \left\langle -s, 1, \frac{\pi}{2} + s \right\rangle$

Ejemplo resuelto: Recovering the position

Find $\mathbf{r}(t)$ if $\mathbf{r}'(t) = \langle 2t, 3, -\sin t \rangle$ and $\mathbf{r}(0) = \langle 1, 0, 2 \rangle$.

Antidifferentiate: $\mathbf{r}(t) = \langle t^2 + C_1, 3t + C_2, \cos t + C_3 \rangle$, one constant per component.
At $t = 0$: $\langle C_1, C_2, 1 + C_3 \rangle = \langle 1, 0, 2 \rangle$, so $C_1 = 1$, $C_2 = 0$, $C_3 = 1$.
Check by differentiating, and by putting in $t = 0$.

Respuesta: $\mathbf{r}(t) = \langle t^2 + 1, 3t, \cos t + 1 \rangle$

Ejemplo resuelto: A definite integral

Evaluate $\displaystyle\int_0^1 \langle t, t^2, 6 \rangle\,dt$.

$\int_0^1 t\,dt = \frac{1}{2}$, $\int_0^1 t^2\,dt = \frac{1}{3}$, $\int_0^1 6\,dt = 6$.

Respuesta: $\left\langle \frac{1}{2}, \frac{1}{3}, 6 \right\rangle$, a vector

Error común

One constant for the whole vector, then forgetting it. An indefinite integral of a vector function needs a constant vector $\mathbf{C} = \langle C_1, C_2, C_3 \rangle$, and an initial condition fixes all three. In the example, $\cos 0 = 1$ means $C_3 = 1$, not ${}2$; plugging in $t = 0$ at the end catches this. Treating $\mathbf{r}'(t_0)$ as a point. It is a direction; the tangent line starts at $\mathbf{r}(t_0)$.

Worked examples

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