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Chapter 11: Problem 50

Convert the point from rectangular coordinates into polar coordinates with \(r \geq 0\) and \(0 \leq \theta<2 \pi\). $$ (-2 \sqrt{10}, 6 \sqrt{10}) $$

Short Answer

Expert verified
The polar coordinates are \((20, 1.893)\).

Step by step solution

01

Identify the Rectangular Coordinates

The given point in rectangular coordinates is \((x, y) = (-2 \sqrt{10}, 6 \sqrt{10})\). We will convert these to polar coordinates \((r, \theta)\).
02

Calculate the Radius (r)

The radius \(r\) in polar coordinates is given by the formula:\[r = \sqrt{x^2 + y^2}\]Substituting the values of \(x\) and \(y\):\[r = \sqrt{(-2 \sqrt{10})^2 + (6 \sqrt{10})^2} = \sqrt{40 + 360} = \sqrt{400} = 20\]Thus, \(r = 20\).
03

Calculate the Angle (\(\theta\))

The angle \(\theta\) is calculated using the following formula:\[\theta = \text{tan}^{-1}\left(\frac{y}{x}\right)\]Substituting the values of \(x\) and \(y\):\[\theta = \text{tan}^{-1}\left(\frac{6 \sqrt{10}}{-2 \sqrt{10}}\right) = \text{tan}^{-1}(-3) \]The tangent of the angle is negative, which places the point in the second quadrant. The reference angle is \(\text{tan}^{-1}(3)\), so:\[\theta = \pi - \text{tan}^{-1}(3)\]Approximating gives:\(\theta \approx \pi - 1.249\approx 1.893\).
04

Write the Polar Coordinates

We have calculated \(r = 20\) and approximated \(\theta \approx 1.893\). Therefore, the polar coordinates of the point are:\[(20, 1.893)\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Rectangular Coordinates
Rectangular coordinates are a part of the Cartesian coordinate system that uses an ordered pair to locate a point on a plane. This pair usually follows the format \( (x, y) \). Here, \( x \) and \( y \) represent the point's horizontal and vertical distances from the origin, respectively.
For instance, in the exercise problem, we have the rectangular coordinates \( (x, y) = (-2 \sqrt{10}, 6 \sqrt{10}) \). This means:
  • The point is \( -2 \sqrt{10} \) units to the left of the y-axis because the x value is negative.
  • The point is \( 6 \sqrt{10} \) units above the x-axis because the y value is positive.
Rectangular coordinates are useful for basic plotting but can become cumbersome when dealing with certain mathematical problems. This is where polar coordinates come in handy, providing a different perspective."},{

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