Problem 43
Question
A polar equation is given. (a) Express the polar equation in parametric form. (b) Use a graphing device to graph the parametric equations you found in part (a). $$ r=2^{\theta / 12}, \quad 0 \leq \theta \leq 4 \pi $$
Step-by-Step Solution
Verified Answer
Convert to parametric: \( x = 2^{\theta/12} \cos \theta \), \( y = 2^{\theta/12} \sin \theta \); Graph this for \( 0 \leq \theta \leq 4\pi \).
1Step 1: Convert to Parametric Equations - X
To express the given polar equation in parametric form, we first need to find the x-coordinate in terms of a parameter \( \theta \). Using the formula for conversion from polar to Cartesian coordinates, we have:\[x = r \cos(\theta)\]Substituting the expression for \( r \), the equation becomes:\[x(\theta) = \left(2^{\theta/12}\right) \cos(\theta)\]
2Step 2: Convert to Parametric Equations - Y
Similarly, find the y-coordinate using the polar-to-Cartesian conversion formula:\[y = r \sin(\theta)\]Substitute the expression for \( r \) to obtain:\[y(\theta) = \left(2^{\theta/12}\right) \sin(\theta)\]
3Step 3: Define the Parameter Range
With the parameter \( \theta \) now associated with the parametric equations:- \( x(\theta) = \left(2^{\theta/12}\right) \cos(\theta) \)- \( y(\theta) = \left(2^{\theta/12}\right) \sin(\theta) \)We set the range for \( \theta \) as given by the problem:\[0 \leq \theta \leq 4\pi\]
4Step 4: Graph the Parametric Functions
Use a graphing device (or software like Desmos, GeoGebra) to plot the parametric equations.1. Enter the parametric equations: - \( x(\theta) = \left(2^{\theta/12}\right) \cos(\theta) \) - \( y(\theta) = \left(2^{\theta/12}\right) \sin(\theta) \)2. Set the parameter \( \theta \) to vary from 0 to \( 4\pi \).3. Observe the graph to see how the spiral behaves.
Key Concepts
Parametric EquationsPolar to Cartesian ConversionGraphing Techniques
Parametric Equations
Parametric equations are a fundamental way to represent a curve in the plane. Instead of describing a curve via a single equation relating the two coordinate variables directly, these equations express each coordinate as a separate function of an independent parameter, often denoted as \( \theta \). This method allows for more flexibility in describing complex curves.
For the given polar equation \( r=2^{\theta / 12} \), the parametric form involves finding expressions for \( x \) and \( y \). Here, we use the conversions:
For the given polar equation \( r=2^{\theta / 12} \), the parametric form involves finding expressions for \( x \) and \( y \). Here, we use the conversions:
- \( x = r \cos(\theta) \)
- \( y = r \sin(\theta) \)
- \( x(\theta) = \left(2^{\theta/12}\right) \cos(\theta) \)
- \( y(\theta) = \left(2^{\theta/12}\right) \sin(\theta) \)
Polar to Cartesian Conversion
Converting between coordinate systems, such as from polar to Cartesian coordinates, is a critical skill in mathematics. This conversion is particularly useful when analyzing curves and plotting them with standard x-y axes. In polar coordinates, the position of a point is determined by a distance from the origin \( r \) and an angle \( \theta \), whereas in Cartesian coordinates, a point is defined by \( x \) and \( y \).
The conversion formulas:
The conversion formulas:
- \( x = r \cos(\theta) \)
- \( y = r \sin(\theta) \)
Graphing Techniques
Once parametric equations are established, graphing them effectively helps visualize the mathematical relationships they represent. This process involves using the parameter's values to plot points, revealing the curve's shape. Graphing devices and software are invaluable tools for this purpose.
For the parametric equations:
For the parametric equations:
- \( x(\theta) = \left(2^{\theta/12}\right) \cos(\theta) \)
- \( y(\theta) = \left(2^{\theta/12}\right) \sin(\theta) \)
Other exercises in this chapter
Problem 42
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Find the intersection points of the pair of ellipses. Sketch the graphs of each pair of equations on the same coordinate axes and label the points of intersecti
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Hyperbolas are called confocal if they have the same foci. (a) Show that the hyperbolas $$\frac{y^{2}}{k}-\frac{x^{2}}{16-k}=1 \quad \text { with } 0
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