Problem 53
Question
A point in rectangular coordinates is given. Convert the point to polar coordinates. $$(-\sqrt{3},-\sqrt{3})$$
Step-by-Step Solution
Verified Answer
The polar coordinate of the point \(-\sqrt{3},-\sqrt{3}\) in rectangular coordinates is \(\sqrt{6}, \frac{5\pi}{4}\).
1Step 1: Identify given coordinates
The point given is \(-\sqrt{3},-\sqrt{3}\) in rectangular coordinates. Let's call the x-coordinate as 'x' and the y-coordinate as 'y'. So, x = -\sqrt{3} and y = -\sqrt{3}.
2Step 2: Calculate the radial coordinate r
In polar coordinates, a point is described using a radial distance 'r' and an angle '\(\theta\)'. The radial coordinate 'r' is calculated as the square root of the sum of the squares of x and y. According to Pythagorean theorem, \(r= \sqrt{x^{2} + y^{2}}\). So, plugging the given x and y into the formula: \(r= \sqrt{(-\sqrt{3})^{2} + (-\sqrt{3})^{2}}\). This simplifies to \(r= \sqrt{6}\).
3Step 3: Find the angle \(\theta\)
Now, the angle '\(\theta\)' in polar coordinates can be found using the equation: \(\theta = \arctan\frac{y}{x}\) . But since the coordinates lie in the third quadrant (as both x and y are negative), we have to add PI to the result to get the correct angle. Plugging in the given y and x into the formula gives: \(\theta = \arctan\frac{-\sqrt{3}}{-\sqrt{3}}\). Solving this yields: \(\theta = \arctan(1) + \pi\). This simplifies to \(\theta = \frac{\pi}{4} + \pi = \frac{5\pi}{4}\).
4Step 4: Write down the polar coordinates
Finally, the polar coordinates are given as the radial distance 'r' and the angle '\(\theta\)'. Hence, the polar coordinate of the point \(-\sqrt{3},-\sqrt{3}\) is \(\sqrt{6}, \frac{5\pi}{4}\).
Key Concepts
Rectangular CoordinatesRadial DistanceAngle ConversionPythagorean Theorem
Rectangular Coordinates
Rectangular coordinates, also known as Cartesian coordinates, describe a point in a two-dimensional plane using two numbers. These numbers represent distances along two perpendicular axes: the x-axis and the y-axis. For the point \(-\sqrt{3}, -\sqrt{3}\), the value of \(-\sqrt{3}\) for both coordinates shows its position relative to the origin (0,0). The x-coordinate \(-\sqrt{3}\) tells us how far the point is from the origin in the horizontal direction, while the y-coordinate \(-\sqrt{3}\) indicates its vertical distance from the origin.
Radial Distance
Radial distance, often denoted as \(r\) in polar coordinates, represents how far a point is from the origin, regardless of direction. This distance is always non-negative and can be calculated using the formula derived from the Pythagorean theorem: \[r = \sqrt{x^2 + y^2}\].
- In our example, for the coordinates \(-\sqrt{3}, -\sqrt{3}\), this formula becomes \[r = \sqrt{(-\sqrt{3})^2 + (-\sqrt{3})^2}\].
- This simplifies to \[r = \sqrt{6}\], indicating the point's radial distance from the origin.
Angle Conversion
In polar coordinates, the angle \(\theta\) describes the direction of the point relative to the positive x-axis. This angle is measured in radians.
- To find \(\theta\), use the arctangent function: \(\theta = \arctan\frac{y}{x}\).
- For the point \(-\sqrt{3}, -\sqrt{3}\), both \(x\) and \(y\) are negative, placing the point in the third quadrant where angles range from \(\pi\) to \(\frac{3\pi}{2}\).
Pythagorean Theorem
The Pythagorean theorem plays a vital role in converting rectangular coordinates to polar coordinates. This theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
- When applying this to rectangular coordinates, the theorem helps calculate the radial distance \(r\) as \(r = \sqrt{x^2 + y^2}\).
- This formula comes directly from the idea that every point on the plane can form a right triangle with the origin, where \(x\) and \(y\) are the triangle's legs, and \(r\) is the hypotenuse.
Other exercises in this chapter
Problem 52
Find the standard form of the equation of the parabola with the given characteristics. Vertex: (-1,2)\(;\) focus: (-1,0)
View solution Problem 52
Use a graphing utility to graph the ellipse. Find the center, foci, and vertices. (Recall that it may be necessary to solve the equation for \(y\) and obtain tw
View solution Problem 53
Use a graphing utility to graph the polar equation. Describe your viewing window. $$r=8 \cos \theta$$
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Use the results of Exercises \(49-52\) to find a set of parametric equations to represent the graph of the line or conic. Line: passes through (0,0) and (3,6)
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