To convert the equation of the line from Cartesian coordinates to polar coordinates, we need to use the relationship between the two coordinate systems. In Cartesian coordinates, the line is given by:
\(y = x - 1\)
In polar coordinates, the relationships are:
- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
Substitute these into the Cartesian equation:
\(r \sin \theta = r \cos \theta - 1\)
Rearranging the terms gives:
\(r \cos \theta - r \sin \theta = 1\)
Factoring out \(r\) from the left side:
\(r (\cos \theta - \sin \theta) = 1\)
This is the polar form of the line. Thus, the correct option is:
\(r(\cos\theta - \sin\theta) = 1\)
Let's verify why this is the correct option and others are not:
- Option 1: \(r = \cos\theta\)
This would imply a circle centered at the origin, which is not equivalent to the given line equation. - Option 2: \(r = \sin\theta\)
Similar to the first option, this would imply a circle, not a line. - Option 3: \(r(\cos\theta + \sin\theta) = 1\)
This is structurally different from the converted equation, as it would imply a line with a different orientation in the polar coordinate system. - Option 4: \(r(\cos\theta - \sin\theta) = 1\)
Correct, as derived from the conversion from Cartesian to polar coordinates.
Thus, the correct answer is \(r(\cos\theta - \sin\theta) = 1\).
Important Questions from Functions Of Single Variable
- Let $f : R \to R$ be a twice-differentiable function and suppose its second derivative
satisfies $f''(x) > 0$ for all $x \in R$. Which of the following statements is/are ALWAYS
correct? - The gradient of $y = 3x^2 \sin(2x)$ at (0.2, 1) is __________ (rounded off to three decimal places).
- If $y = x^x$, then $\frac{dy}{dx}$ is
Given the function $$f(x) = |x| + |x - 1|,$$ For all the real values of x, which one of the following statements is CORRECT ?
- Given $x$ is real, identify all the even-functions among the following:
