Question

Describe the graphs of the equations and provide alternative equations in the specified coordinate systems.

Change $\varphi =\frac{\mathrm{\pi }}{4}$to the rectangular and spherical systems.

Short answer

The equation in rectangular coordinates is$x^2+y^2-z^2=0$

Step-by-step solution

Step 1:Given information

The given expression is$\varphi =\frac{\mathrm{\pi }}{4}$

Simplification (a) 

To convert $\varphi =\frac{\mathrm{\pi }}{4}$into rectangular coordinates

We have

$r=\sqrt{x^2+y^2+z^2},\tan \theta =\frac{y}{x},\cos \varphi =\frac{z}{\sqrt{x^2+y^2+z^2}}$

$\Rightarrow \cos \varphi =\frac{z}{\sqrt{x^2+y^2+z^2}}$

$\Rightarrow \varphi ={\cos }^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$

Substituting $\Rightarrow \varphi ={\cos }^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$in $\varphi =\frac{\mathrm{\pi }}{4}$

$\Rightarrow \frac{\mathrm{\pi }}{4}={\cos }^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$

$\Rightarrow \cos \frac{\mathrm{\pi }}{4}=\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$

$\Rightarrow \left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)=\frac{1}{\sqrt{2}}$

$\Rightarrow \left(\frac{z^2}{x^2+y^2+z^2}\right)=\frac{1}{2}$

$\Rightarrow x^2+y^2+z^2=2z^2$

$\Rightarrow x^2+y^2-z^2=0$

Hence,The equation in rectangular coordinates is$x^2+y^2-z^2=0$