Lydia Carey

2022-06-22

What is the polar form of $(1,18)$ ?

Anika Stevenson

Beginner2022-06-23Added 19 answers

Step 1

Given Cartesian coordinates $(x,y)$ in Quadrant I

$r=\sqrt{{x}^{2}+{y}^{2}}$

and

$\theta =\mathrm{arctan}\left(\frac{y}{x}\right)$

Given Cartesian coordinates $(x,y)$ in Quadrant I

$r=\sqrt{{x}^{2}+{y}^{2}}$

and

$\theta =\mathrm{arctan}\left(\frac{y}{x}\right)$

abbracciopj

Beginner2022-06-24Added 4 answers

Step 1

Polar format: $(r,\text{}\theta )$

$r=\sqrt{{x}^{2}+{y}^{2}}$

$\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$

apply both formulas when going from Cartesian $\to $ polar

$\sqrt{{1}^{2}+{18}^{2}}=\sqrt{325}\approx 18.0$

$\theta ={\mathrm{tan}}^{-1}\left(\frac{18}{1}\right)={\mathrm{tan}}^{-1}\left(18\right)\approx 1.5radians$

Thus our answer of:

Polar format of (1,18) Cartesian is: (18, 1.5)

Polar format: $(r,\text{}\theta )$

$r=\sqrt{{x}^{2}+{y}^{2}}$

$\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$

apply both formulas when going from Cartesian $\to $ polar

$\sqrt{{1}^{2}+{18}^{2}}=\sqrt{325}\approx 18.0$

$\theta ={\mathrm{tan}}^{-1}\left(\frac{18}{1}\right)={\mathrm{tan}}^{-1}\left(18\right)\approx 1.5radians$

Thus our answer of:

Polar format of (1,18) Cartesian is: (18, 1.5)