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Aleks04 [339]
3 years ago
7

Plot the point whose polar coordinates are given. Then find two other pairs of polar coordinates of this point, one with r >

0, and one with r < 0.
(a) (2, 5pi/6)
(b) (1, -2pi/3)
(c) (-1, 5pi/4)
Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
5 0

Answer:

The other pairs are:

(a)\ (2, \frac{5\pi}{6}) \to  (2, \frac{17\pi}{6}) and (-2, \frac{23\pi}{6})

(b)\ (1, -\frac{2\pi}{3}) \to (1, \frac{4\pi}{3}) and (-1, \frac{7\pi}{3})

(c)\ (-1, \frac{5\pi}{4}) \to (-1, \frac{3\pi}{4} ) and (1, \frac{7\pi}{4})

See attachment for plots

Step-by-step explanation:

Given

(a)\ (2, \frac{5\pi}{6})

(b)\ (1, -\frac{2\pi}{3})

(c)\ (-1, \frac{5\pi}{4})

Solving (a): Plot a, b and c

See attachment for plots

Solving (b): Find other pairs for r > 0 and r < 0

The general rule is that:

The other points can be derived using

(r, \theta) = (r, \theta + 2n\pi)

and

(r, \theta) = (-r, \theta + (2n + 1)\pi)

Let n =1 ---- You can assume any value of n

So, we have:

(r, \theta) = (r, \theta + 2n\pi)

(r, \theta) = (r, \theta + 2*1*\pi)

(r, \theta) = (r, \theta + 2\pi)

(r, \theta) = (-r, \theta + (2n + 1)\pi)

(r, \theta) = (-r, \theta + (2*1 + 1)\pi)

(r, \theta) = (-r, \theta + (2 + 1)\pi)

(r, \theta) = (-r, \theta + 3\pi)

(a)\ (2, \frac{5\pi}{6})

r = 2\ \ \ \ \theta = \frac{5\pi}{6}      

So, the pairs are:

(r, \theta) = (r, \theta + 2\pi)

(2, \frac{5\pi}{6}) = (2, \frac{5\pi}{6} + 2\pi)

Take LCM

(2, \frac{5\pi}{6}) = (2, \frac{5\pi+12\pi}{6})

(2, \frac{5\pi}{6}) = (2, \frac{17\pi}{6})

And

(r, \theta) = (-r, \theta + 3\pi)

(2, \frac{5\pi}{6}) = (-2, \frac{5\pi}{6} + 3\pi)

Take LCM

(2, \frac{5\pi}{6}) = (-2, \frac{5\pi+18\pi}{6})

(2, \frac{5\pi}{6}) = (-2, \frac{23\pi}{6})

The other pairs are:

(2, \frac{17\pi}{6}) and (-2, \frac{23\pi}{6})

(b)\ (1, -\frac{2\pi}{3})

r = 1\ \ \ \theta = -\frac{2\pi}{3}      

So, the pairs are:

(r, \theta) = (r, \theta + 2\pi)

(1, -\frac{2\pi}{3}) = (1, -\frac{2\pi}{3} + 2\pi)

Take LCM

(1, -\frac{2\pi}{3}) = (1, \frac{-2\pi+6\pi}{3})

(1, -\frac{2\pi}{3}) = (1, \frac{4\pi}{3})

And

(r, \theta) = (-r, \theta + 3\pi)

(1, -\frac{2\pi}{3}) = (-1, -\frac{2\pi}{3} + 3\pi)

Take LCM

(1, -\frac{2\pi}{3}) = (-1, \frac{-2\pi+9\pi}{3})

(1, -\frac{2\pi}{3}) = (-1, \frac{7\pi}{3})

The other pairs are:

(1, \frac{4\pi}{3}) and (-1, \frac{7\pi}{3})

(c)\ (-1, \frac{5\pi}{4})

r = -1 \ \ \ \ \theta = \frac{-5\pi}{4}

So, the pairs are

(r, \theta) = (r, \theta + 2\pi)

(-1, \frac{-5\pi}{4}) = (-1, \frac{-5\pi}{4} + 2\pi)

Take LCM

(-1, \frac{-5\pi}{4}) = (-1, \frac{-5\pi+8\pi}{4} )

(-1, \frac{-5\pi}{4}) = (-1, \frac{3\pi}{4} )

And

(r, \theta) = (-r, \theta + 3\pi)

(-1, \frac{-5\pi}{4}) = (-(-1), \frac{-5\pi}{4}+ 3\pi)

Take LCM

(-1, \frac{-5\pi}{4}) = (1, \frac{-5\pi+12\pi}{4})

(-1, \frac{-5\pi}{4}) = (1, \frac{7\pi}{4})

So, the other pairs are:

(-1, \frac{3\pi}{4} ) and (1, \frac{7\pi}{4})

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