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Convert the polar equation to rectangular form and identify the type of curve represented.

[tex]\theta=\frac{\pi}{3}[/tex]

Part: [tex]$0 / 2$[/tex]

Part 1 of 2


Sagot :

To convert the polar equation [tex]\(\theta = \frac{\pi}{3}\)[/tex] into rectangular form and identify the type of curve represented, we will follow these steps:

### Step 1: Understanding the Polar Equation

The given polar equation is:
[tex]\[ \theta = \frac{\pi}{3} \][/tex]
This means that the angle [tex]\(\theta\)[/tex] is fixed at [tex]\(\frac{\pi}{3}\)[/tex] radians.

### Step 2: Polar to Rectangular Coordinates

In polar coordinates, (r, [tex]\(\theta\)[/tex]) can be converted to rectangular coordinates (x, y) using the following relationships:
[tex]\[ x = r \cos(\theta) \][/tex]
[tex]\[ y = r \sin(\theta) \][/tex]

### Step 3: Substituting the Fixed Angle

For [tex]\(\theta = \frac{\pi}{3}\)[/tex], we use the known trigonometric values:
[tex]\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \][/tex]
[tex]\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \][/tex]

Thus, substituting these values into the equations for x and y, we get:
[tex]\[ x = r \cdot \frac{1}{2} \][/tex]
[tex]\[ y = r \cdot \frac{\sqrt{3}}{2} \][/tex]

### Step 4: Eliminating [tex]\(r\)[/tex]

To eliminate [tex]\(r\)[/tex], we can divide the equation for [tex]\(y\)[/tex] by the equation for [tex]\(x\)[/tex]:
[tex]\[ \frac{y}{x} = \frac{r \cdot \frac{\sqrt{3}}{2}}{r \cdot \frac{1}{2}} = \frac{\sqrt{3}}{1} = \sqrt{3} \][/tex]

This simplifies to:
[tex]\[ y = \sqrt{3} \cdot x \][/tex]

### Step 5: Rectangular Form and Curve Type

Hence, the rectangular form of the given polar equation is:
[tex]\[ y = \sqrt{3} \cdot x \][/tex]

This equation represents a straight line passing through the origin (0,0) with a slope of [tex]\(\sqrt{3}\)[/tex]. Thus, the type of curve is a straight line.

Therefore, the rectangular form of the polar equation [tex]\(\theta = \frac{\pi}{3}\)[/tex] is:
[tex]\[ y = \sqrt{3} \cdot x \][/tex]
And the type of curve represented is a straight line.