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.

## Expert Answer

Consider a function:

\[ r Â ( Â t Â ) Â = Â \sqrt { 9Â –Â t ^ 2 } i + t ^ 2 j â€“ 5 t k \]

\[ r ( t ) = < 9Â –Â t ^ 2Â Â , t ^ 2 Â , â€“ 5 tÂ > \]

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:

\[Â 9Â –Â tÂ ^ 2 = 0Â \]

\[Â tÂ ^ 2Â = 9Â \]

\[Â tÂ = \pm 3 Â \]

**Domain** in interval notation:

\[ ( – \infty , Â – 3) \cup ( + 3 , \infty ) \]

The **component j** of the given vector is as follows:

\[Â tÂ ^Â 2 =Â 0Â \]

Taking square root on both sides of the equation:

\[Â t Â =Â 0 \]

\[Â { Â Â t : t \in R }Â \]

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:

\[Â – 5 tÂ =Â 0Â \]

\[Â tÂ =Â 0Â \]

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

**Domain** in interval notation:

\[Â { Â Â t : t \in R }Â \]

## Numerical Solution

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

## Example

\[Â f (Â tÂ )Â =Â \frac { 7 y } { y + 9 } \]

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:

\[Â yÂ +Â 9Â =Â 0Â \]

Rearranging the above equation:

\[ Â yÂ \neqÂ – 9 \]

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:

\[ ( – \infty , – 9 ) \cup ( – 9 , \infty ) \]Â

*Image/Mathematical drawings are created in Geogebra**.*