Problem 20
Question
Draw a sketch of the graph of the given equation.\(r=4 \sin 5 \theta\) (five- leafed rose)
Step-by-Step Solution
Verified Answer
The graph of \(r=4 \sin 5 \theta\) is a five-petaled rose.
1Step 1: Understand the Polar Equation
The given equation is in polar form: \[r = 4 \sin 5\theta\]This is a polar equation representing a rose curve with multiple petals.
2Step 2: Identify Properties of the Rose Curve
The general form for a rose curve is \(r = a \sin(n\theta)\). Here, \(a = 4\) and \(n = 5\). Because \(n = 5\) is odd, the curve will have exactly 5 petals.
3Step 3: Determine Petal Locations
To find where the petals are located, solve for the angles \(\theta\) where \(r\) reaches maximum value or is zero. The maximum value of \(r\) is 4 when the sine function equals 1, occurring at angles where \sin(5\theta) = 1.This happens at: \[5\theta = \(2k+1\)\pi/2, \text{for integers k}\theta = (2k+1)\pi/(10)\]. This results in 5 angles evenly spaced.
4Step 4: Sketch the Axes
Draw the polar coordinate system with a central origin point. Label circles radiating outward which could help in measuring distance \(r\).
5Step 5: Plot Points and Draw Petals
Since there are 5 evenly spaced angles, plot points at angles \[\pi/10, 3\pi/10, 5\pi/10, 7\pi/10, 9\pi/10\].At each angle, the radial distance \(r\) will reach up to 4 units from the origin, drawing each petal symmetrically.
Key Concepts
rose curvepolar coordinatesmultiple petals
rose curve
A 'rose curve' is a special type of graph that you get from certain polar equations. It's named because the resulting graph looks a lot like a flower with multiple petals. This is true for any polar equation of the form \(r = a \sin(n\theta)\) or \(r = a \cos(n\theta)\).
The 'a' value influences the size of the petals, while the 'n' value affects the number of petals. Here, the example \(r=4\sin(5\theta)\) is clearly a rose curve equation. The value of 'a' is 4, which determines the length of the petals. Because 'n' is 5, this means there will be 5 petals, since 'n' is an odd number.
If 'n' were an even number, the number of petals would actually be doubled. So, for example, if 'n' were 4, the curve would have 8 petals. But with an odd number like 5, the number of petals remains 5. Understanding the basic structure of the rose curve makes it easier to sketch and study in polar coordinates.
The 'a' value influences the size of the petals, while the 'n' value affects the number of petals. Here, the example \(r=4\sin(5\theta)\) is clearly a rose curve equation. The value of 'a' is 4, which determines the length of the petals. Because 'n' is 5, this means there will be 5 petals, since 'n' is an odd number.
If 'n' were an even number, the number of petals would actually be doubled. So, for example, if 'n' were 4, the curve would have 8 petals. But with an odd number like 5, the number of petals remains 5. Understanding the basic structure of the rose curve makes it easier to sketch and study in polar coordinates.
polar coordinates
In polar coordinates, points are located based on their distance from a central point (called the pole or origin) and the angle from a reference direction (usually the positive x-axis).
A point in polar coordinates is written as \(r, \theta\). Here, 'r' is the radial distance from the origin to the point, and 'theta' is the angle in radians.
For example, in the equation \(r = 4 \sin(5\theta)\), 'r' changes depending on the angle 'theta'. When 'theta' changes, it determines the radial distance 'r', which in turn traces out the curve.
Polar coordinates are particularly useful for plotting curves like the rose curve because they naturally describe circular and radial features more efficiently than Cartesian coordinates. Instead of plotting on an x-y grid, we're plotting on a radial grid with circles and angles.
A point in polar coordinates is written as \(r, \theta\). Here, 'r' is the radial distance from the origin to the point, and 'theta' is the angle in radians.
For example, in the equation \(r = 4 \sin(5\theta)\), 'r' changes depending on the angle 'theta'. When 'theta' changes, it determines the radial distance 'r', which in turn traces out the curve.
Polar coordinates are particularly useful for plotting curves like the rose curve because they naturally describe circular and radial features more efficiently than Cartesian coordinates. Instead of plotting on an x-y grid, we're plotting on a radial grid with circles and angles.
multiple petals
One of the fascinating aspects of rose curves is their petals. The equation \(r = 4 \sin(5\theta)\) forms 5 petals because the 'n' value is 5. These petals are spaced evenly around the circle.
To find the exact positions of these petals, we solve for 'theta' when 'r' reaches its maximum value, which is 4 in this case. This happens when \(\sin(5\theta) = 1\).
When \(\sin(5\theta) = 1\), 5\theta must be equal to an odd multiple of \(\pi/2\). Therefore, \(\theta = (2k+1)\pi/(10)\) for integers 'k'. This spacing ensures that the 5 petals are symmetrically placed around the circle.
By plotting these points and connecting them smoothly, we get a beautiful five-petaled flower-like curve. This symmetry and periodicity make rose curves not only important in mathematics but also visually appealing and naturally occurring in many phenomena.
To find the exact positions of these petals, we solve for 'theta' when 'r' reaches its maximum value, which is 4 in this case. This happens when \(\sin(5\theta) = 1\).
When \(\sin(5\theta) = 1\), 5\theta must be equal to an odd multiple of \(\pi/2\). Therefore, \(\theta = (2k+1)\pi/(10)\) for integers 'k'. This spacing ensures that the 5 petals are symmetrically placed around the circle.
By plotting these points and connecting them smoothly, we get a beautiful five-petaled flower-like curve. This symmetry and periodicity make rose curves not only important in mathematics but also visually appealing and naturally occurring in many phenomena.
Other exercises in this chapter
Problem 19
Find a polar equation of the graph having the given cartesian equation.\(\left(x^{2}+y^{2}\right)^{2}=4\left(x^{2}-y^{2}\right)\)
View solution Problem 20
Find the area of the region swept out by the radius vector of the spiral \(r=a \theta\) during its second revolution which was not swept out during its first re
View solution Problem 22
Draw a sketch of the graph of the given equation.\(r=3 \cos 3 \theta\) (three- leafed rose)
View solution Problem 22
Find a polar equation of the graph having the given cartesian equation.\(y=\frac{2 x}{x^{2}+1}\)
View solution