This problem aims to find the cartesian equation for the curve and after that identify the curve. To better understand the problem, you should be familiar with cartesian coordinate systems, polar coordinates, and conversion from polar to cartesian coordinates.
A two-dimensional coordinate system in which a point on a plane is determined by a distance from a pole (reference point) and an angle from the reference plane, is known as the polar coordinate. On the other hand, spherical coordinates are the 3 coordinates that determine the location of a point in a 3-dimensional trajectory. We can convert cartesian coordinates to polar coordinates using the equations:
\[ x = r\cos\theta \]
\[ y = r\sin\theta \]
Where $r$ is the distance from the reference point, and can be found using $r = \sqrt{x^2 + y^2}$,
and $\theta$ is the angle with the plane, which can be calculated as $\theta = \tan^{-1}{\dfrac{y}{x}}$.
Expert Answer
We know that $r$ and $\theta$ are called polar coordinates of $P$ such that $P(r,\theta).
Now we are given a polar equation of the curve that is:
\[ r = 5\cos\theta \]
To convert the above equation into the form of $x^2 + y^2 = r^2$, we will be multiplying both sides by $r$:
\[ r^2 = 5r\cos\theta \]
First, we will transform the above polar equation from polar to cartesian coordinates.
Transformation of polar to Cartesian coordinates can be done using the concept,
\[x^2 + y^2 = r^2, \space x = r\cos\theta \]
Therefore, the given curve in the cartesian coordinates can be written as:
\[ x^2 + y^2 = 5x \]
Rewriting the equation as:
\[ x^2 + y^2 – 5x = 0 \]
Applying the technique for completing the square:
\[ x^2 + y^2 – 5x + \dfrac{25}{4} – \dfrac{25}{4} = 0 \]
\[ (x – \dfrac{5}{2})^2 + y^2 = \dfrac{25}{4} \]
This equation denotes a circle that is centered at a point $(\dfrac{5}{2},0)$ with radius $\dfrac{5}{2}$.
Numerical Result
The polar equation $r = 5 \cos \theta$ transformed into cartesian coordinates as $(x – \dfrac{5}{2})^2 + y^2 = \dfrac{25}{4}$, which represents a circle with center point $(\dfrac{5}{2},0)$ and radius $\dfrac{5}{2}$.
Example
Identify the curve by figuring out the cartesian equation for $r^2 \cos2 \theta = 1$.
We know that $r$ and $\theta$ are polar coordinates of $P$, such that $P(r,\theta).
We are given a polar equation of the curve that is:
\[r^2 \cos2 \theta = 1\]
First, we will transform the above polar equation from polar to cartesian coordinates.
Transformation of polar to Cartesian coordinates can be done using the concept,
\[x^2 + y^2 = r^2, \space x = r\cos\theta, \space y = r\sin\theta \]
Therefore,
\[r^2\cos2\theta = 1\]
Using the trigonometric formula for $\cos2\theta$, that is:
\[ \cos2\theta = \cos^2\theta – \sin^2\theta \]
Rewriting the equation as:
\[r^2(\cos^2\theta – \sin^2\theta) = 1\]
\[r^2\cos^2\theta – r^2\sin^2\theta = 1\]
\[(r\cos\theta)^2 – (r\sin\theta)^2 = 1\]
Plugging the values of $ x = r\cos\theta, \space y = r\sin\theta $ gives:
\[ x^2 + y^2 = 1 \]
Therefore, the cartesian equation $ x^2 + y^2 = 1$ represents a hyperbola.