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transform rectangular equation to polar equations $y = 3x + 2$ $y^2 = 3…

Question

transform rectangular equation to polar equations
$y = 3x + 2$
$y^2 = 3 - x^2$

Explanation:

Response

To transform the given rectangular equations to polar equations, we use the relationships between rectangular coordinates \((x, y)\) and polar coordinates \((r, \theta)\):
\(x = r\cos\theta\), \(y = r\sin\theta\), and \(r^{2}=x^{2}+y^{2}\).

Equation 1: \(y = 3x + 2\)

Step 1: Substitute \(x = r\cos\theta\) and \(y = r\sin\theta\)

Substitute \(x\) and \(y\) in the equation:
\(r\sin\theta = 3(r\cos\theta)+2\)

Step 2: Simplify (optional, to solve for \(r\))

Rearrange to isolate \(r\):
\(r\sin\theta - 3r\cos\theta = 2\)
Factor out \(r\):
\(r(\sin\theta - 3\cos\theta)=2\)
Thus, \(r=\frac{2}{\sin\theta - 3\cos\theta}\)

Equation 2: \(y^{2}=3 - x^{2}\)

Step 1: Rearrange the equation

Add \(x^{2}\) to both sides:
\(x^{2}+y^{2}=3\)

Step 2: Substitute \(r^{2}=x^{2}+y^{2}\)

Using the polar coordinate identity \(r^{2}=x^{2}+y^{2}\), substitute into the equation:
\(r^{2}=3\)

Step 3: Solve for \(r\)

Take the square root of both sides (assuming \(r\geq0\) for polar coordinates):
\(r = \sqrt{3}\)

Final Polar Equations:
  • For \(y = 3x + 2\): \(\boldsymbol{r=\frac{2}{\sin\theta - 3\cos\theta}}\) (or the unsimplified form \(r\sin\theta = 3r\cos\theta + 2\))
  • For \(y^{2}=3 - x^{2}\): \(\boldsymbol{r=\sqrt{3}}\) (or \(r^{2}=3\))

Answer:

To transform the given rectangular equations to polar equations, we use the relationships between rectangular coordinates \((x, y)\) and polar coordinates \((r, \theta)\):
\(x = r\cos\theta\), \(y = r\sin\theta\), and \(r^{2}=x^{2}+y^{2}\).

Equation 1: \(y = 3x + 2\)

Step 1: Substitute \(x = r\cos\theta\) and \(y = r\sin\theta\)

Substitute \(x\) and \(y\) in the equation:
\(r\sin\theta = 3(r\cos\theta)+2\)

Step 2: Simplify (optional, to solve for \(r\))

Rearrange to isolate \(r\):
\(r\sin\theta - 3r\cos\theta = 2\)
Factor out \(r\):
\(r(\sin\theta - 3\cos\theta)=2\)
Thus, \(r=\frac{2}{\sin\theta - 3\cos\theta}\)

Equation 2: \(y^{2}=3 - x^{2}\)

Step 1: Rearrange the equation

Add \(x^{2}\) to both sides:
\(x^{2}+y^{2}=3\)

Step 2: Substitute \(r^{2}=x^{2}+y^{2}\)

Using the polar coordinate identity \(r^{2}=x^{2}+y^{2}\), substitute into the equation:
\(r^{2}=3\)

Step 3: Solve for \(r\)

Take the square root of both sides (assuming \(r\geq0\) for polar coordinates):
\(r = \sqrt{3}\)

Final Polar Equations:
  • For \(y = 3x + 2\): \(\boldsymbol{r=\frac{2}{\sin\theta - 3\cos\theta}}\) (or the unsimplified form \(r\sin\theta = 3r\cos\theta + 2\))
  • For \(y^{2}=3 - x^{2}\): \(\boldsymbol{r=\sqrt{3}}\) (or \(r^{2}=3\))