**\[Cylinder\ x^2+y^2=4\]**

**\[Surface\ z=xy\]**

The aim of this question is to find the **vector function** of the **curve** that is generated when a **cylinder** is **intersected** by a **surface**.

The basic concept behind this article is the **Vector-Valued Function** and representation of different **geometrical figures** in **parametric equations**.

A **vector-valued function** is defined as a **mathematical function** consisting of **one or more variables** having a range, which is a **set of vectors** in **multi-dimensions**. We can use a **scalar** or a **vector parameter** as an **input** for the **vector-valued function,** whereas its **output** will be a **vector**.

For **two dimensions**, the **vector-valued function** is:

\[r(t)=x(t)\hat{i}+y(t)\hat{j}\]

For **three dimensions**, the **vector-valued function** is:

\[r(t)=x(t)\hat{i}+y(t)\hat{j}+z(t)\hat{k}\]

Or:

\[r(t)\ =\ \langle x(t),\ y(t),\ z(t) \rangle \]

## Expert Answer

The **Equation for Cylinder**:

\[x^2+y^2=4\]

The **Equation for Surface**:

\[z=xy\]

When a **plane surface intersects** a **three-dimensional cylindrical** **figure**, the **curve of intersection** created will be in a **three-dimensional plane** in the form of a **circle**.

Therefore, the equation of a **standard circle** with **Center** $(0,\ 0)$ is derived by considering the position coordinates of **circle centers** with their **constant radius** $r$ as follows:

\[x^2+y^2=r^2\]

Where:

$R=$ **Radius of Circle**

$(x,\ y)=$ **Any point on Circle**

As per **Cylindrical Coordinate System**, the **parametric equations** for $x$ and $y$ are:

\[x(t)=rcos(t)\]

\[y(t)=rsin(t)\]

Where:

$t=$ **Counter-clockwise angle** from the **x-axis** in the **x,y plane** and having a **range** of:

\[0\ \le\ t\ \le\ 2\pi\]

As the **Equation for Cylinder** is $x^2+y^2=4$, so the **radius** $r$ will be:

\[x^2+y^2\ =\ {4\ =(2)}^2\]

Hence:

\[r\ =\ 2\]

By substituting the value of $r\ =\ 2$ in **parametric equations** for $x$ and $y$, we get:

\[x(t)\ =\ r\ cos(t)\]

\[y(t)\ =\ r\ sin(t)\]

By substituting the value of $x$ and $y$ in $z$, we get:

\[z(t)\ =\ x(t)\ \times\ y(t)\]

\[z\ =\ 2\ cos(t)\ \times\ 2\ sin(t)\]

By simplifying the equation:

\[z\ =\ 4\ sin(t)\ cos(t)\]

So the **vector function** will be represented as follows:

\[r(t)\ =\ \langle x(t),\ y(t),\ z(t)\rangle\]

\[r(t)\ =\ \langle\ 2\ cos(t),\ 2\ sin(t)\ \ ,\ 4\ sin(t)cos(t)\ \rangle\]

## Numerical Result

The **curve of intersection** of **cylinder** and** surface** will be represented by a **vector function** as follows:

Then that represents as follows:

\[r(t)\ =\ \langle\ 2\ cos(t),\ 2\ sin(t)\ \ ,\ 4\ sin(t)cos(t)\ \rangle\]

## Example

A **cylinder** $x^2+y^2\ =\ 36$ and** surface** $4y+z=21$ intersect each other and form a **curve of intersection**. Find its **vector function**.

**Solution**

The **Equation for Cylinder**:

\[x^2+y^2\ =\ 36\]

The **Equation for Surface**:

\[4y+z=21\]

\[z=21\ -\ 4y\]

As the **Equation for Cylinder** is $x^2+y^2\ =\ 36$, so the **radius** $r$ will be:

\[x^2+y^2\ =\ {36\ =(6)}^2\]

Hence:

\[r\ =\ 6\]

By substituting the value of $r\ =\ 6$ in **parametric equations** for $x$ and $y$, we get:

\[x(t)\ =\ 6\ cos(t)\]

\[y(t)\ =\ 6\ sin(t)\]

By substituting the value of $x$ and $y$ in $z$, we get:

\[z=21\ -\ 4y\]

\[z=21\ -\ 4(6\ sin(t))\]

\[z=21\ -\ 24\ sin(t)\]

So, the **vector function** will be:

\[r(t)\ =\ \langle\ 6\ cos(t),\ 6\ sin(t)\ \ ,\ 21\ -\ 24\ sin(t)\ \rangle\]