# Vector Differentiation and Vector Integration: Vector Calculus

## Vector Differentiation and Integration: An Intro

Vector functions can be readily subjected to computations of differentiation and integration in order to derive different components of space curves. Our first article on the subject of vector calculus laid out the various aspects of vector calculus. This included a review on vectors, examination of certain types of computations such as dot and cross products, investigated vector functions and the space curves they define, briefly covered differentiation and integration of these functions, and applied all of this towards defining arc length and curvature. We then broke these topics down individually in greater depth. For example, we discussed thoroughly the proofs of the dot products and cross products and worked through several examples of computations using these techniques. In another article, we extrapolated on vector functions, how they are created, and modeling space curves with these functions. Let us expand upon vector differentiation and vector integration here.

## Vector Differentiation

##### Perspectives of Vector Differentiation

Suppose we have a vector ‘r’ and we desire to take the derivative of this function. As with the differentiation of real-valued functions, the differentiation of vector functions follows a comparative pattern:

\frac{dr}{dt}=r'(t)=\lim_{h\to0}\frac{r(t+h)-r(t)}{h}

This implies that the derivative of a vector is a tangent that exists when the value ‘h’ approaches 0. In fact, we can define the tangent function ‘T’ accordingly:

T(t) = \frac{r'(t)}{|r'(t)|}

It is important to note that taking the derivative of a vector function is just as simple as differentiation in real valued functions. To do so requires only for the derivative of the component functions to be differentiated such that:

r'(t)=< f'(t), g'(t), h'(t) >
##### Proof of Vector Differentiation

Can we prove the previous relationship between the derivative of the vector function and the derivatives of the component vectors to be true? In fact, we may do so using the orignial equation for the derivative of a vector function as follows:

r'(t)=\lim_{\Delta t\to0}[r(t+\Delta t)-r(t)] \newline =\lim_{\Delta t\to0}[< f(t+\Delta t), g(t+\Delta t), h(t+\Delta t) > -< f(t), g(t), h(t) >] \newline =< \lim_{t\to0} \frac{f(t+\Delta t)-f(t)}{\Delta t}, \lim_{t\to0} \frac{g(t+\Delta t)-g(t)}{\Delta t}, \lim_{t\to0} \frac{h(t+\Delta t)-h(t)}{\Delta t} >\newline = < f'(t), g'(t), h'(t) >

This proof utilizes the general definition of vector differentiation which addresses the limits of the vector function. The limit of the vector function is the limit of the component functions in the vector. These limits define the derivatives of the component functions, proving that the derivative of the vector function is in fact the derivative of the vector component functions.

##### Differentiating Vector Functions

Suppose we are given a vector function of the following form:

r(t) = < 1+t^3, te^{-1}, sin2t >

We can take this information given, and use it to discern the derivative of the vector function. We can also utilize the derivative to determine the tangent derivative at any point along the space curve. First, let us differentiate:

r(t) = < 1+t^3, te^{-1}, sin2t >\newline r'(t)=<3t^2, e^{-1},2cos(2t)>

We took the derivative of the vector function by differentiating each component vector of the function. Let us graph both of these to see how they compare:

Now, suppose we desire to discern the tangent vector that exists when t=0. To determine this, all we must do is substitute the value of zero in for ‘t’ in the derivative function:

r'(t)=<3t^2, e^{-1},2cos(2(0))>\newline r'(t)=<3(0)^2, e^{-1},2cos(2(0))>\newline T(t)=\frac{r'(t)}{|r'(t)|}\newline T(0)=\frac{<3(0)^2, e^{-1},2cos(2(0))>}{\sqrt{(0)+(.368)^2+4}}\newline T(0)=\frac{ < 0, .368, 2 > }{2.03357}\newline T(0)=<0, .1809, .9835 >

As can be seen here, computation of the tangent vector produces a new vector that is tangential to the point of movement on the space curve.

## Vector Integration

As was the case with vector function differentiation, the integration of vector functions is not too disimilar from integration of real value functions. Another point of consistency lies in the fact that integration of the vector function requires only that the component functions of the vector themselves be integrated. Thus, we arrive at a technique that looks as follows:

\int^b_a r(t) dt = < \int^b_a f(t) dt,\int^b_a g(t) dt,\int^b_a h(t) dt >

Suppose we have a vector function comprised of three components such that:

r(t)= < 2sin(t), -cos(t), t^2 >

Then integrating this function requires us to simply integrate each component in the following fashion:

\int_a^b r(t)dt= \int_a^b< 2sin(t), -cos(t), t^2 > dt\newline =<2\int_a^bsin(t)dt, -\int_a^bcos(t)dt, \int_a^b t^2dt>\newline =< -cos(t), -sin(t), \frac{t^3}{3} >

As can be observed, the integral of the vector function was readily computed by discerning the integral of each component. Let us graph these together to observe how they compare: