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Describe in words the surface whose equation is given. φ = π/6

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.

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