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Magazine Chapter 1: Problem 245
For the cardioid \(r=1+\sin \theta,\) find the slope of the tangent line when \(\theta=\frac{\pi}{3}\)
Short Answer
Expert verified
The slope of the tangent line is \( -\frac{1}{3} \).
Step by step solution
01
Convert to Cartesian Coordinates
The cardioid equation in polar coordinates is given as \( r = 1 + \sin \theta \). To find the slope, we first convert the equation to Cartesian coordinates. Use the relationships \( x = r \cos \theta \) and \( y = r \sin \theta \). So, \( x = (1 + \sin \theta) \cos \theta \) and \( y = (1 + \sin \theta) \sin \theta \).
02
Express Cartesian Derivatives
Find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \). For \( x \), differentiate \( (1+\sin \theta) \cos \theta \) with respect to \( \theta \), yielding \( \frac{dx}{d\theta} = -\sin \theta (1 + \sin \theta) + \cos \theta \cos \theta \). For \( y \), differentiate \( (1 + \sin \theta) \sin \theta \), resulting in \( \frac{dy}{d\theta} = \cos \theta (1 + \sin \theta) + \sin \theta \cos \theta \).
03
Calculate Slope as Derivative Ratio
The slope of the tangent line for the cardioid is \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \). Substituting the expressions from the previous step, calculate the ratio to find \( \frac{dy}{dx} \).
04
Evaluate at Specific Angle
Evaluate \( \frac{dy}{dx} \) at \( \theta = \frac{\pi}{3} \). Substitute \( \theta = \frac{\pi}{3} \) into the expressions for \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) and compute the specific values for the derivatives. Simplify using trigonometric identities as needed.
05
Determine Final Slope Value
Compute the slope using the evaluated derivatives: \( \frac{dy}{dx} = \frac{\cos \left(\frac{\pi}{3}\right) \[1 + \sin \left(\frac{\pi}{3}\right)\] + \sin \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}\right)}{-\sin \left(\frac{\pi}{3}\right) \[1 + \sin \left(\frac{\pi}{3}\right) \] + \cos^2 \left(\frac{\pi}{3}\right)} \). Substitute and simplify to get the final slope.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cardioid
A cardioid is a distinctive heart-shaped curve in mathematics, which can be described using polar coordinates. The term "cardioid," derived from the Greek word for heart, reflects this charming shape. In polar form, a cardioid may appear as \( r = 1 + \sin \theta \) or a similar equation featuring a sine or cosine component.
Here's what you need to know about cardioids:
Here's what you need to know about cardioids:
- They possess a single cusp (a pointed end) where the curve goes back on itself.
- Cardioids can be graphically represented using polar plots, revealing their classic, symmetric heart shape.
- The entire curve is swept out as \( \theta \) varies from \( 0 \) to \( 2\pi \).
Slope of Tangent Line
The slope of the tangent line to a curve in polar coordinates gives insight into the curve's steepness at a particular point. To find this for a cardioid such as \( r = 1 + \sin \theta \), you'll need some calculus tools.
Finding the slope involves:
Finding the slope involves:
- First, converting the polar equation to its Cartesian form using \( x = r \cos \theta \) and \( y = r \sin \theta \).
- Then, calculating the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), which correspond to rates of change in Cartesian coordinates.
- Finally, using these derivatives to compute the slope by finding the ratio \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).
Cartesian Coordinates
Cartesian coordinates provide a foundational system in mathematics to describe points in a plane using two numbers, typically denoted as \( (x, y) \).
Here's how Cartesian coordinates play a role in analyzing curves like cardioids:
Here's how Cartesian coordinates play a role in analyzing curves like cardioids:
- They allow for easy plotting and analysis of curves by representing radial equations in a more familiar \( x \) and \( y \) coordinate system.
- By transforming polar curves into Cartesian equations, we can leverage algebraic methods for deeper investigations.
- This conversion also facilitates the use of derivative calculus to find slopes, tangents, and various other analytical properties of curves.
