Problem 245 - Calculus Volume 3

Question

For the cardioid \(r=1+\sin \theta,\) find the slope of the tangent line when \(\theta=\frac{\pi}{3}\).

Step-by-Step Solution

Answer

The slope is -1.

Step 1: Express slope in polar coordinates

For a polar equation of the form \( r = f(\theta) \), the slope of the tangent line at a given point \((r, \theta)\) is found using the formula:\[\text{slope} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}\]Here, \( f(\theta) = 1 + \sin\theta \). Our goal is to first determine \( f'(\theta) \).

Step 2: Differentiate the polar function

Differentiate \( f(\theta) = 1 + \sin\theta \) with respect to \( \theta \):\[f'(\theta) = \cos\theta\]

Step 3: Evaluate at the given angle

Substitute \(\theta = \frac{\pi}{3}\) into \(f(\theta)\), \(f'(\theta)\), and calculate trigonometric values:- \( f\left(\frac{\pi}{3}\right) = 1 + \sin\frac{\pi}{3} = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2} \)- \( f'\left(\frac{\pi}{3}\right) = \cos\frac{\pi}{3} = \frac{1}{2} \)

Step 4: Substitute into the slope formula

Plug in the values of \(f'(\theta)\), \(f(\theta)\), and \(\theta = \frac{\pi}{3}\) into the slope formula:\[\text{slope} = \frac{\left(\frac{1}{2}\right)\sin\frac{\pi}{3} + \left(\frac{2 + \sqrt{3}}{2}\right)\cos\frac{\pi}{3}}{\left(\frac{1}{2}\right)\cos\frac{\pi}{3} - \left(\frac{2 + \sqrt{3}}{2}\right)\sin\frac{\pi}{3}}\]Evaluate:- \( \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \)- \( \cos\frac{\pi}{3} = \frac{1}{2} \)Recalculate:- Numerator: \( \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{2 + \sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} + \frac{2 + \sqrt{3}}{4} = \frac{2 + 2\sqrt{3}}{4} \)- Denominator: \( \frac{1}{2} \cdot \frac{1}{2} - \frac{2 + \sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} - \frac{\sqrt{3}(2 + \sqrt{3})}{4} \)

Step 5: Simplify the fraction

Simplify the expression to find the slope:Calculate the numerator:\[ \frac{2 + 2\sqrt{3}}{4} = \frac{1 + \sqrt{3}}{2} \]Calculate the denominator:\[ \frac{1}{4} - \frac{2\sqrt{3} + 3}{4} = \frac{1 - 2\sqrt{3} - 3}{4} = \frac{-2 - 2\sqrt{3}}{4} = \frac{-1 - \sqrt{3}}{2} \]The slope is:\[ \frac{\frac{1 + \sqrt{3}}{2}}{\frac{-1 - \sqrt{3}}{2}} = -1 \]

Step 6: Conclusion: Final result

Therefore, the slope of the tangent line to the cardioid \(r = 1 + \sin \theta\) at \(\theta = \frac{\pi}{3}\) is \(-1\).

Key Concepts

Cardioid

A cardioid is a fascinating shape in mathematics and looks somewhat like a heart. It gets its name from the French word "cardioide," meaning heart. The term is derived because of its characteristic shape, which looks like a stylized heart or a vector arrow. The equation for a cardioid can be written in polar coordinates, where the distance from the origin to a point on the curve varies as a function of the angle to that point. For example, the cardioid we are examining is described by the polar equation \(r = 1 + \sin \theta\). This means the radius \(r\) from the origin to the curve depends on the angle \(\theta\). Since \(\sin\theta\) takes values between -1 and 1, \(r\) will range from 0 to 2, creating the cardioid's distinctive looping shape.

Tangent Line

The tangent line to a curve is a straight line that just touches the curve at a particular point. It represents the direction in which the curve is heading at that point. In the context of our cardioid, we are interested in the slope of this tangent line at a specific angle \(\theta\). Calculating this slope involves understanding how the radius of the curve changes with respect to the angle.

• The tangent line gives a linear approximation to the curve at that point, offering insights into the curve's behavior.
• Finding the slope of this line is crucial for understanding the local geometry of the curve.
• For the problem, we evaluated this slope at \(\theta = \frac{\pi}{3}\), leading to a slope of -1, indicating the line descends as \(\theta\) increases.

Remember, the slope of the tangent line can tell us a lot about how the curve behaves near that point, including whether the curve is curving upwards or downwards.

Derivative in Polar Coordinates

In polar coordinates, the derivative is used to determine the slope of the tangent line to the curve at a certain angle. This derivative requires differentiating the polar equation's radial function \(r = f(\theta)\) with respect to \(\theta\). Using the derivative, we find how the curve changes with every tiny increase in angle and hence find the slope of the tangent line.

• The formula \(\frac{f'\left(\theta\right)\sin\theta + f\left(\theta\right)\cos\theta}{f'\left(\theta\right)\cos\theta - f\left(\theta\right)\sin\theta}\) gives us the slope of the tangent line.
• For the cardioid \(r = 1 + \sin\theta\), we found \(f'(\theta) = \cos\theta\).
• This expression incorporates trigonometric functions because the position on the curve depends on angles.

Calculating derivatives in polar coordinates is essential because it accounts for how the shape is oriented and changes with angles.

Trigonometric Values

Working in polar coordinates often means dealing with trigonometric values like sine and cosine, which are crucial for calculations involving angles. When we have to calculate derivatives or slopes in polar coordinates, sine and cosine allow us to relate linear and angular changes.

• At \(\theta = \frac{\pi}{3}\), key trigonometric values are \(\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\) and \(\cos\frac{\pi}{3} = \frac{1}{2}\).
• These values are commonly encountered because they correspond to standard angles on the unit circle.
• Accurate calculation of trigonometric values ensures that the slope calculations and other derivations maintain precision.

Understanding how to derive these values quickly is vital for solving problems involving polar coordinates efficiently.