Graphs of Functions and Derivatives 131Example 1
Determine the symmetry, if any, of the graph ofr= 2 +4 cosθ.
Step 1: Since 2+4 cos(−θ)= 2 +4 cosθ, the graph is symmetric about the polar axis.
Step 2: 2 +4 cos(π−θ)= 2 −4 cosθ, so the graph is not symmetric about the line
θ=
π
2.
Step 3: Since 2+4 cos(θ+π)= 2 + 4 [cosθcosπ−sinθsinπ]= 2 −4 cosθ, the graph is
not symmetric about the pole.
Example 2
Determine the symmetry, if any, of the graphr= 3 −3 sinθ.
Step 1: Since 3−3 sin(−θ)= 3 +3 sinθis not equal tor= 3 −3 sinθ, the graph is not
symmetric about the polar axis.
Step 2: 3 −3 sin(π−θ)= 3 −3 sinθ, so the graph is symmetric about the lineθ=π/2.
Step 3: Since 3−3 sin(θ+π)= 3 −3[sinθcosπ+sinπcosθ]= 3 +3 sinθ, the graph is
not symmetric about the pole.
Example 3
Determine the symmetry, if any, of the graph ofr=5 cos(4θ).
Step 1: Since 5 cos(4(−θ))=5 cos 4θ, the graph is symmetric about the polar axis.
Step 2: 5 cos(4(π−θ))=5 cos(4π− 4 θ) which, by identity, is equal to 5[cos 4πcos 4θ+
sin 4πsinθ] or 5 cos 4θ, the graph is symmetric about the lineθ=π/2.
Step 3: Since 5 cos 4(θ+π)=5 cos(4θ+ 4 π)=5 cos(4θ), the graph is symmetric about the
pole.
Vectors
A vector represents a displacement of both magnitude and direction. The length,r,of
the vector is its magnitude, and the angle,θ, it makes with thex-axis gives its direction.
The vector can be resolved into a horizontal and a vertical component.x=‖r‖cosθand
y=‖r‖sinθ.
A unit vector is a vector of magnitude 1. Ifi=〈1, 0〉is the unit vector parallel to the
positivex-axis, that is, a unit vector with direction angleθ=0, andj=〈0, 1〉is the unit
vector parallel to they-axis, with an angleθ=
π
2
, then any vector in the plane can be
represented asxi+yjor simply as the ordered pair〈x,y〉. The magnitude of the vector is
‖r‖=
√
x^2 +y^2 , and the direction can be found from tanθ=
y
x. Since tan−^1
(y
x)
will returnvalues in quadrant I or quadrant IV, if the terminal point of the vector falls in quadrant II
or quadrant III, the direction angle will be equal to tan−^1
(y
x)
+π.