/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Express the following polar coor... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 19

Express the following polar coordinates in Cartesian coordinates. \(\left(-4, \frac{3 \pi}{4}\right)\)

Short Answer

Expert verified
Answer: The Cartesian coordinates equivalent to the given polar coordinates are \((2\sqrt{2}, -2\sqrt{2})\).

Step by step solution

01

Identify the given polar coordinates

The given polar coordinates are \((-4, \frac{3\pi}{4})\). Here, \(r = -4\) and \(\theta = \frac{3\pi}{4}\).
02

Apply the conversion formulas

To convert the polar coordinates to Cartesian coordinates, we will use the following formulas: \(x = r \cos(\theta)\) \(y = r \sin(\theta)\)
03

Find the value of x

Using the conversion formula for \(x\), we have: \(x = (-4) \cos\left(\frac{3\pi}{4}\right)\) The cosine of \(\frac{3\pi}{4}\) is \(-\frac{\sqrt{2}}{2}\). So, we end up with: \(x = (-4) \left(-\frac{\sqrt{2}}{2}\right) = 2\sqrt{2}\)
04

Find the value of y

Using the conversion formula for \(y\), we have: \(y = (-4) \sin\left(\frac{3\pi}{4}\right)\) The sine of \(\frac{3\pi}{4}\) is \(\frac{\sqrt{2}}{2}\). So, we end up with: \(y = (-4) \left(\frac{\sqrt{2}}{2}\right) = -2\sqrt{2}\)
05

Write the final Cartesian coordinates

Now that we have the \(x\) and \(y\) values, we can write the Cartesian coordinates as \((x, y)\) or \((2\sqrt{2}, -2\sqrt{2})\). So, the Cartesian coordinates equivalent to the given polar coordinates are: \((2\sqrt{2}, -2\sqrt{2})\)

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Key Concepts

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

Polar Coordinates
Polar coordinates are a way to describe the position of a point in a plane using two values:
  • the distance from the origin (often represented as \(r\))
  • the angle from the positive x-axis (represented by \(\theta\))
The pair \((r, \theta)\) is unique because it reflects how far and in which direction a point is, compared to the center of a coordinate system.
With the initial problem, the polar coordinates given are \((-4, \frac{3\pi}{4})\). This means:
  • The point is 4 units away from the origin, but moving in the opposite direction since \(r\) is negative
  • The direction or angle is \(\frac{3\pi}{4}\) radians from the positive x-axis
It's important to note the effects of using a negative \(r\), as it indicates the point is in the direction opposite of \(\theta\). This provides a concise and elegant way to convey information in navigation and physics problems.
Cartesian Coordinates
Cartesian coordinates are used to define a point in space through two coordinate axes intersecting at the origin:
  • The x-coordinate tells how far along the horizontal axis the point is
  • The y-coordinate informs how far along the vertical axis the point exists
For the provided exercise, the Cartesian coordinates were found to be \((2\sqrt{2}, -2\sqrt{2})\). This represents:
  • An x-value of \(2\sqrt{2}\), indicating the point鈥檚 distance to the right of the origin
  • A y-value of \(-2\sqrt{2}\), showing that the point is downward from the origin
This method is particularly useful in grid-like systems and is commonly used because of its straightforward rectangular layout, making it convenient for most mathematical and engineering calculations.
Coordinate Transformation
Coordinate transformation involves converting between different measures of describing a point's location, such as from polar to Cartesian coordinates. To accomplish this:
  • The formulas \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\) are vital
These equations help translate the circular nature of polar coordinates into the linear framework of Cartesian coordinates. In the exercise's case:
  • \(x = (-4) \cos(\frac{3\pi}{4}) = 2\sqrt{2}\)
  • \(y = (-4) \sin(\frac{3\pi}{4}) = -2\sqrt{2}\)
By using trigonometric functions such as sine and cosine, the angle component of polar coordinates is mapped into direct x and y components.
This conversion process is critical in fields requiring precise physical placement, motion analysis, or solving complex equations where a coordinate transformation can simplify the task.

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Most popular questions from this chapter

An ellipse (discussed in detail in Section 10.4 ) is generated by the parametric equations \(x=a \cos t, y=b \sin t.\) If \(0 < a < b,\) then the long axis (or major axis) lies on the \(y\) -axis and the short axis (or minor axis) lies on the \(x\) -axis. If \(0 < b < a,\) the axes are reversed. The lengths of the axes in the \(x\) - and \(y\) -directions are \(2 a\) and \(2 b,\) respectively. Sketch the graph of the following ellipses. Specify an interval in t over which the entire curve is generated. $$x=4 \cos t, y=9 \sin t$$

Completed in 1937, San Francisco's Golden Gate Bridge is \(2.7 \mathrm{km}\) long and weighs about 890,000 tons. The length of the span between the two central towers is \(1280 \mathrm{m}\) the towers themselves extend \(152 \mathrm{m}\) above the roadway. The cables that support the deck of the bridge between the two towers hang in a parabola (see figure). Assuming the origin is midway between the towers on the deck of the bridge, find an equation that describes the cables. How long is a guy wire that hangs vertically from the cables to the roadway \(500 \mathrm{m}\) from the center of the bridge?

Equations of the form \(r^{2}=a \sin 2 \theta\) and \(r^{2}=a \cos 2 \theta\) describe lemniscates (see Example 7 ). Graph the following lemniscates. \(r^{2}=-2 \sin 2 \theta\)

The region bounded by the parabola \(y=a x^{2}\) and the horizontal line \(y=h\) is revolved about the \(y\) -axis to generate a solid bounded by a surface called a paraboloid (where \(a>0\) and \(h>0\) ). Show that the volume of the solid is \(\frac{3}{2}\) the volume of the cone with the same base and vertex.

A plane traveling horizontally at \(100 \mathrm{m} / \mathrm{s}\) over flat ground at an elevation of \(4000 \mathrm{m}\) must drop an emergency packet on a target on the ground. The trajectory of the packet is given by $$x=100 t, \quad y=-4.9 t^{2}+4000, \quad \text { for } t \geq 0,$$ where the origin is the point on the ground directly beneath the plane at the moment of the release. How many horizontal meters before the target should the packet be released in order to hit the target?
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