To find: The equivalent polar equation for the given rectangular-coordinate equation. Given: x = r cos theta y = r sin theta b. From rectangular to polar: r= pm sqrt{x^{2} + y^{2}} cos theta= frac{x}{r}, sin theta= frac{y}{r}, tan theta= frac{x}{y} Calculation: Given: equation in rectangular-coordinate is y = x. Converting into equivalent polar equation - y=x Put x=r cos theta, y=r sin theta, Rightarrow r sin theta=r cos theta Rightarrow frac{sin theta}{cos theta}=1 Rightarrow tan theta=1 Thus, desired equivalent polar equation would be theta=1

Question
Alternate coordinate systems
asked 2020-11-22
To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
\(\displaystyle\ {x}=\ {r}{\cos{\theta}}\)
\(\displaystyle\ {y}=\ {r}{\sin{\theta}}\)
b. From rectangular to polar:
\(\displaystyle{r}=\pm\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}\)
\(\displaystyle{\cos{\theta}}={\frac{{{x}}}{{{r}}}},{\sin{\theta}}={\frac{{{y}}}{{{r}}}},{\tan{\theta}}={\frac{{{x}}}{{{y}}}}\)
Calculation:
Given: equation in rectangular-coordinate is \(\displaystyle{y}={x}\).
Converting into equivalent polar equation -
\(\displaystyle{y}={x}\)
Put \(\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},\)
\(\displaystyle\Rightarrow\ {r}{\sin{\theta}}={r}{\cos{\theta}}\)
\(\displaystyle\Rightarrow\ {\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}={1}\)
\(\displaystyle\Rightarrow\ {\tan{\theta}}={1}\)
Thus, desired equivalent polar equation would be \(\displaystyle\theta={1}\)

Answers (1)

2020-11-23
Concept used:
Conversion formulafor coordinate systems are given as -
a. From polar to rectangular:
\(\displaystyle\ {x}=\ {r}{\cos{\theta}}\)
\(\displaystyle\ {y}=\ {r}{\sin{\theta}}\)
b. From rectangular to polar:
\(\displaystyle{r}=\pm\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}\)
\(\displaystyle{\cos{\theta}}={\frac{{{x}}}{{{r}}}},{\sin{\theta}}={\frac{{{y}}}{{{r}}}},{\tan{\theta}}={\frac{{{x}}}{{{y}}}}\)
Calculation:
Given: equation in rectangular-coordinate is \(\displaystyle{y}={x}\).
Converting into equivalent polar equation -
\(\displaystyle{y}={x}\)
Put \(\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}}\)
\(\displaystyle\Rightarrow\ {r}{\sin{\theta}}={r}{\cos{\theta}}\)
\(\displaystyle\Rightarrow\ {\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}={1}\)
\(\displaystyle\Rightarrow\ {\tan{\theta}}={1}\)
Thus, desired equivalent polar equation would be \(\displaystyle\theta={1}\)
0

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