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Chapter 6: Problem 23

A point in rectangular coordinates is given. Convert the point to polar coordinates. $$(-\sqrt{3},-\sqrt{3})$$

Short Answer

Expert verified
The polar coordinates equivalent to the point (-\sqrt{3},-\sqrt{3}) in Cartesian form are \((2\sqrt{3}, \frac{5蟺}{4})\)

Step by step solution

01

Calculation of r

The r-coordinate in polar form (distance from the origin) is calculated using the formula \( r = \sqrt{x^2 + y^2} \). So, if the cartesian coordinates are (-\sqrt{3},-\sqrt{3}), insert these values into the formula, yielding \( r = \sqrt{(-\sqrt{3})^2 + (-\sqrt{3})^2} = \sqrt{3+3} = 2\sqrt{3} \).
02

Calculation of 胃

We use the arctan function to find 胃, where \( 胃 = arctan(\frac{y}{x}) \). Always pay attention to whether the point is in the first, second, third, or fourth quadrant. Given the point (-\sqrt{3},-\sqrt{3}), we have \( 胃 = arctan(\frac{-\sqrt{3}}{-\sqrt{3}}) = arctan(1) \). Arctan(1) normally gives the angle \( 胃 = \frac{蟺}{4} \) radians, but since the point lies in the third quadrant (both coordinates are negative), the true angle is \( 胃 = \frac{蟺}{4} + 蟺 = \frac{5蟺}{4} \) radians.
03

Write the Final Answer

The polar coordinates are therefore given by the pair \((r, 胃)\), or in this case \((2\sqrt{3}, \frac{5蟺}{4})\).

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Most popular questions from this chapter

In Exercises 25-28, use a graphing utility to graph the polar equation. Identify the graph. $$ r=\frac{5}{-4+2 \cos \theta} $$

\(r=4(1-\sin \theta)\)

In Exercises 17-40, sketch the graph of the polar equation using symmetry, zeros, maximum \(r\)-values, and any other additional points. \(r=5\)

In Exercises 33-48, find a polar equation of the conic with its focus at the pole. $$ \begin{array}{lll} {\text { Conic }} & \text { Eccentricity } & \text { Directrix } \\ \text { Parabola } & e=1 & y=-2 \\ \end{array} $$

\(r=5 \sin 2 \theta\)
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