Problem 50
Question
Use a graphing utility to (a) graph the polar equation, (b) draw the tangent line at the given value of \(\boldsymbol{\theta}\), and (c) find \(d y / d x\) at the given value of \(\theta\). (Hint: Let the increment between the values of \(\theta\) equal \(\pi / 24 .)\) $$ r=3-2 \cos \theta, \theta=0 $$
Step-by-Step Solution
Verified Answer
The short answer to this exercise would be: The graph of the polar equation \(r=3-2 \cos(θ)\) is a limaçon-like structure. The tangent line at \(θ=0\) is exactly the initial line. The derivative \(dy/dx\) at \(θ=0\) could be calculated by transforming the polar equation into Cartesian form, finding their derivatives and then taking the ratio of them at the point \(θ = 0\).
1Step 1: Graphing the polar equation
Enter the polar equation \(r=3-2 \cos \theta\) into the graphing utility. Ensure the polar coordinates mode is selected. This will plot the polar equation.
2Step 2: Plotting the point where \(θ=0\)
To plot a point, substitute the value of \(\theta\) into the equation to get the radius. So, \(r= 3 - 2\cos(0) = 1\). On the polar grid, mark the point at 1 unit along the initial line (\(θ = 0\)).
3Step 3: Drawing the tangent line at \(\theta=0\)
The tangent to a curve at a point is a straight line that just 'touches' the curve at that point. In polar coordinates, since all lines through the origin cut the polar curve in the direction of increasing \(θ\), for \(\theta= 0\), the tangent line is the initial line itself.
4Step 4: Compute \(dy/dx\)
We could transform the polar equation into Cartesian form by substituting \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). This gives us \(x = (3 - 2 \cos(\theta) )\cdot \cos(\theta)\) and \(y = (3 - 2 \cos(\theta)) \cdot \sin(\theta)\). We differentiate both equations with respect to \(\theta\) and then divide \(dy/d\theta\) by \(dx/d\theta\) to get the rate of change \(dy/dx\) at \(\theta = 0\).
Key Concepts
Tangent LinesDerivatives in Polar CoordinatesGraphing Utilities
Tangent Lines
In mathematics, a tangent line is a line that lightly touches a curve at a single point but doesn't cross it at that point. Think of placing a pencil on a round ball; the pencil barely brushes the surface without digging into it. In polar coordinates, which involve angles and radii instead of just x and y like in Cartesian coordinates, defining a tangent line can be a bit tricky.
For a polar curve given by the equation \(r = f(\theta)\), the tangent line at a given angle \(\theta\) can sometimes be straightforward, like when \(\theta = 0\). At this point, the tangent line aligns with the initial line. This means it follows the path radiating directly from the origin in the direction of the angle used. The key is understanding that in polar coordinates, the tangent line's relationship with the curve reflects changes in angle, not just positions like in Cartesian coordinates.
For a polar curve given by the equation \(r = f(\theta)\), the tangent line at a given angle \(\theta\) can sometimes be straightforward, like when \(\theta = 0\). At this point, the tangent line aligns with the initial line. This means it follows the path radiating directly from the origin in the direction of the angle used. The key is understanding that in polar coordinates, the tangent line's relationship with the curve reflects changes in angle, not just positions like in Cartesian coordinates.
Derivatives in Polar Coordinates
Calculating derivatives in polar coordinates might sound challenging, but it's a crucial concept for analyzing how curves change. Unlike the Cartesian system, where motion along x or y is direct, polar coordinates involve motion that winds around a central origin.
To find the derivative \(dy/dx\) in polar coordinates, we express our polar equation \(r = f(\theta)\) in terms of \(x\) and \(y\) coordinates. Using the relationships \(x = r \cos\theta\) and \(y = r \sin\theta\), we get functions for \(x(\theta)\) and \(y(\theta)\). By differentiating these with respect to \(\theta\), we get \(dx/d\theta\) and \(dy/d\theta\).
Interestingly, \(dy/dx\) can then be found by dividing \(dy/d\theta\) by \(dx/d\theta\). In essence, this tells us how the height of the curve changes with respect to its width as it twists around the origin. It's a neat way to translate circular movement into familiar horizontal and vertical shifts.
To find the derivative \(dy/dx\) in polar coordinates, we express our polar equation \(r = f(\theta)\) in terms of \(x\) and \(y\) coordinates. Using the relationships \(x = r \cos\theta\) and \(y = r \sin\theta\), we get functions for \(x(\theta)\) and \(y(\theta)\). By differentiating these with respect to \(\theta\), we get \(dx/d\theta\) and \(dy/d\theta\).
Interestingly, \(dy/dx\) can then be found by dividing \(dy/d\theta\) by \(dx/d\theta\). In essence, this tells us how the height of the curve changes with respect to its width as it twists around the origin. It's a neat way to translate circular movement into familiar horizontal and vertical shifts.
Graphing Utilities
Graphing utilities are powerful tools that help transform theoretical math equations into visual representations. They effectively bridge the gap between paper calculations and visual understanding. Whether using software programs or calculators with graphing capabilities, these tools can handle complex equations.
To use a graphing utility for a polar equation like \(r = 3 - 2 \cos \theta\), one typically selects the polar mode within the tool. Simply inputting the equation provides a dynamic graph that can highlight how the radius changes with \(\theta\). Additionally, graphing utilities allow for detailed analysis by letting users mark specific points, such as where \(\theta = 0\).
To use a graphing utility for a polar equation like \(r = 3 - 2 \cos \theta\), one typically selects the polar mode within the tool. Simply inputting the equation provides a dynamic graph that can highlight how the radius changes with \(\theta\). Additionally, graphing utilities allow for detailed analysis by letting users mark specific points, such as where \(\theta = 0\).
- Accurate graphing with preset increments like \(\pi/24\) can make exploration easier.
- Visual feedback can help in understanding the curve's behavior and any symmetries or patterns.
- These utilities often provide additional features, such as calculating derivatives or marking tangent lines, which can save time and improve comprehension.
Other exercises in this chapter
Problem 50
Explain why finding points of intersection of polar graphs may require further analysis beyond solving two equations simultaneously
View solution Problem 50
Find the arc length of the curve on the given interval. $$ x=t, \quad y=\frac{t^{5}}{10}+\frac{1}{6 t^{3}} \quad 1 \leq t \leq 2 $$
View solution Problem 51
On November \(27,1963,\) the United States launched Explorer \(18 .\) Its low and high points above the surface of Earth were approximately 119 miles and 123,00
View solution Problem 51
Which integral yields the arc length of \(r=3(1-\cos 2 \theta)\) ? State why the other integrals are incorrect. (a) \(3 \int_{0}^{2 \pi} \sqrt{(1-\cos 2 \theta)
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