A vector function is a function that takes one or more variables and returns a vector. We’ll spend most of this section looking at vector functions of a single variable as most of the places where vector functions show up, there will be vector functions of single variables. We will however briefly look at vector functions of two variables, three variable and n-variable.

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A function is a rule that assign to each element in the Domain and element in the Range. Vector valued function is a function whose domain is a set of real number and Range is set of vectors. In three-dimensional, for every number t in the domain of r having a unique vector in V_{3} denoted by r(t).

If f(t), g(t) and h(t) are the components of the vector r(t), then f, g, h are real valued functions called component function of $\vec{r}$ and it is written as r(t) = [f(t), g(t), h(t)] = f(t)i + g(t)j + h(t)k, where t is independent variable.

If r(t) = (f(t), g(t), h(t)), then

$\lim_{x\to a} r(t) = (\lim_{x\to a} f(t), \lim_{x\to a} g(t), \lim_{x\to a} h(t))$

provided limits of the component functions exist.Also, limits of vector function obey same rules as that of the limit of real-valued functions.### Solved Examples

**Question 1: **Find $\lim_{t\to 0} r(t) = (1 + t)i + tj + \frac{\sin t}{t} k$

** Solution: **
**Question 2: **Describe the curve defined by the vector function r(t) = (t, 3 + 2t, 2 + t)

** Solution: **
**Question 3: **Sketch the curve for Helix.

** Solution: **
**Question 4: **Find a vector equation and parametric equations for the line segment that joins the point X(2, -1, 3) to point Y(1, 3, -2).

** Solution: **
In mathematics, the vector valued function is a function whose domain is the set of all real numbers and the range is the set of vectors. This is also called the vector function. A vector function of a single variable in R^{2}, R^{3} and R^{n} are in the following form:

$\vec{r} = [f(t), g(t)]$ , $\vec{r} = [f(t), g(t), h(t)]$ and $\vec{r} = [f(t), g(t), h(t), k(t), ...... n(t)]$ where, f(t), g(t), h(t), ..... n(t) are component functions and t is independent variable.

### Derivative of Vector Function:

Derivation of vector function with single variable is given as follows:

$\vec{r} (t) = (f'(t), g'(t), h'(t)) = f'(t)i + g'(t)j + h'(t)k$

Most of the basic facts that we know about derivatives still hold. However, here are some facts about derivatives of vector functions:

Let $\vec{u}(t)$ and $\vec{v}(t)$ be differentiable vector-valued functions, let k be a scalar and let c be a constant vector. Then

### Sketch Vector Function:

A vector-valued function of a real variable can be written in component form as f(t) = f_{1}(t)i + f_{2}(t)j + f_{3}(t)k or in the form f(t) = (f_{1}(t), f_{2}(t), f_{3}(t)) for some real-valued functions f_{1}(t), f_{2}(t), f_{3}(t),
called the component functions of f. The first form is often used when
emphasizing that f(t) is a vector, and the second form is useful when
considering just the terminal points of the vectors.

The main idea that we want to discuss is that of graphing and identifying the graph given by a vector function. Before we do that, however we should talk briefly about the domain of a vector function. The domain of a vector function is the set of all the values of t for which all the component functions are defined.

### Structure of Harmonic Multi- Vector Function:

Let F be a meromorphic function. Then an expression F(x) = f(g(x)), where f is meromorphic and g is factorization of F with f and g. F is said to be non-factorisable or prime for every presentation of the form F(x) = f(g(x)), either f or g are linear. If every representation of F says f is rational and g is polynomial (f is linear whenever g is transcendental, g is linear whenever f is transcendental). If the factors are restricted to entire functions, the factorization is said to be in entire sense.

Factorization theory of Bi-complex meromorphic functions play a vital role since the parallel definitions of factorization of meromorphic functions of one complex variable do not work. Sometimes, many results from one variable theory hold in bi-complex situation, whereas some fails. This is the difference between the two situations.

In general, for any bi-complex transcendental meromorphic function F, there exists a bi-complex meromorphic function G, such that, G F is prime even

if set: { a â† T : F + a $\Phi$ is not prime} is empty for any non-constant fractional linear bi-complex function $\Phi$.### Solved Examples

**Question 1: **Compute $\vec{r'} (t)$ for $\vec{r}(t) = t^4 i + 3t j - 5 k$

** Solution: **
**Question 2: **Find the derivative of f(x,y,z) = 2x^{2 } - z - 1 at the point (1, 2, 3). The direction of A = 2i - j + 4k.

** Solution: **
**Step 1: **$\frac{d}{du} f = \Delta f \hat{u}$

**Step 2:** $\Delta f = 4xi - 1k$

**Step 3:** $\delta f$ at $(1, 2, 3)$ = $4i - k$

**Step 4:** $\hat{u} = \frac{A}{|A|} = \frac{2i - j - 4k}{\sqrt{21}}$

**Step 5:** $\Delta f . \hat{u} = (4i - k) . \frac{2i - j - 4k}{\sqrt{21}}$

**Question 3: **Determine the domain of the following function. $\vec{r} = (\sqrt{t + 1}, \ln (3 - t))$

** Solution: **
**Question 4: **Find the domain of the function $f(x,y) = \frac{1}{x - y}$

** Solution: **
**Question 5: **Find a vector function that represents the curve of intersection of the cylinder x^{2} + y^{2 }= 1 and plane y + z = 2.

** Solution: **

If f(t), g(t) and h(t) are the components of the vector r(t), then f, g, h are real valued functions called component function of $\vec{r}$ and it is written as r(t) = [f(t), g(t), h(t)] = f(t)i + g(t)j + h(t)k, where t is independent variable.

If r(t) = (f(t), g(t), h(t)), then

$\lim_{x\to a} r(t) = (\lim_{x\to a} f(t), \lim_{x\to a} g(t), \lim_{x\to a} h(t))$

provided limits of the component functions exist.Also, limits of vector function obey same rules as that of the limit of real-valued functions.

Since the limit of r is the vector whose components are the limits of the component functions of r,

$\lim_{t\to 0} r(t) = \lim_{t\to 0} (1 + t)i + \lim_{t\to 0} tj + \lim_{t\to 0} \frac{\sin t}{t} k$

= $i + k$

A vector function r is continuous at 'a' if $\lim_{x\to a} r(x) = r(a)$. r is continuous at 'a' if and only if its component functions f, g, h are continuous at 'a'. There is a close relation between vector function and space curve.

Let f, g and h are continuous real-valued function on an interval I. Then, the set C of all point (x, y, z) in space, where, x = f(t), y = g(t), z = h(t) and t varies in interval I is called a space curve.

t is parameter and x = f(t), y = g(t), z = h(t) are called parametric equations of c. If vector function r(t) is the position vector of the point P on c. Thus, any continuous vector function r defines a space curve c.

$\lim_{t\to 0} r(t) = \lim_{t\to 0} (1 + t)i + \lim_{t\to 0} tj + \lim_{t\to 0} \frac{\sin t}{t} k$

= $i + k$

A vector function r is continuous at 'a' if $\lim_{x\to a} r(x) = r(a)$. r is continuous at 'a' if and only if its component functions f, g, h are continuous at 'a'. There is a close relation between vector function and space curve.

Let f, g and h are continuous real-valued function on an interval I. Then, the set C of all point (x, y, z) in space, where, x = f(t), y = g(t), z = h(t) and t varies in interval I is called a space curve.

t is parameter and x = f(t), y = g(t), z = h(t) are called parametric equations of c. If vector function r(t) is the position vector of the point P on c. Thus, any continuous vector function r defines a space curve c.

Given parametric equations are $x = t, y = 3 + 2t, z = 2 + t$ which recognize that parametric equation of line passing through the point (0,3,2) and parallel to the vector (1,2,1) can be written as r = r_{0 }+ tv, where r_{0} = (0,3,2) and v = (1,2,1) for x = f (t), y = g(t), z = h(t).

We know vector equation of the helix is $r(t) = \cos t i + \sin t j + t k$ and parametric equations for this curve are $x = \cos t, y = \sin t, z = t$

Since, x^{2 }+ y^{2 }= cos^{2}t + sin^{2}t = 1, curve must lie on the circular cylinder x^{2} + y^{2 }= 1. Point (x, y, z) lies directly above the point (x, y, 0) which moves counterclockwise around the given circle in xy - plane (vector equation r(t) = (cos t, sin t, 0) is the projection onto xy - plane) Here, z = t, the curve spirals

upward around the cylinder as t increases.

The corkscrew shape of helix is familiar from its occurrence in coiled spring.

Since, x

upward around the cylinder as t increases.

The corkscrew shape of helix is familiar from its occurrence in coiled spring.

The vector equation for the line segment that joins the tip of vector r_{0} to the tip of vector r_{1:}

r(t) = (1 - t)r_{0} + tr_{1}, tâ†[0,1]

To obtain vector equation of line segment, we take r_{0 } = (2, -1, 3) and r_{1 }= (1, 3, -2)

r(t) = (1 - t)(2, -1, 3) + t(1, 3, -2)

r(t) = (2 - t, 4t - 1, 3 - 5t)

So, the parametric equations are x = 2 - t, y = 4t - 1, z = 3 - 5t

r(t) = (1 - t)r

To obtain vector equation of line segment, we take r

r(t) = (1 - t)(2, -1, 3) + t(1, 3, -2)

r(t) = (2 - t, 4t - 1, 3 - 5t)

So, the parametric equations are x = 2 - t, y = 4t - 1, z = 3 - 5t

$\vec{r} = [f(t), g(t)]$ , $\vec{r} = [f(t), g(t), h(t)]$ and $\vec{r} = [f(t), g(t), h(t), k(t), ...... n(t)]$ where, f(t), g(t), h(t), ..... n(t) are component functions and t is independent variable.

Derivation of vector function with single variable is given as follows:

$\vec{r} (t) = (f'(t), g'(t), h'(t)) = f'(t)i + g'(t)j + h'(t)k$

Most of the basic facts that we know about derivatives still hold. However, here are some facts about derivatives of vector functions:

Let $\vec{u}(t)$ and $\vec{v}(t)$ be differentiable vector-valued functions, let k be a scalar and let c be a constant vector. Then

- $\frac{\mathrm{d} }{\mathrm{d} t}(\vec{u}(t) + \vec{v}(t)) = \vec{u'}(t) + \vec{v'}(t)$
- $\frac{\mathrm{d} }{\mathrm{d} t} (k\vec{u}(t))' = k \vec{u'} (t)$
- $\frac{\mathrm{d} }{\mathrm{d} t}(c) = 0$
- $\frac{\mathrm{d} }{\mathrm{d} t} f(t) \vec{u}(t) = f'(t) \vec{u}(t) + f(t) \vec{u'}(t)$
- $\frac{\mathrm{d} }{\mathrm{d} t}(\vec{u}(t) . \vec{v}(t)) = \vec{u}(t) . \vec{v'}(t) + \vec{v}(t) . \vec{u'}(t)$
- $\frac{\mathrm{d} }{\mathrm{d} t} (\vec{u}(t) x \vec{v} (t)) = \vec{u}(t) x \vec{v'} (t) + \vec{v} (t) x \vec{u'}(t)$
- $\frac{\mathrm{d} }{\mathrm{d} t}(\vec{u}(t) \vec{f(t)}) = \vec{u'}(t)f(t) + \vec{f'(t)}\vec{u}(t)$

A vector-valued function of a real variable can be written in component form as f(t) = f

The main idea that we want to discuss is that of graphing and identifying the graph given by a vector function. Before we do that, however we should talk briefly about the domain of a vector function. The domain of a vector function is the set of all the values of t for which all the component functions are defined.

Let F be a meromorphic function. Then an expression F(x) = f(g(x)), where f is meromorphic and g is factorization of F with f and g. F is said to be non-factorisable or prime for every presentation of the form F(x) = f(g(x)), either f or g are linear. If every representation of F says f is rational and g is polynomial (f is linear whenever g is transcendental, g is linear whenever f is transcendental). If the factors are restricted to entire functions, the factorization is said to be in entire sense.

Factorization theory of Bi-complex meromorphic functions play a vital role since the parallel definitions of factorization of meromorphic functions of one complex variable do not work. Sometimes, many results from one variable theory hold in bi-complex situation, whereas some fails. This is the difference between the two situations.

In general, for any bi-complex transcendental meromorphic function F, there exists a bi-complex meromorphic function G, such that, G F is prime even

if set: { a â† T : F + a $\Phi$ is not prime} is empty for any non-constant fractional linear bi-complex function $\Phi$.

$\vec{r'} (t) = 4t^3i + 3j + 0k$

= $4t^3i + 3j$

= $4t^3i + 3j$

The first component is defined for t $\geq$ -1

Second component is only defined for t < 3

Putting all of these together gives the domain [-1,3).

This is the interval for which both components are defined.

Second component is only defined for t < 3

Putting all of these together gives the domain [-1,3).

This is the interval for which both components are defined.

In this case, f(x,y) is all of R^{2} except points (x,y) for which x = y. The Domain is set D = {(x,y): x not equal to y} and Range is all real number except 0.

Let us move into looking at the graph of vector functions. In order to graph a vector function, all we do is think of the vector returned by the vector function as a position vector for points on the graph.

$\vec{r} = (a, b, c)$ is a vector that $\vec{r}$ is a vector that starts at the origin and ends at the point (a,b,c). So, in order to sketch the graph of a vector function, we need to plug in some values of t and then, plot points that correspond to the resulting position vector.

Let us move into looking at the graph of vector functions. In order to graph a vector function, all we do is think of the vector returned by the vector function as a position vector for points on the graph.

$\vec{r} = (a, b, c)$ is a vector that $\vec{r}$ is a vector that starts at the origin and ends at the point (a,b,c). So, in order to sketch the graph of a vector function, we need to plug in some values of t and then, plot points that correspond to the resulting position vector.

Given equation of cylinder and plane are x^{2} + y^{2 }= 1 and y + z = 2 respectively.

The figure shows the curve of intersection C, which is ellipse

The figure shows the curve of intersection C, which is ellipse

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