Problem 46 - The Calculus with Analytic Geometry

Question

Find an equation of each of the tangent lines to the given curve at the pole.Find a polar equation of the tangent line to the curve \(r=-6 \sin \theta\) at the point \(\left(6, \frac{3}{2} \pi\right)\).

Step-by-Step Solution

Answer

The polar equation of the tangent line is \ y = -6.

Step 1: Polar to Cartesian Conversion

Convert the polar coordinates to Cartesian coordinates. Given point is \((6, \frac{3}{2}\frac{\theta}{\text{π}})\). Using formulas \(x = r \times \text{cos}(\theta)\) and \(y = r \times \text{sin}(\theta)\), we get:\[x = 6 \times \text{cos}(\frac{3}{2}\text{π}) = 6 \times 0 = 0\]\[y = 6 \times \text{sin}(\frac{3}{2}\text{π}) = 6 \times (-1) = -6\]\Thus, the Cartesian coordinates are \((0, -6)\).

Step 2: Differentiate Polar Equation

Differentiate the polar equation \(r = -6 \text{sin}(\theta)\) to find the slope of the tangent. The formula for the slope \(\frac{dy}{dx}\) in polar coordinates is \(\frac{\frac{dr}{d\theta} \text{sin}(\theta) + r \text{cos}(\theta)}{\frac{dr}{d\theta} \text{cos}(\theta) - r \text{sin}(\theta)}\):\[r = -6 \text{sin}(\theta)\]\[\frac{dr}{d\theta} = -6 \text{cos}(\theta)\]\[\frac{dy}{dx} = \frac{-6 \text{cos}(\theta) \text{sin}(\theta) + (-6 \text{sin}(\theta)) \text{cos}(\theta)}{-6 \text{cos}(\theta) \text{cos}(\theta) - (-6 \text{sin}(\theta)) \text{sin}(\theta)} = \frac{-12 \text{sin}(\theta) \text{cos}(\theta)}{-6 (\text{cos}^2(\theta) - \text{sin}^2(\theta))} = \frac{2 \text{sin}(\theta) \text{cos}(\theta)}{\text{cos}^2(\theta) - \text{sin}^2(\theta)}\]\

Step 3: Evaluate at Specific Point

Substitute \(\theta = \frac{3}{2} \text{π}\) into \( \frac{dy}{dx}\). Since, \(\text{sin}(\frac{3}{2}\text{π}) = -1\) and \(\text{cos}(\frac{3}{2}\text{π}) = 0\):\[\frac{dy}{dx} = \frac{2 \times (-1) \times 0}{0^2 - (-1)^2} = 0\]\Thus, the slope of the tangent line at the point is 0.

Step 4: Equation of Tangent Line

Since the slope \(m = 0\) and the Cartesian coordinates of the point are \( (0, -6) \, the equation of the tangent line can be written as \y = -6\).

Key Concepts

Polar to Cartesian Conversion

When working with curves in polar coordinates, it’s often helpful to convert points to Cartesian coordinates. This makes it easier to visualize the problem. Polar coordinates \(r, \theta\) can be transformed into Cartesian coordinates using the equations:

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

For example, given the point \(6, \frac{3}{2} \pi\), we find:

• \(x = 6 \cos(\frac{3}{2} \pi) = 6 \times 0 = 0\)
• \(y = 6 \sin(\frac{3}{2} \pi) = 6 \times (-1) = -6\)

Thus, the Cartesian coordinates of the point are \(0, -6\). Converting points into Cartesian coordinates sets a strong foundation for further analysis, like finding the slope of the tangent line.

Differentiation in Polar Coordinates

To understand the slope of a curve in polar coordinates, differentiation is key. When you have a polar equation \(r(\theta)\), you can convert derivatives into Cartesian form with appropriate formulas. To find \frac{dy}{dx}\, use:

• \(\frac{dy}{dx} = \frac{\frac{dr}{d\theta} \sin(\theta) + r \cos(\theta)}{\frac{dr}{d\theta} \cos(\theta) - r \sin(\theta)}\)

Let's differentiate the given polar equation \(r = -6 \sin(\theta)\):

• Find the derivative: \(\frac{dr}{d\theta} = -6 \cos(\theta)\)
• Apply it in the formula:
  • \(\frac{dy}{dx} = \frac{-6 \cos(\theta) \sin(\theta) + (-6 \sin(\theta)) \cos(\theta)}{-6 \cos(\theta) \cos(\theta) - (-6 \sin(\theta)) \sin(\theta)}\)
  • Combine like terms: \(\frac{dy}{dx} = \frac{-12 \sin(\theta) \cos(\theta)}{-6 (\cos^2(\theta) - \sin^2(\theta))} = \frac{2 \sin(\theta) \cos(\theta)}{\cos^2(\theta) - \sin^2(\theta)}\)

The derivative step is essential, as it prepares us for evaluating the slope at specific points.

Slope of Tangent Line

Once you have \(\frac{dy}{dx}\) from the differentiation, find the slope of the tangent line at a specific point by plugging in the angle \(\theta \). For instance, to find the slope at \(\left(6, \frac{3}{2} \pi\right)\):

• Since \(\sin(\frac{3}{2} \pi) = -1\) and \(\cos(\frac{3}{2} \pi) = 0\), substitute these values into \(\frac{dy}{dx}\)
• This results in \(\frac{dy}{dx} = \frac{2 \times (-1) \times 0}{0^2 - (-1)^2} = 0\)

Hence, the slope of the tangent line at this point is 0. Knowing the slope tells us if the line is horizontal, vertical, or something else. In this case, a slope of 0 indicates a horizontal tangent line.

Equation of Tangent Line

After identifying the slope, use it with the point’s Cartesian coordinates to find the equation of the tangent line. Given a slope (m) and a point (x, y), the line’s equation can be formulated using:

• \(y - y_1 = m(x - x_1)\)

Here, the slope \(m = 0\) and the Cartesian coordinates are \(0, -6\). This results in:

• \(y - (-6) = 0(x - 0)\)
• Simplified: \(y = -6\)

Finally, the equation of the tangent line is simply \(y = -6\). This straightforward equation signifies that the tangent line is horizontal, passing through the y-coordinate \(-6\).