The aim of the question is to learn how to **visualize a given equation** by **comparing with the standard shape equations**.

The **equation of the cone** (for example) is given by the following formula:

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

Similarly, the e**quation of the circle** (in xy-plane) is given by the following formula:

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

Where x, y, z are the **cartesian coordinates** and R is the **radius of the circle**.

## Expert Answer

**Given:**

\[ \phi \ = \ \dfrac{ \pi }{ 6 } \]

The **cartesian coordinates** can be calculated using the following formulas:

\[ x \ = \ R \ cos( \theta ) \ sin( \phi ) \ = \ \dfrac{ 1 }{ \sqrt{ 2 } } R \ cos( \theta ) \]

\[ y \ = \ R \ sin( \theta ) \ sin( \phi ) \ = \ \dfrac{ 1 }{ \sqrt{ 2 } } R \ sin( \theta ) \]

\[ z \ = \ R \ cos( \phi ) \ = \ \dfrac{ 1 }{ \sqrt{ 2 } } R \]

Lets find $ x^2 \ + \ y^2 $:

\[ x^2 \ + \ y^2 \ = \ \bigg ( \dfrac{ 1 }{ \sqrt{ 2 } } R \ cos( \theta ) \bigg )^2 \ + \ \bigg ( \dfrac{ 1 }{ \sqrt{ 2 } } R \ sin( \theta ) \bigg )^2 \]

\[ x^2 \ + \ y^2 \ = \ \dfrac{ 1 }{ 2 } R^2 \ \bigg ( cos^2( \theta ) \ + \ sin^2( \theta ) \bigg ) \]

Since $ cos^2( \theta ) \ + \ sin^2( \theta ) \ = \ 1 $:

\[ x^2 \ + \ y^2 \ = \ \dfrac{ 1 }{ 2 } R^2 \]

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

**The above equation represents a cone centered at the origin along the z-axis.**

To find the direction of this cone, we solve the above equation for z:

\[ z \ = \ \pm \sqrt{ x^2 + y^2 } \]

Since **R is always positive, z must also be always positive:**

\[ z \ = \ + \sqrt{ x^2 + y^2 } \]

Hence, the **cone is located along the positive z-axis**.

## Numerical Result

The given equation represents **a cone** with **vertex at the origin** directed **along the positive z-axis**.

## Example

**Describe the following equation in words:**

\[ \phi \ = \ \dfrac{ \pi }{ 2 } \]

The **cartesian coordinates** of this equation are:

\[ x \ = \ R \ cos( \theta ) \ sin( \phi ) \ = \ R \ cos( \theta ) \]

\[ y \ = \ R \ sin( \theta ) \ sin( \phi ) \ = \ R \ sin( \theta ) \]

\[ z \ = \ R \ cos( \phi ) \ = \ 0 \]

Lets find $ x^2 \ + \ y^2 $:

\[ x^2 \ + \ y^2 \ = \ \bigg ( R \ cos( \theta ) \bigg )^2 \ + \ \bigg ( \dfrac{ 1 }{ \sqrt{ 2 } } R \ sin( \theta ) \bigg )^2 \]

\[ x^2 \ + \ y^2 \ = \ R^2 \ \bigg ( cos^2( \theta ) \ + \ sin^2( \theta ) \bigg ) \]

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

The above equation represents **a circle centered at the origin in the xy-plane with radius R**.