A rose has n petals if n is odd and 2n petals if n is

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A rose has n petals if n is odd and 2n petals if n is even. r=a cos ( ) , r=a sin ( ) . Roses are graphs of polar equations of the forms,
r = 4cos(4 ) Since n is even, number of petals are 2*4 = 8
r = 4cos(5 ) Since n is odd, number of petals are 5
Lemniscates. The word lemniscates comes from the Latin word for ribbon Lemniscates are graphs of polar equations of the form r 2 =a 2 cos(2 ), r 2 = a 2 sin(2 )
r 2 = 4cos(2 )
There exist many kinds of interesting spirals. Spirals. r= ± θ r=e θ r= a θ .
TANGENTS TO POLAR CURVES To find a tangent line to a polar curve r = f ( θ ) , we regard θ as a parameter and write its parametric equations as: x = r cos θ = f ( θ ) cos θ y = r sin θ = f ( θ ) sin θ
TANGENTS TO POLAR CURVES contd. Method for finding slopes of parametric curves we have; sin cos cos sin dy d dy dy d dx dx d r r dy d d dx dr dx r d d
TANGENTS TO POLAR CURVES We locate horizontal tangents by finding the points where dy / = 0 (provided that dx / ≠ 0). Likewise, we locate vertical tangents at the points where dx / = 0 (provided that dy / ≠ 0).
TANGENTS TO POLAR CURVES Notice that, if we are looking for tangent lines at the pole, then r = 0 and it simplifies to: (Note that there are many tangents with different slope at origin) tan if 0 dy dr dx d
TANGENTS TO POLAR CURVES For instance, in Example 9, we found that r = cos 2 θ = 0 when θ = π /4 or 3 π /4. This means that the lines θ = π /4 and θ = 3 π /4 (or y = x and y = x ) are tangent lines to r = cos 2 θ at the origin.
Example 10 a.For the cardioid r = 1 + sin θ of Example 7, find the slope of the tangent line when θ = π /3. b. Find the points on the cardioid where the tangent line is horizontal or vertical.
Example 10(a) Using the formula with r = 1 + sin θ , we have: 2 sin cos cos sin cos sin (1 sin )cos cos cos (1 sin )sin cos (1 2sin ) cos (1 2sin ) 1 2sin sin (1 sin )(1 2sin ) dr r dy d dr dx r d
Example 10(a) The slope of the tangent at the point where θ = π /3 is: 3 1 2 cos( /3)(1 2sin( /3)) (1 sin( /3))(1 2sin( /3) (1 3) (1 3 / 2)(1 3) 1 3 1 3 1 (2 3)(1 3) 1 3 dy dx 
Example 10(b) Observe that: dy d cos (1 2sin ) 0 when 2 , 3 2 , 7 6 , 11 6 dx d (1 sin )(1 2sin ) 0 when 3 2 , 6 , 5 6
Example 10(b) contd. Hence, there are horizontal tangents at the points (2, π /2), (½, 7 π /6), (½, 11 π /6) and vertical tangents at (3/2, π /6), (3/2, 5 π /6) When θ = 3 π /2, both dy / and dx / are 0. So, we must be careful.
Example 10(b) contd. Using l’Hospital’s Rule, we have: (3 / 2) (3 /2) (3 /2) (3 /2) (3 /2) lim 1 2sin cos lim lim 1 2sin 1 sin 1 cos lim 3 1 sin 1 sin lim 3 cos dy dx      
Example 10(b) contd. By symmetry, (3 / 2) lim dy dx  
Example 10(b) contd. Thus, there is a vertical tangent line at the pole.
Intersection Points. We know that a given point can be represented in multiple ways in polar coordinates. It may happen that only one of these polar coordinates satisfies a given curve equation while a different polar coordinates of the same point satisfies another curve equation.
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