Differentiation / integration of vectors

For each point $t$ belonging to the subset $D$ of the real $R$, given the real functions $x(t),y(t),z(t)$, one vector you can think of $x(t), y(t), z(t))$．This vector $\boldsymbol{F}(t)$ is called one variable vector-valued function or vector function from $D$ to $R^{3}$ and denoted by

$$\boldsymbol{F}(t) = (x(t),y(t),z(t))$$

or

$$\boldsymbol{F}(t) = x(t)\:\boldsymbol{i} + y(t)\:\boldsymbol{j} + z(t)\:\boldsymbol{k}$$

Often $\boldsymbol{F}(t)$ is geometrically a mapping from the real axis $t$ to the vector connecting the origin and the point $(x (t), y (t), z (t))$. It will be treated.

Example 2..1  

For $${\boldsymbol{F}(t) = t\cos{t} \:\boldsymbol{i} + t\sin{t} \:\boldsymbol{j} + t^2 \:\boldsymbol{k}}$$，find the trace of $\boldsymbol{F}(t)$ .

Answer The components of $\boldsymbol{F}(t)$ are $x = t\cos{t}, y = t\sin{t}, z = t^2$. Then $z = x^2 + y^2$ and the trace of $\boldsymbol{F}(t)$, $(x (t), y (t), z (t))$ is on the paraboloid $z = x^2 + y^2$.

Definition 2..1  

In the vector function $\boldsymbol{F}(t)$, if $t \rightarrow t_{0}$ and $\boldsymbol{F}(t) \rightarrow {\ bf L}$, then $\boldsymbol{The limit value of F}(t)$ is said to be ${\ bf L}$, which is expressed as follows．

$$\lim_{t \rightarrow t_{0}}\boldsymbol{F}(t) = {\bf L}$$

Since the definition of the limit value is the same as for the one-variable function, you probably expect the definition of continuity to be the same as for the one-variable function. In fact, that's the case.

Definition 2..2  

When $\lim_{t \rightarrow t_{0}} \boldsymbol{F}(t) = \boldsymbol{F}(t_{0})$ holds, the vector function $\boldsymbol{F}(t)$ it is said that $t = t_{0}$ is continuous. Also, when all $t$ in the interval $[a, b]$ are continuous, $\boldsymbol{F}(t)$ is said to be continuous in the interval $[a, b]$, which is expressed as $\boldsymbol{F}(t) \in C[a, b]$.

In this way, various definitions in a one-variable function are inherited by a vector function.

Definition 2..3  

A vector function ${F}(t)$ is said to be differentiable at $t = t_{0}$ if

$$\lim_{t \rightarrow t_{0}}\frac{\boldsymbol{F}(t) - \boldsymbol{F}(t_{0})}{t - t_{0}} = \boldsymbol{A}$$

exists for $t = t_{0}$. Also, the limit $\boldsymbol{A}$ is called a coefficient of derivative at $t_{0}$ and denoted by $\boldsymbol{F}^{\prime}(t_{0})$．

We will study in the next chapter, the direction of the vector $\boldsymbol{F}^{\prime}(t_{0})$ is the direction of the tangent line to the curve $\boldsymbol{F}(t)$.

As the sum of vectors and the scalar multiple are defined by the sum of the corresponding components and the scalar multiple, the calculation of the limit value, derivative coefficient, and indefinite integral of the vector function is performed by the limit value, derivative coefficient, of the component of the vector function. It can be obtained from the indefinite integral.

Theorem 2..1  

Let $\boldsymbol{F}(t) = x(t)\:\boldsymbol{i} + y(t)\:\boldsymbol{j} + z(t)\:\boldsymbol{k}$. Then the followings are true．

$${(a) \ \lim_{t \rightarrow t_{0}}\boldsymbol{F}(t) = \lim_{t \righ... ...ow t_{0}}y(t)\:\boldsymbol{j} + \lim_{t \rightarrow t_{0}}z(t)\:\boldsymbol{k}}$$

$${(b) \ \boldsymbol{F}(t) \ \mbox{is continuous} \ \Leftrightarrow x(t), y(t), z(t) \ \mbox{are continuous}}$$

$${(c) \ \boldsymbol{F} \in C'[a,b] \Rightarrow \boldsymbol{F}^{\pri... ...bol{i} + y^{\prime}(t_{0})\:\boldsymbol{j} + z^{\prime}(t_{0})\:\boldsymbol{k}}$$

$${(d) \ \boldsymbol{F} \in C[a,b] \Rightarrow \int_{a}^{b}\boldsymb... ...i} + \int_{a}^{b} y(t)dt\:\boldsymbol{j} + \int_{a}^{b} z(t)dt\:\boldsymbol{k}}$$

Proof (a)

$$\vert\boldsymbol{F}(t) - {\bf L}\vert = \sqrt{(x(t) - l_{1})^2 + (y(t) - l_{2})^2 + (z(t) - l_{3})^2 }$$

Then，if $x(t) \rightarrow l_{1}, y(t) \rightarrow l_{2}, z(t) \rightarrow l_{3}$，then $\boldsymbol{F}(t) \rightarrow {\ bf L}$．Also，if $\boldsymbol{F}(t) \rightarrow {\ bf L}$，

$$\sqrt{(x(t) - l_{1})^2 + (y(t) - l_{2})^2 + (z(t) - l_{3})^2 } \rightarrow 0$$

Note that everything in the square root is the sum of squares. Then

$$(x(t) - l_{1})^2 \rightarrow 0, (y(t) - l_{2})^2 \rightarrow 0, (z(t) - l_{3})^2 \rightarrow 0$$

Thus, $x(t) \rightarrow l_{1}, y(t) \rightarrow l_{2}, z(t) \rightarrow l_{3}$ implies

$$\lim_{t \rightarrow t_{0}}\boldsymbol{F}(t) = \lim_{t \rightarrow... ...row t_{0}}y(t)\:\boldsymbol{j} + \lim_{t \rightarrow t_{0}}z(t)\:\boldsymbol{k}$$

Proof of (b)，(c)，(d) are your exercise. ．

Example 2..2  

For $${\boldsymbol{F}(t) = t^{2}\:\boldsymbol{i} + t\:\boldsymbol{j} + t^{3}\:\boldsymbol{k}}$$， find $\boldsymbol{F}^{\prime}(t)$.

Answer Differentiate eachcomponents, we have

$$\boldsymbol{F}^{\prime}(t) = 2t\:\boldsymbol{i} + \:\boldsymbol{j} + 3t^{2}\:\boldsymbol{k}$$

If $\boldsymbol{F}(t),$$G$$(t)$ are differentiable for $t$, then the followings are true.

$$\begin{array}{ll} (1) & (\boldsymbol{F}(t) \pm \mbox{\boldmat... ...ymbol{F}(t) \times \mbox{\boldmath $G$}^{\prime}(t) \end{array}$$

Example 2..3  

For $\boldsymbol{F} = 5t^2\:\boldsymbol{i} + t\:\boldsymbol{j} - t^2\:\boldsymbol{k}, \$   $G$$= \sin{t}\:\boldsymbol{i} - \cos{t}\:\boldsymbol{j}$，find the followings:． (1) $(\boldsymbol{F}\cdot$$G$$)'$ (2) $(\boldsymbol{F} \times$   $G$$)'$ Answer $F' = (5t^2)'\:\boldsymbol{i} + (t)'\:\boldsymbol{j} + - (t^2)' \:\boldsymbol{k}... ... (\cos{t})'\:\boldsymbol{j} = \cos{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j}$.

(1)

$$(\boldsymbol{F}\cdot \mbox{\boldmath$G$} )' = \boldsymbol{F}'\cdot \mbox{\boldmath$G$} + \boldsymbol{F}\cdot \mbox{\boldmath$G$} '$$

$$= (10t \:\boldsymbol{i} + \:\boldsymbol{j} - 2t \:\boldsymbol{k})\c... ... - t^2\:\boldsymbol{k})\cdot(\cos{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j})$$

$$= (5t^2 - 1)\cos{t} + 11\sin{t}$$

(2)

$$(\boldsymbol{F}\times \mbox{\boldmath$G$} )' = \boldsymbol{F}'\times \mbox{\boldmath$G$} + \boldsymbol{F}\times \mbox{\boldmath$G$} '$$

$$= (10t \:\boldsymbol{i} + \:\boldsymbol{j} - 2t \:\boldsymbol{k})\t... ... t^2\:\boldsymbol{k})\times (\cos{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j})$$

$$= \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ...oldsymbol{k}\\ 5t^2 & t & -t^2\\ \cos{t} & \sin{t} & 0 \end{array}\right\vert$$

$$= (t^2 \sin{t} - 2t \cos{t})\:\boldsymbol{i} - (t^2 \cos{t} + 2t \sin{t})\:\boldsymbol{j} + (5t^2 \sin{t} - \sin{t} - 11t\cos{t})\:\boldsymbol{k}$$

Example 2..4  

Suppose that the magnitude of $\boldsymbol{F}(t)$, $\vert\boldsymbol{F}(t)\vert$ is constant，show that $\boldsymbol{F}(t)$ and $\boldsymbol{F}^{\prime}(t)$ are orthogonal for all $t$．

Answer Let $\vert\boldsymbol{F}(t)\vert = c$. Then $\vert\boldsymbol{F}(t)\vert^2 = \boldsymbol{F}(t) \cdot\boldsymbol{F}(t) = c^{2}$. Thus, by the derivative of the vector function, we have

$$(\boldsymbol{F}(t) \cdot\boldsymbol{F}(t))^{\prime} = 2\boldsymbol{F}^{\prime}(t) \cdot\boldsymbol{F}(t) = 0$$

Thus the inner product is 0 and $\boldsymbol{F}(t)$ and $\boldsymbol{F}^{\prime}(t)$ are orthogonal.

For every vector functions $F(t),G(T)$ and a constant $\alpha$，a constant vector $\boldsymbol{C}$, we have the followings:

$$\begin{array}{ll} (1) & \int (\boldsymbol{F} + \mbox{\boldmat... ...oldsymbol{F}}{dt}\times \mbox{\boldmath $G$}\;dt\\ \end{array}$$

Example 2..5  

For every vector function $\boldsymbol{F} = \boldsymbol{F}(t)$，prove that $\int \boldsymbol{F}\cdot\boldsymbol{F}'\;dt = \frac{1}{2}\boldsymbol{F}\cdot\boldsymbol{F}$． Answer $\int \boldsymbol{F}\cdot\boldsymbol{F}'\;dt = \boldsymbol{F}\cdot\boldsymbol{F} - \int\boldsymbol{F}'\cdot\boldsymbol{F}\;dt$．Therefore,， $\int \boldsymbol{F}\cdot\boldsymbol{F}'\;dt = \frac{1}{2}\boldsymbol{F}\cdot\boldsymbol{F}$.

Question 2..1  

Find $\int_{2}^{3}\boldsymbol{F}\cdot\frac{d\boldsymbol{F}}{dt}\;dt$, provide $\boldsymbol{F}(2) = 2\:\boldsymbol{i} -\boldsymbol{j} + 2\:\boldsymbol{k},\ \boldsymbol{F}(3) = 4\:\boldsymbol{i} - 2\:\boldsymbol{j} + 3\:\boldsymbol{k}$