Week 8. Polar curves · 2019. 4. 15. · Week 8. Polar curves In ru or: Jer´ emie Bettinelli (´ [email protected]) Tutorial Assi ants: –Nicolas Brigouleix (groups

Analysis MAA – Bachelor – Year Course webpage: http://www.normalesup.org/˜bettinel/ens/bachelor/maa.html

Week . Polar curves

In ru�or: Jeremie Bettinelli ([email protected])Tutorial Assi ants:

– Nicolas Brigouleix (groups &, [email protected])– Ludovic Cesbron (groups &, [email protected]).

The solutions to the exercises that have not been solved during tutorials will be available on the course webpage.

Basic exercises

Under and and ma er these exercises in order to pass the course.

:::::::Exercise

::1. h Lines and circles g Give a polar parameterization of the following curves.

x

y

dθθ

M(θ)

x

y

θ

M(θ)

x

y

r

x

y

θ

θ

M(θ)

r

x

y

θ

d

r

Hint: It’s a trap!

Solution to exercise 1. For in ance,

a) ρ(θ) =d

cos(θ −θ), θ ∈

(θ − π ,θ + π

). In particular, the vertical line x = d is parameterized

by θ ∈ (−π ,π ) 7→ d

cos(θ)and the horizontal line y = d is parameterized by θ ∈ (,π) 7→ d

sin(θ).

b) This is a trap! It is not possible: for θ , θ mod π, we have ρ(θ) = . By continuity, ρ ≡ .

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c) ρ(θ) = r, θ ∈R (or any interval containing some [a,a+ π) or (a,a+ π]).

d) ρ(θ) = r cos(θ −θ), θ ∈ R (or any interval containing some [a,a+π) or (a,a+π]). Note thatthe circle is visited twice on a period of ρ.

e) This is another trap! It is not possible: ρ should be defined on some interval [θ −θ,θ +θ]

of length less than π. As a result, for each θ ∈ [θ −θ,θ+θ], the lineL (O,~uθ) can cross the

curve only once.

Fun fa�. The mapping z ∈ C 7→ z∈ C ∪ {∞} is called an inversion. We see from a) and d) that this

mapping transforms lines avoiding O into circles passing through O. There even exi s a device that canperform this operation:

:::::::Exercise

:::2. If (I,ρ) is a polar curve with image Γ , and θ ∈ R, λ > are real numbers, give a pa-

rameterization of the image of Γ through the rotation about the pole O of angle θ and scaled by the

fa�or λ.

x

y

x

y

θ

×λ

For a point A ∈R, can you give a polar parameterization of A+ Γ ?



Solution to exercise 2. The polar curve θ ∈ I 7→ λρ(θ −θ) works.

Remember that, with Cartesian parametric curves, it was easy to do a translation; with polar curves, it iseasy to do a rotation.

The curve A+ Γ does not always have a polar parameterization and, when it has, it is not easy to

find it from a polar parameterization of Γ (see Exercise ).

Exercise 3. Study the polar curve given by ρ(θ) = sin(θ).

Solution to exercise 3. The domain of definition is not explicit; we take R, on which ρ is well defined.

Interval of udy. The fun�ion ρ is π -periodic so that we can udy the polar curve on an interval

of length π . We then complete the pi�ure by subsequent rotations about O of angle π (recall that

~uθ+ πis the image of ~uθ through a rotation about O of angle π ). As π ∈ πZ, the pi�ure is

completed after two rotations.

Moreover, ρ is odd, so that we udy ([, π ],ρ) and complete the pi�ure by refle�ion across the

vertical axis.

Finally, ρ(π −θ) = ρ(θ), so that we udy ([, π ],ρ) and complete the pi�ure by refle�ion across

the lineL (O,~u π).

Sign and variations. We have ρ′(θ) = cos(θ).

θ

ρ′

ρ

π

+

Particular times. The time is the only time where M = O. The tangent is the horizontal axis

(L (O,~u)).

Sketch.



x

y

π

Multiple points. There is no need to udy multiple points of ([,π),ρ) as we know their locations

thanks to the isometries. More precisely, there is is only one multiple point: O = M() = M(π ) =

M(π ) is a triple point.

Important exercises

Ma er these exercises for a good grade.

Exercise 4. Study the polar curve given by ρ(θ) =√

cos(θ).

Solution to exercise 4. The fun�ion ρ is defined on [−π ,π ] mod π. A priori, we need to udy each

polar curve ([kπ − π , kπ+ π ],ρ) for k ∈Z.

Interval of udy. The fun�ion ρ is π-periodic so that the image of ([(k+)π− π , (k+)π+ π ],ρ) is

deduced from the image of ([kπ− π , kπ+ π ],ρ) for k ∈Z through a rotation about O of angle π. It is

thus sufficient to udy the polar curve ([−π ,π ],ρ); we then obtain the other curves by applying the

above rotation.

Moreover, ρ is even, so that we udy ([, π ],ρ) and complete the pi�ure by refle�ion across the

horizontal axis.

Sign and variations. The fun�ion ρ is not differentiable at π . On [, π ), we have ρ′(θ) =−sin(θ)√

cos(θ).



θ

ρ′

ρ

π

− −∞

Particular times. The time π is the only time where M = O. The tangent is the fir bise�or

(L (O,~u π)).

Sketch.

x

y

π

−π

Beware that this pi�ure represents the two polar curves ([−π ,π ],ρ) and ([π − π ,π+ π

],ρ).

:::::::Exercise

::5. Study the polar curve given by ρ(θ) = sin

(θ

).

Solution to exercise 5. The domain of definition is not explicit; we take R, on which ρ is well defined.

Interval of udy. The fun�ion ρ is π-periodic so that we can udy the polar curve on an interval

of length π. We then complete the pi�ure by one rotation aboutO of angle π, that is, one rotation

about O of angle π.

Moreover, ρ is odd, so that we udy ([, π ],ρ) and complete the pi�ure by refle�ion across the

vertical axis.


the second bise�orL (O,~u π

).

Sign and variations. We have ρ′(θ) = cos

(θ

). (We put the values at π

and π as they are easily

computable and will help make the sketch more precise.)



θ

ρ′

ρ

π

π

π

+

√ +

+

√

Particular times. The time is the only time where M = O. The tangent is the horizontal line

(L (O,~u)).

Sketch. Let us fir draw the curve on [, π ] and then the whole curve through the above isome-

tries.

x

y

, π

π

ππ

x

y

Multiple points. There is no need to udy multiple points of ([,π),ρ) as we know their locations

thanks to the isometries. More precisely,

• O =M() =M(π ) =M(π) =M(π ) is a quadruple point;

• there are double points:(±√ ,±

√

),(,±

√

)and

(±√ ,

).

:::::::Exercise

::6.

. Study the polar curve given by ρ(θ) =√

+ sin(θ) +√− sin(θ)

.

Hint: Using cosines in ead of sines, try to simplify the expression of ρ as much as possible.



. There is something peculiar about this curve, do you see what?

Solution to exercise 6.

. As θ ∈R 7→ + sin(θ) and θ ∈R 7→ − sin(θ) are nonnegative fun�ions that never cancel at

the same time, the fun�ion ρ is well defined on R.

Interval of udy. The fun�ion ρ is π-periodic so that we obtain the whole curve by udy-

ing the polar curve on any interval of length π.

Moreover, ρ is even, so that we udy ([,π],ρ) and complete the pi�ure by refle�ion across

the horizontal axis.

Next, ρ(π −θ) = ρ(θ), so that we udy ([, π ],ρ) and complete the pi�ure by refle�ion across

the vertical axis.


the fir bise�orL (O,~u π).

Sketch. Let us fir try to simplify the expression of ρ. For θ ∈ [, π ],

ρ(θ)

=√+ sin(θ) +

√− sin(θ)

=√+ cos

(π − θ

)+√− cos

(π − θ

)=

√cos

(π −θ

)+√sin

(π −θ

)=

(cos

(π

)cos

(π −θ

)+ sin

(π

)sin

(π −θ

))= cos(θ) .

In the end, we obtain ρ(θ) = cos(θ) . We have seen in Exercise that this is the polar curve

of part of the vertical line with equation x = . Alternatively, write ρ(θ)cos(θ) =

, that is,

x(θ) = . We can sketch this curve right away.

x

y

O

π



. What is remarkable is that this curve has singular times whose locations are not the pole O,

namely the four corners of the square. This is not contradi�ory with the course as the polar

curve is not differentiable at these times (recall that, if a polar curve is differentiable, all its

singular times have location the pole O).

Exercise 7. h Conics g La week, we have seen the definition of conics via two focuses (or a focus

and dire�rix for the parabola). In fa�, we can also define all of them with one focus F and a dire�rix

D, thank to an extra parameter e ∈R?+ called the eccentricity, which, roughly speaking, measures how far

from a circle the conic is. A conic is the set of all points P of the plane – this is also called the locus of

points P – such that

P F = edi (P ,D) . ()

If e = , we recover the definition of la week for the parabola. If e , , let us put the focus at (−c,) and

the dire�rix at x = − ce .

x

y

− ce

di (P ,D) P

P F

D

F

−c

. Show that P = (x,y) satisfies () if and only if

x

a+ ε

y

b= ,

for some a, b ∈R?+, where ε = if < e < and ε = − if e > . This shows the equivalence with la week’s definitions: we have an ellipse when < e < and a hyperbola when e > .

. Give a polar parameterization of the conic with focus the pole F =O, eccentricity e > and dire�rix

the following line D:



x

y

dθ

D

Hint: Remember that the di ance from the point (x, y) to the line of Cartesian equation ax+by + c =

is|ax + by + c|√a + b

.

. Find the asymptotes of the hyperbola.

x

y D

e =

Solution to exercise 7.



. The point P = (x,y) satisfies () if and only if

P F = edi (P ,D) ⇐⇒ (x+ c) + y = e(x+

c

e)

⇐⇒ x +��XXcx+ c + y = ex +��XXcx+c

e

⇐⇒ x(− e) + y =c

e− c

⇐⇒ e

cx +

e

c(− e)y =

⇐⇒ x

a+ ε

y

b= ,

where a :=ec

, b :=c√|− e|e

, and ε = Sign(− e).

. The dire�rix D is the set of points (x,y) such that (x,y) ·~uθ = d, that is, admits the Cartesian

equation cos(θ)x + sin(θ)y − d = . The point P =(ρ(θ)cos(θ),ρ(θ)sin(θ)

)thus satisfies () if

and only if∣∣∣ρ(θ)∣∣∣ = e

∣∣∣cos(θ)ρ(θ)cos(θ) + sin(θ)ρ(θ)sin(θ)− d∣∣∣ = e

∣∣∣ρ(θ)cos(θ −θ)− d∣∣∣ ,

that is,

ρ(θ) = ±e(ρ(θ)cos(θ −θ)− d

)⇐⇒ ρ(θ) =

εde+ εecos(θ −θ)

,

where ε = ±. In fa�, both equations are equivalent through the mapping (ρ,θ) 7→ (−ρ,θ +π).

As a result, we can keep either parameterizations, for in ance

• ρ : θ ∈ [,π] 7→ de+ ecos(θ −θ)

when < e < (ellipse);

• ρ : θ ∈ (θ −π,θ +π) 7→ d+ cos(θ −θ)

when e = (parabola);

• when e > (hyperbola), we have one polar curve per branch:

ρ : θ ∈ (θ − a,θ + a) 7→ de+ ecos(θ −θ)

,

ρ : θ ∈ (θ + a,π+θ − a) 7→de

+ ecos(θ −θ),

where a := arccos(−e ) is the canceling time of t ∈ [,π] 7→ + ecos(t).

. As θ → θ − a, ρ(θ) → ±∞. We know from the course that we need to look at the limit of



ρ(θ)sin(θ − (θ − a)

). We can write

ρ(θ)sin(θ − (θ − a)

)=d sin

(θ − (θ − a)

)cos(θ −θ)− cos(a)

=−d sin

(θ − (θ − a)

)sin

(θ−θ−a

)sin

(θ−θ+a

)∼ d

sin(a)=

de√e −

.

We have as asymptote the line L(

de√e−

~vθ−a,~uθ−a). Similarly, as θ → θ + a, we find an

asymptoteL(−de√e−

~vθ+a,~uθ+a).

More involved exercises (optional)

Ma er these exercises in order to exceed expe�ations.

Exercise 8. Study the polar curve given by ρ(θ) =cos(θ) + sin(θ) +

.

Solution to exercise 8. The fun�ion ρ is defined on (−π ,−π ) mod π and (−π ,

π ) mod π. A priori,

we need to udy each corresponding polar curves.

Interval of udy. The fun�ion ρ is π-periodic so that we obtain the whole image by udying

the two polar curves ((−π ,−π ),ρ) and ((−π ,

π ),ρ).

We may observe that ρ(π −θ) = ρ(θ) but, unfortunately, this transformation is not easy to apply. . .

Sign and variations. For θ ∈ (−π ,π ) \ {−π }, we have

ρ′(θ) =−sin(θ)

(sin(θ) +

)− cos(θ)

(cos(θ) +

)(sin(θ) +

)= −+ sin(θ) + cos(θ)(

sin(θ) + )

= −+√cos

(θ − π

)(sin(θ) +

) < ,

so that ρ is always decreasing. On the two intervals of udy, it moreover cancels at ±π

.



θ

ρ′

ρ

−π

−π −π

π

π

− − − −

+∞

−∞

+∞

−∞

Particular times. At ±π

, the location is the pole. The tangents are the linesL (O,~u± π ).

Extremities. Let us art with asymptotic dire�ions.

• As θ↘−π , ρ(θ)→ +∞ : the polar curve admits as asymptotic dire�ion the ray R(O,~u− π ).

• As θ↗−π , ρ(θ)→−∞ : the polar curve admits as asymptotic dire�ion the ray R(O,−~u− π ).

• As θ↘−π , ρ(θ)→ +∞ : the polar curve admits as asymptotic dire�ion the ray R(O,~u− π ).

• As θ↗ π , ρ(θ)→−∞ : the polar curve admits as asymptotic dire�ion the ray R(O,−~u π

).

In order to look for asymptotes, it is enough to udy ρ in a neighborhood of −π and of −π by

periodicity (we momentarily change the interval of udy).

• Let θ = −π + h. As h→ ,

ρ(θ)sin(h) =cos

(−π + h

)+

sin(−π + h

)+

sin(h)

=√cos(h) + sin(h) +

−cos(h) +√sin(h) +

sin(h)

∼+√

√

= +√.

We thus have an asymptote: the lineL((+

√

)~v− π ,~u− π

).

• Let θ = −π + h. As h→ ,

ρ(θ)sin(h) =cos

(−π + h

)+

sin(−π + h

)+

sin(h)

=−√cos(h) + sin(h) + −cos(h)−√sin(h) +

sin(h)

∼−√−√

= −√.



We thus have an asymptote: the lineL((−

√

)~v− π ,~u−

π

).

Double points. From the previous udy (and a rough draft of the plot), we see that the curve of

((−π ,π ),ρ) admits a double point visited at some times θ ∈ (−π ,

π ) and θ ∈ (π ,

π ). We thus

mu have

θ = θ +π and ρ(θ) = −ρ(θ) .

We thus need to solve

ρ(θ +π) = −ρ(θ) ⇐⇒ −cos(θ) + −sin(θ) +

= −cos(θ) + sin(θ) +

⇐⇒ −cos(θ)sin(θ) + = cos(θ)sin(θ)−

⇐⇒ sin(θ) =

⇐⇒ θ ∈{π ,π

}mod π

⇐⇒ θ ∈{π ,π

}mod π.

The only such time θ ∈ (−π ,π ) satisfying θ + π ∈ (π ,

π ) is θ = π

. The double point of

((−π ,π ),ρ) is thus M

(π

)=M

(π

).

We furthermore observe that the curves of ((−π ,−π ),ρ) and ((−π ,

π ),ρ) have two interse�ion

points. The fir one is the pole, the other one is visited at some times θ ∈ (−π ,π ) and θ ∈

(−π ,−π ) We thus mu have

θ = θ −π and ρ(θ) = −ρ(θ) .

From the previous udy, we find that θ = π . The interse�ion point of both curves that is not the

pole is thus M(π

)=M

(−π

).

Sketch. In order to make the sketch more precise, compute the locations and tangents when they

are easyly computable: say at times −π , , π , π , π , π. There are no rules for choosing these extra

points, ju feeling. . .



x

y

−π+

π−

−π−

−π+


Documents

Week 8. Polar curves · 2019. 4. 15. · Week 8. Polar curves In ru or: Jer´ emie Bettinelli (´ [email protected]) Tutorial Assi ants: –Nicolas Brigouleix (groups