Question

Find the points at which the following polar curves have a horizontal or vertical tangent line.The cardioid \(r=1+\sin \theta\)

Short answer

The cardioid has horizontal tangents at \((0,0)\) and vertical tangents at \((0,2)\).

Step-by-step solution

Understand the Polar Equation and Derivatives

The polar equation given is \(r = 1 + \sin\theta\). To find horizontal and vertical tangents, we need to convert into Cartesian coordinates using \(x = r \cos\theta\) and \(y = r \sin\theta\). A horizontal tangent corresponds to \( \frac{dy}{d\theta} = 0\) and \( \frac{dx}{d\theta} eq 0\), while a vertical tangent corresponds to \( \frac{dx}{d\theta} = 0\) and \( \frac{dy}{d\theta} eq 0\).

Convert Polar to Cartesian Coordinates

Substitute \(r = 1 + \sin\theta\) into \(x\) and \(y\):\[x = (1 + \sin\theta) \cos\theta = \cos\theta + \cos\theta \sin\theta\]\[y = (1 + \sin\theta) \sin\theta = \sin\theta + \sin^2\theta\]

Find Derivatives \( \frac{dx}{d\theta}\) and \( \frac{dy}{d\theta}\)

Differentiate \(x\) and \(y\) with respect to \(\theta\):\[\frac{dx}{d\theta} = -\sin\theta + \cos(2\theta) \]\[\frac{dy}{d\theta} = \cos\theta + 2\sin\theta \cos\theta = \cos\theta + \sin(2\theta)\]

Set \( \frac{dy}{d\theta} = 0\) for Horizontal Tangents

Solve \(\cos\theta + \sin(2\theta) = 0\).Using the identity \(\sin(2\theta) = 2\sin\theta \cos\theta\), we have:\[\cos\theta + 2\sin\theta \cos\theta = 0 \]\[\cos\theta(1 + 2\sin\theta) = 0\]This gives \(\cos\theta=0\) or \(\sin\theta = -\frac{1}{2}\).

Solutions for \(\cos\theta = 0\)

If \(\cos\theta = 0\), \(\theta = \frac{\pi}{2}\) or \(\theta = \frac{3\pi}{2}\). Calculate \(r\) for these angles:\(r = 1 + \sin\left(\frac{\pi}{2}\right) = 2\) and \(r = 1 + \sin\left(\frac{3\pi}{2}\right) = 0\). Points are: \((0, 0)\) as there is no movement at \(\theta = \frac{3\pi}{2}\), and the point \(\left(0, 2\right)\).

Solutions for \(\sin\theta = -\frac{1}{2}\)

If \(\sin\theta = -\frac{1}{2}\), \(\theta = \frac{7\pi}{6}\) or \(\theta = \frac{11\pi}{6}\). Calculate \(r\) for these angles:\(r = 1 - \frac{1}{2} = \frac{1}{2}\).Points are: \(\left(\frac{1}{2} \cos\frac{7\pi}{6}, \frac{1}{2} \sin\frac{7\pi}{6}\right)\) and \(\left(\frac{1}{2} \cos\frac{11\pi}{6}, \frac{1}{2} \sin\frac{11\pi}{6}\right)\).

Set \( \frac{dx}{d\theta} = 0\) for Vertical Tangents

Solve \(-\sin\theta + \cos(2\theta) = 0\).Using identity \(\cos(2\theta) = \cos^2\theta - \sin^2\theta\), this equation does not provide additional solutions not covered by Step 4, meaning \(\frac{dy}{d\theta} = 0\) hold for vertical tangents.

Key concepts

Polar Coordinates

In mathematics, polar coordinates provide a different way to represent points in a plane. Unlike Cartesian coordinates, which use a grid system of x and y, polar coordinates are based on a circle and angles. This system uses two values:

• the radius, \(r\), which is the distance from the origin to the point
• the angle, \(\theta\), which is measured from the positive x-axis

Polar coordinates are particularly useful for curves and shapes where radial symmetry makes them easier to describe. A common curve usually represented in polar coordinates is a cardioid, often expressed as \( r = 1 + \sin\theta \). By using polar coordinates, we can explore curves in new ways, making them essential for calculus related to curves like tangent lines.

Horizontal Tangents

Horizontal tangents occur at specific points on a polar curve where the slope is zero. In polar coordinates, these points can be identified by converting the given polar equations into Cartesian forms.
After transformation, you calculate the derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\). For a point to have a horizontal tangent line, the derivative of \(y\) with respect to \(\theta\), \(\frac{dy}{d\theta}\), must equal zero, while \(\frac{dx}{d\theta}\) must not equal zero.
By analyzing these conditions, we find critical angles \(\theta\) and their corresponding \(r\) values. For the curve \(r = 1 + \sin\theta\), one might solve \(\cos\theta + \, 2\sin(\theta)\cos(\theta) = 0\).
Simplifying, this becomes \(\cos\theta(1 + \sin\theta) = 0\), solving for \(\cos\theta = 0\), such as \(\theta = \pi/2\) and \(\theta = 3\pi/2\). The resulting coordinates indicate where horizontal tangents occur.

Vertical Tangents

Vertical tangents on polar curves are points where the curve is rising or falling directly. These points are identified where the slope becomes undefined, which occurs when \(\frac{dx}{d\theta} = 0\).
Here, the condition for a vertical tangent can be expressed in terms of the derivatives you've calculated. For \(r = 1 + \sin\theta\), you solve the equation \(-\sin\theta + \cos(2\theta) = 0\).
Solving this often confirms or provides new angles \(\theta\) where vertical tangents appear but requires ensuring \(\frac{dy}{d\theta} eq 0\).
In cases such as the cardioid \(r = 1 + \sin\theta\), it is common that most solutions from horizontal tangents work for vertical tangents, showing the points where the slope shifts between undefined and zero.

Derivatives of Polar Equations

To find tangent lines of polar curves, we need to convert polar equations to a Cartesian form and explore their derivatives. Calculus with polar equations involves differentiating these to find the rate of change with respect to the angle \(\theta\).
The expressions \(x = r\cos\theta\) and \(y = r\sin\theta\) are differentiated to obtain \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\). These derivatives tell us how quickly \(x\) and \(y\) change as \(\theta\) changes and are key to analyzing tangents:

• Horizontal tangents: \(\frac{dy}{d\theta} = 0\)
• Vertical tangents: \(\frac{dx}{d\theta} = 0\)

These derivatives help uncover valuable insights about the nature and behavior of curves represented dynamically by polar equations.