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 equation 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.