Question

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

Short answer

Horizontal tangent at (2, 2); vertical tangent at (0, 0).

Step-by-step solution

Convert Polar Equations to Cartesian Coordinates

The given polar equation is \( r = 4 \cos \theta \). \ We convert it to Cartesian coordinates using the formulas \( x = r \cos \theta \) and \( y = r \sin \theta \). In polar coordinates, \( r = x/\cos \theta \) and \( r = 4 \cos \theta \). Equating the two expressions for \( r \), \( x = 4 \cos^2 \theta \). Therefore, \( x^2 + y^2 = 4x \).

Find the Derivatives

The equation from the conversion step is \( x^2 + y^2 = 4x \). To find the derivative, we implicitly differentiate with respect to \( x \): \( 2x + 2y \frac{dy}{dx} = 4 \). Solving for \( \frac{dy}{dx} \), we get \( \frac{dy}{dx} = \frac{2(y - 2)}{x} \).

Find Horizontal Tangents

Horizontal tangents occur when \( \frac{dy}{dx} = 0 \). Solving \( 2(y - 2) = 0 \), we find \( y = 2 \). Substituting \( y = 2 \) in the original equation \( x^2 + y^2 = 4x \), we get \( x^2 + 4 = 4x \) which simplifies to \( (x - 2)^2 = 0 \). Hence, \( x = 2 \). Therefore, the point is \( (2, 2) \).

Find Vertical Tangents

Vertical tangents occur when \( \frac{dy}{dx} \) is undefined. This is when the denominator of \( \frac{dy}{dx} = \frac{2(y - 2)}{x} \) is zero, i.e., \( x = 0 \). Substitute \( x = 0 \) in the original equation to find \( y \). \( 0^2 + y^2 = 0 \), hence \( y = 0 \). Therefore, the point is \( (0, 0) \).

Verify Polar Points for Correctness

The polar form points corresponding to points in Cartesian form are \( (r, \theta) = (2\sqrt{2}, \pi/4) \) for \( (2, 2) \) and \( (0, \theta) \) for \( (0, 0) \). Both points satisfy the equation \( r = 4 \cos \theta \). Therefore, the points are correct.

Key concepts

Horizontal Tangents

When dealing with curves in polar coordinates, a horizontal tangent is a line that runs left to right, parallel to the x-axis. To find points where the curve has a horizontal tangent, we look for places where the slope of the tangent line is zero. This corresponds to the derivative of the y-coordinate with respect to the x-coordinate, \( \frac{dy}{dx} \), being equal to zero.
In the given problem with the polar equation \( r = 4 \cos \theta \), we first convert it to the Cartesian equation \( x^2 + y^2 = 4x \). When we compute the derivative using implicit differentiation, the condition for a horizontal tangent becomes \( 2(y - 2) = 0 \), which simplifies to \( y = 2 \).
Substituting \( y = 2 \) into our converted equation, we solve for \( x \) to find the horizontal tangent points in the Cartesian system. This results in \( (x - 2)^2 = 0 \), meaning that \( x = 2 \). Therefore, the horizontal tangent occurs at the point \( (2, 2) \).
This means that at \( (2, 2) \), the slope is zero so the tangent line is perfectly flat.

Vertical Tangents

Vertical tangents are lines that run up and down, parallel to the y-axis. They occur when the slope of the tangent line is undefined, which happens when the denominator of the derivative \( \frac{dy}{dx} \) is zero. In our polar problem, we convert the equation into Cartesian coordinates, giving us \( x^2 + y^2 = 4x \), which yields the derivative \( \frac{dy}{dx} = \frac{2(y - 2)}{x} \).
For the slope to be undefined, we set the denominator equal to zero. Thus, we have \( x = 0 \).
After setting \( x = 0 \) and substituting into the equation \( x^2 + y^2 = 4x \), we find \( y = 0 \). Therefore, the point \( (0, 0) \) is where the curve has a vertical tangent.
At this point, the tangent line stands perfectly vertical, signifying no real slope due to the direction being straight up.

Implicit Differentiation

Implicit differentiation is used when functions are not given in the traditional y = f(x) form, but rather in an equation involving both x and y. In this exercise, the process begins with converting the polar coordinates to Cartesian form, resulting in \( x^2 + y^2 = 4x \).
This implicit equation cannot be solved for just y or just x straightforwardly, so we differentiate both sides with respect to x. This leads to \( 2x + 2y \frac{dy}{dx} = 4 \).
Next, by isolating \( \frac{dy}{dx} \), we derive \( \frac{dy}{dx} = \frac{2(y-2)}{x} \).
This result is then used to analyze the presence of horizontal and vertical tangents by setting the numerator and denominator to zero respectively.
Implicit differentiation is essential for problems involving curves like this one, where x and y are intertwined, providing a pathway to uncover important characteristics of the curve.