How do you graph $r = 2 + tan (θ)$ $r = 2 + tan(theta)$ ?

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How do you graph $r = 2 + tan (θ)$ $r = 2 + tan(theta)$ ?

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Answer 1 · 2016-04-29T05:07:05

The form is r = $r_{1} + r_{2}$ $r_1 + r_2$ , where $r_{1} = 2$ $r_1 = 2$ for a circle, with center at pole and radius = 2. $r_{2} = tan θ$ $r_2 = tan theta$ . Radially, $(r, θ)$ $(r, theta)$ is distant $tan θ$ $tan theta$ , from $(2, θ)$ $(2, theta)$ on the circle. .

Explanation:

The asymptotes to this 4-branch curve are $θ = \frac{π}{2}$ $theta = pi/2$ and the

opposite $θ = \frac{3 π}{2}$ $theta=(3pi)/2$ .

Due to infinite discontinuities at $θ = \frac{π}{2} and θ = \frac{3 π}{2}$ $theta = pi/2 and theta=(3pi)/2$ , .the

four branches for $θ \in [0, \frac{π}{2}), (\frac{π}{2}, π], [π, \frac{3 π}{2}) and (\frac{3 π}{2}, π]$ $theta in [0, pi/2), (pi/2, pi], [pi, (3pi)/2) and ((3pi)/2, pi]$ are traced in the four quadrants, in the order, ist, 3rd,

2nd and 4th. They brace the circle r =2 at (2, 0) and (2, $π$ $pi$ ).

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How do you graph $r = 2 + tan (θ)$ $r = 2 + tan(theta)$ ?

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How do you graph r=2+tan(θ)r = 2 + tan(theta)?

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Explanation:

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How do you graph $r = 2 + tan (θ)$ $r = 2 + tan(theta)$ ?