Question

Find the points at which the following polar curves have a horizontal or vertical tangent line.\(r^{2}=4 \cos (2 \theta)\)

Short answer

Points: (2, 0), (-2, 0); Horizontal tangents where \( \theta = \pi/2, 3\pi/2 \), vertical tangents where \( \theta = 0, \pi \).

Step-by-step solution

Convert Polar to Cartesian

The given polar equation is \( r^2 = 4 \, \cos(2\theta) \). To find where the curve has horizontal or vertical tangents, we first express it in Cartesian coordinates. We use the identities: \( r^2 = x^2 + y^2 \), \( x = r \, \cos\theta \), and \( y = r \, \sin\theta \). Substitute the expressions for polar parameters: \( x = \sqrt{4 \, \cos(2\theta)} \, \cos(\theta) \) and \( y = \sqrt{4 \, \cos(2\theta)} \, \sin(\theta) \).

Find Derivatives

To find tangent lines, compute the derivatives \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \), giving us \( dy/dx = \frac{dy/d\theta}{dx/d\theta} \). Using \( dy/d\theta = r' \sin\theta + r \cos\theta \) and \( dx/d\theta = r' \cos\theta - r \sin\theta \), with \( r' = 0 \) for this case, it simplifies to \( dy/d\theta = r \cos\theta \) and \( dx/d\theta = -r \sin\theta \).

Set Dy/Dx For Horizontal and Vertical Tangents

A horizontal tangent occurs when \( dy/dx = 0 \), requiring \( dy/d\theta = 0 \) which simplifies to \( \cos\theta = 0 \), giving \( \theta = \pi/2, 3\pi/2 \). A vertical tangent occurs when \( dx/d\theta = 0 \), leading to \( -r \sin\theta = 0 \), simplifying to \( \sin\theta = 0 \), giving \( \theta = 0, \pi \).

Substitute Back to Find Specific Points

Substitute the \( \theta \) values back into the polar equation to find the specific points. For \( \theta = \pi/2, 3\pi/2 \), \( r = 0 \). For \( \theta = 0, \pi \), substitute into \( r^2 = 4 \cos(2\theta) \): at \( \theta = 0 \), \( r = 2 \); at \( \theta = \pi \), \( r = -2 \). Convert to Cartesian: \( (x, y) = (2, 0) \) and \( (-2, 0) \).

Key concepts

Horizontal Tangents

Horizontal tangents occur in a curve when the slope of the tangent line is zero. For polar curves, this means that the derivative \( \frac{dy}{dx} \) equals zero.
To find where this happens, set the derivative \( \frac{dy}{d\theta} \) equal to zero, since \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \).
In the original problem, which involves the curve \( r^2 = 4 \cos(2\theta) \), we calculate the derivatives. The expression for \( \frac{dy}{d\theta} = r \cos(\theta) \) should be equated to zero for horizontal tangents. This translates to \( \cos(\theta) = 0 \).
Thus, the points \( \theta = \pi/2 \) and \( 3\pi/2 \) correspond to where the curve may have horizontal tangents. Substituting back into the polar equation to find the radii, we find these to be points \((0, 0)\) in Cartesian coordinates.

Vertical Tangents

Vertical tangents are the places on a curve where the tangent line has an undefined slope. For polar curves, this occurs when the derivative \( \frac{dy/dx} \) is undefined, meaning the denominator, \( \frac{dx}{d\theta} \), equals zero.
In our equation \( r^2 = 4 \cos(2\theta) \), we find \( \frac{dx}{d\theta} = -r \sin(\theta) \).
Setting \( \frac{dx}{d\theta} = 0 \), we find that \( \sin(\theta) = 0 \).
Therefore, the points \( \theta = 0 \) and \( \pi \) are where the curve may have vertical tangents. Calculate the radius \( r \) at these angles using the polar equation:
\( r = 2 \) and \( -2 \) respectively, giving the Cartesian points \( (2, 0) \) and \( (-2, 0) \).

Polar to Cartesian Conversion

Polar to Cartesian conversion is crucial for understanding curves represented in polar coordinates by expressing them in the more familiar Cartesian form. This involves utilizing relationships between the coordinates:

• \( x = r \cos\theta \)
• \( y = r \sin\theta \)

By substituting these into the expression for \( r \) from the polar equation, we transition the function to the \( (x, y) \) plane.
Consider the given equation \( r^2 = 4 \cos(2\theta) \). Via the identities \( r^2 = x^2 + y^2 \), \( x = r \cos\theta \), and \( y = r \sin\theta \), the transformation allows us to reinterpret the polar curve in Cartesian terms, providing a clearer path to finding derivatives and tangent points. This conversion simplifies further calculations and aids in visual understanding of the curve's behavior.