# Find the domain of the vector function. (Enter your answer using interval notation).

This question aims to find the domain of a vector-valued function and the answer should be expressed in an interval notation.

A vector-valued function is a mathematical function that consists of more than one variable that has a range of multi-dimensional vectors. The domain of a vector-valued function is the set of real numbers and its range consists of a vector. Vector or scalar-valued functions can be inserted.

These types of functions play a big role in calculating different curves both in two-dimensional and three-dimensional space.

Acceleration, velocity, displacement, and distance of any variable can be easily found by making vector-valued functions and applying line functions and contours to these functions both in an open and closed field.

Consider a function:

$r ( t ) = \sqrt { 9 – t ^ 2 } i + t ^ 2 j – 5 t k$

$r ( t ) = < 9 – t ^ 2 , t ^ 2 , – 5 t >$

The set of all real numbers is the domain of rational numbers and the denominator must be a non-zero number. Put the function equal to zero to find the restriction of the domain of rational numbers.

Taking the square on both sides of the equation:

$9 – t ^ 2 = 0$

$t ^ 2 = 9$

$t = \pm 3$

Domain in interval notation:

$( – \infty , – 3) \cup ( + 3 , \infty )$

The component j of the given vector is as follows:

$t ^ 2 = 0$

Taking square root on both sides of the equation:

$t = 0$

${ t : t \in R }$

The domain component is all real numbers so it is not restricted to any number.

The component k of the given vector is as follows:

$– 5 t = 0$

$t = 0$

The domain of this component is all real numbers so it is not restricted to any number.

Domain in interval notation:

${ t : t \in R }$

## Numerical Solution

The domain of a given vector-valued function is  $( – \infty , – 3) \cup ( + 3 , \infty )$  for component i and for other components, the domain is all real numbers without any restriction.

## Example

$f ( t ) = \frac { 7 y } { y + 9 }$

The set of all real numbers is the domain of rational numbers and the denominator must be a non-zero number. Put denominator equal to zero to find the restriction of the domain of rational numbers.

By setting the denominator equal to zero, we get:

$y + 9 = 0$

Rearranging the above equation:

$y \neq – 9$

Hence, – 9 is a number at which the domain becomes restricted. The domain of the given function must lie to the left or right side of this number.

Interval notation:

$( – \infty , – 9 ) \cup ( – 9 , \infty )$

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