DETERMINING WHETHER A POINT LIES ON A POLAR CURVE 1

DETERMINING WHETHER A POINT LIES ON A
POLAR CURVE
SATISH K. PANDEY
1. Polar Coordinates Of A Point
We know that a point [r, θ] in polar coordinate system does not have
a unique representation, that is, the polar coordinates assigned to a
point are not unique. Many pairs [r, θ] can represent the same point.
As for example the point represented by [1, π/4] can also be represented
by [1, π/4 + 2π] or by [−1, π/4 + π]. In general, we saw that
[r, θ] = [r, θ + 2nπ] = [−r, θ + (2n + 1)π] for all integers n.
2. Showing That A Point Lies On A Polar Curve
The fact that a single point has many pairs of polar coordinates can
cause complications. In particular, it means that a point [r1 , θ1 ] can
lie on a curve given by a polar equation although the coordinates r1
and θ1 do not satisfy the equation. As for example, the coordinates of
[2, π/2] do not satisfy the equation r = 2 cos 2θ:
r=2
but
2 cos 2θ = 2 cos(2π/2) = 2 cos π = −2.
However, it can be easily shown that the point [2, π/2] does lie on the
curve r = 2 cos 2θ, for [2, π/2] = [−2, 3π/2] and the coordinates of
[−2, 3π/2] do satisfy the equation:
3π
) = 2 cos 3π = −2.
2
In general, a point P [r1 , θ1 ] lies on a curve given by a polar equation if
it has at least one polar coordinate representaion [r, θ] with coordinates
that satisfy the equation. We illustrate this fact by one more example.
r = −2
and
2 cos 2θ = 2 cos(2
Example 2.1. Show that the point [1, − 65 π] lies on the polar curve
r2 = sin 3θ.
Date: March 5, 2014.
1
Proof. It is easy to see that the coordinates of [1, − 65 π] do not satisfy
the equation r2 = sin 3θ:
5
5
r2 = 1
but
sin 3θ = sin(−3 π) = − sin( π) = −1.
6
2
5
Neverthless, [1, − 6 π] = [−1, π/6] and the coordinates of [−1, π/6] do
not satisfy the equation r2 = sin 3θ:
π
π
r2 = 1
and
sin 3θ = sin(3 ) = sin( ) = 1.
6
2
3. How To Assert That A Point Does Not Lie On A Polar
Curve?
The problem creeps in when we want to assert that a point does not
lie on a curve given by a polar equation. Having said that a single
point has many (actually infinitely many) pairs of polar coordinates, if
we want to show that a point [r, θ] does not lie on a curve, we need to
show that none of it’s representations satisfy the equation of the polar
curve. Of course, it is not feasible to check all possible representations
by plugging in the values of their coordinates in the equation of the
polar curve. So the nagging question is:
Is there a way to assert that a point [r, θ] does not lie on a curve given by
a polar equation without checking all the representations of that point?
In particular, the question is: If few of the infinitely many representations of a given point [r, θ] do not satisfy the polar equation of the
curve, does that allow us to assert that the point does not lie on the
curve? Let us describe this problem by an example.
Problem 3.1. Determine whether the point [1, π/4] lies on the curve
r2 = cos 2θ.
Observe that the coordinates [1, π/4] do not satisfy the equation
r = cos 2θ :
π
π
r2 = 1
but
cos 2θ = cos(2 ) = cos( ) = 0.
4
2
π
Next we naturally think of the coordinates [−1, 5 4 ]; for [1, π/4] =
[−1, 5 π4 ]. However, the coordinates [−1, 5 π4 ] too do not satisfy the equation r2 = cos 2θ :
5π
5π
r2 = 1
but
cos 2θ = cos(2 ) = cos( ) = 0.
4
2
So, does that mean the point [1, π/4] does not lie on the curve? The
answer is YES! The point does not lie on the curve. In other words,
2
to show that a point [r1 , θ1 ] does not lie on the curve r2 = cos 2θ, it
suffices to show that the coordinates [r1 , θ1 ] and [−r1 , π + θ1 ] do not lie
on the curve r2 = cos 2θ.
We wish to generalize the above condition for a bigger class of polar
curves. We claim that
Proposition 3.2. If r = ρ(θ) is a polar curve involving only the
trigonometric functions sin x, cos x, tan x, cot x, sec x, and csc x then in
order to show that a point [r1 , θ1 ] does not lie on the curve r = ρ(θ), it
suffices to show that the coordinates [r1 , θ1 ] and [−r1 , π + θ1 ] do not lie
on the curve r = ρ(θ).
Proof. This is not a rigorous proof and follows from a simple observation. If f is one of the above mentioned trigonometric functions
then, [f (Kθ + C)] = [f (K(2nπ + θ) + C)] for every θ. As for example, sin(Kθ + C) = sin(K(2nπ + θ) + C); for sin(K(2nπ + θ) + C) =
sin(2Knπ+Kθ+C) = sin(Kθ+C) because sine function is periodic and
2nπ serves as its period for any integer n. In fact all trigonometric function are periodic with 2π being their period. This ensures that if [r1 , θ1 ]
does not lie on the curve r = ρ(θ) then [r1 , θ1 + 2nπ] will also not lie on
it. Also, if in addition [−r1 , π + θ1 ] does not lie on the curve r = ρ(θ)
then, by the previous argument, [−r1 , π + θ1 + 2nπ] = [−r1 , (2n + 1)π]
will also not lie on it. This ensures that none of the representations of
the point [r1 , θ1 ] satisfies the equation of the curve and hence the point
does not lie on the curve.
Remark 3.3. Note that the above proposition has been shown to be
true only for polar curves involving trigonometric functions.

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