Chapter 10: Problem 18
Find the slopes of the curves in Exercises at the given points. Sketch the curves along with their tangent lines at these points. $$\text { Cardioid } \quad r=-1+\sin \theta ; \quad \theta=0, \pi$$
Short Answer
Expert verified
The slopes at \(\theta = 0\) and \(\theta = \pi\) are both undefined (vertical tangents).
Step by step solution
01
Convert from Polar to Cartesian Coordinates
The cardioid is given by the equation \(r = -1 + \sin \theta\). To find the slopes, we first convert the equation to Cartesian coordinates.The conversions are:\[x = r \cos(\theta), \quad y = r \sin(\theta)\]Substituting \(r = -1 + \sin \theta\), we get:\[x = (-1 + \sin \theta) \cos \theta, \quad y = (-1 + \sin \theta) \sin \theta\].
02
Differentiate to Find Derivatives
To find the slope of the tangent line at a given point, we need to find the derivative \(\frac{dy}{dx}\).Using the chain rule and the derivative forms for \(x\) and \(y\) in terms of \(\theta\),we obtain:\[\frac{dx}{d\theta} = \sin(\theta) \sin(\theta) + (-1 + \sin(\theta)) \cos(\theta) (-\sin(\theta))\]\[\frac{dy}{d\theta} = \sin(\theta) \cos(\theta) + (-1 + \sin(\theta)) \sin(\theta) \cos(\theta)\]Finally, apply the formula:\[\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\].
03
Calculate the Slope at \(\theta = 0\)
At \(\theta = 0\), substitute \(\theta = 0\) into \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\).\(\sin(0) = 0\) and \(\cos(0) = 1\), so:\[\frac{dx}{d\theta} = 0 \cdot 0 + (-1 + 0) \cdot 1 \cdot (-0) = 0\]\[\frac{dy}{d\theta} = 0\cdot 1 + (-1 + 0)\cdot 0\cdot 1 = 0\]Thus, the slope \(\frac{dy}{dx} = \frac{0}{0}\) is undefined at \(\theta=0\), indicating a vertical tangent.
04
Calculate the Slope at \(\theta = \pi\)
At \(\theta = \pi\), substitute \(\theta = \pi\) into \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\).\(\sin(\pi) = 0\) and \(\cos(\pi) = -1\), so:\[\frac{dx}{d\theta} = 0 \cdot 0 + (-1 + 0) (-1) (0) = 0\]\[\frac{dy}{d\theta} = 0 \cdot (-1) + (-1 + 0)(0 \cdot (-1)) = 0\]Thus, the slope \(\frac{dy}{dx} = \frac{0}{0}\) is also undefined at \(\theta=\pi\), indicating a vertical tangent.
05
Sketch the Curves and Tangent Lines
With the slopes being vertical (undefined) at both given points, sketch the cardioid in the Cartesian plane.At both \(\theta = 0\) and \(\theta = \pi\), plot points where \(r\) would be \(-1\) on the horizontal and horizontal reflected line (i.e., \((x,y) = (-1,0)\) and \((x,y) = (1,0)\)).Draw vertical lines at these points to represent the tangent lines.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Coordinates
Polar coordinates are a unique way to represent points on a plane using a combination of angles (\(\theta\)) and distances (\(r\)) from a central origin. This coordinate system is especially useful when dealing with curves and forms like circles and spirals.
Unlike the Cartesian system, which utilizes x and y coordinates to determine a location, the polar system provides an angle and a radial distance:
Unlike the Cartesian system, which utilizes x and y coordinates to determine a location, the polar system provides an angle and a radial distance:
- \(r\): The distance from the origin to a point.
- \(\theta\): The angle from the positive x-axis to the line connecting the point to the origin.
Slope of Tangent Line
The slope of a tangent line to a curve at a given point informs us how steep the curve is visually at that point. It represents the instantaneous rate of change of the curve's y-coordinate as the x-coordinate changes. In simple terms, it tells us how vertical or horizontal the line appears to be.
For a curve represented by polar coordinates, the slope of the tangent line can be found after transforming the polar equation into Cartesian coordinates. Once transformed, the derivative \( \frac{dy}{dx} \) gives the slope. However, if the calculation results in \( \frac{0}{0} \), the slope is undefined, signifying a vertical tangent line.
In the case of the cardioid, at the angles \(\theta = 0\) and \(\theta = \pi\), the derivatives yield an undefined slope. Consequently, the tangents at these points are vertical lines.
For a curve represented by polar coordinates, the slope of the tangent line can be found after transforming the polar equation into Cartesian coordinates. Once transformed, the derivative \( \frac{dy}{dx} \) gives the slope. However, if the calculation results in \( \frac{0}{0} \), the slope is undefined, signifying a vertical tangent line.
In the case of the cardioid, at the angles \(\theta = 0\) and \(\theta = \pi\), the derivatives yield an undefined slope. Consequently, the tangents at these points are vertical lines.
Derivative
The derivative is a fundamental concept in calculus, used to determine the rate at which a function changes. When we differentiate a function, we're effectively calculating its slope at any given point. In the context of curves, the derivative helps define the tangent's slope.
To calculate the derivative of a function like our cardioid, first express it in terms of x and y, then apply the chain rule. This involves performing derivatives separately on functions of the angle \(\theta\):
To calculate the derivative of a function like our cardioid, first express it in terms of x and y, then apply the chain rule. This involves performing derivatives separately on functions of the angle \(\theta\):
- \( \frac{dx}{d\theta} \): The derivative of the x-coordinate with respect to \(\theta\).
- \( \frac{dy}{d\theta} \): The derivative of the y-coordinate with respect to \(\theta\).
Cartesian Coordinates
Cartesian coordinates use the familiar x and y axes to map points on a plane. Unlike the polar system, which uses angles and distances, Cartesian coordinates give a clear, direct placement based on a fixed grid.
Converting from polar to Cartesian coordinates involves using the transformation formulas:
Converting from polar to Cartesian coordinates involves using the transformation formulas:
- \(x = r \cos(\theta)\)
- \(y = r \sin(\theta)\)