MODULE 1_Calculus_part 1_Presentation.pdf

Subject: Mathematics-1 for ALL Stream
Subject code: BMATC/E/M/S101
Acharya Institute of Technology
Department of Mathematics
Bangalore - 560107
April 17, 2023
Department of Mathematics (AIT) Mathematics-1 for ALL Stream ( 1/22) April 17, 2023 1 / 22

Module-1
DIFFERENTIAL CALCULUS-1
Polar Curves
Definition
Cartesian Curve: The curve whose coordinates are (x, y) Cartesian
system is called Cartesian curve.
Definition
Polar Curve: The curve whose coordinates are (r, θ) polar system is
called polar curve.

Relationship between the Cartesian coordinates (x, y) and
the polar coordinates (r, θ):
From the right angled triangle OQP we
have
cosθ = OQ
OP
= x
r
⇒ x = rcosθ
sinθ = QP
OP
= y
r
⇒ y = rsinθ
r =
p
(x2 + y2)
θ = tan−
1(y
x
)

Angle between radius vector and tangent:
(With usual notation, prove that tanϕ = r dθ
dr
) (Jan-2020, July-2018)
Let P(r, θ) be any point on the curve
r = f (θ)
∴ XÔP = θ and OP = r.
Let PL be the tangent to the curve at
P subtending an angle ψ with the pos-
itive direction of the initial line (x-axis)
and ϕ be the angle between the radius
vector OP and the tangent PL.
That is OP̂L = ϕ
From the figure we have
ψ = ϕ + θ
(we know that an exterior angle is
equal to the sum of the interior oppo-
site angles)

⇒ tanψ = tan(ϕ + θ)
tanψ =
tanϕ + tanθ
1 − tanϕtanθ
(1)
Let (x, y) be the cartesian coordinates of P so that we have,
x = rcosθ, y = rsinθ
Since r is a function of θ, we can as well regard these as parametric
equations in terms of θ.
We also know from the geometrical meaning of the derivative that
tanψ = dy
dx
=slope of the tangent PL
⇒ tanψ =
dy
dθ
dx
dθ
=
d
dθ
(rsinθ)
d
dθ
(rcosθ)
= rcosθ+r′sinθ
−rsinθ+r′cosθ
where r′
= dr
dθ
Dividing both the numerator and denominator by r′
cosθ we get,

tanψ =
rcosθ
r′cosθ
+ r′sinθ
r′cosθ
−rsinθ
r′cosθ
+ r′cosθ
r′cosθ
=
r
r′ + tanθ
1 − r
r′ · tanθ
(2)
Comparing equations (1) and (2) we have
tanϕ = r
r′ = r
dr
dθ
= r dθ
dr
tanϕ = r dθ
dr
Equivalently we can write it in the form
1
tanϕ
= 1
r
dr
dθ
or
cotϕ = 1
r
dr
dθ
.

Length of the perpendicular from the pole to the tangent:
(With usual notation, prove that for the curve r = f (θ),
1
p2 = 1
r2 + 1
r4 (dr
dθ
)2
) (Sep-2020.Jan-2018)
Let O be the pole and OL be the initial
line. Let P(r, θ) be any point on the curve
r = f (θ) and hence we have LÔP = θ and
OP = r. Draw ON=p (say) perpendicular
from the pole to the tangent at P and let
ϕ be the angle made by the radius vector
with the tangent.
From the figure we have ON̂P = 900
and
LÔP = θ.
Now from the right angled triangle ONP
sinϕ = ON
OP
= p
r
p = rsinϕ .............(1)

Squaring equation (1) and taking the reciprocal we have,
1
p2 = 1
r2 · 1
sin2ϕ
= cosec2ϕ
r2 = 1
r2 (1 + cot2
ϕ) = 1
r2 [1 + 1
r2 (dr
dθ
)2
]
1
p2 = 1
r2 + 1
r4 (dr
dθ
)2
...............(2)
Let 1
r
= u
Differentiating with respect to θ we have
− 1
r2 (dr
dθ
) = (du
dθ
)
Squaring on both sides we get
1
r4 (dr
dθ
)2
= (du
dθ
)2
By substituting this in equation (2) we get
1
p2 = u2
+ (du
dθ
)2
.

Angle of intersection of two polar curves:
We know that the angle of intersection
of any two curves is equal to the an-
gle between the tangents drawn at the
point of intersection of the two curves.
From the figure we have the angle be-
tween the two tangents is equal to
ϕ = ϕ2 − ϕ1.
∴ The acute angle of the intersection
of the curves is equal to |ϕ2 − ϕ1| .
that is ϕ = |ϕ2 − ϕ1|
or tanϕ = | tanϕ2−tanϕ1
1+tanϕ1tanϕ2
|

Note: 1. If ϕ = |ϕ2 − ϕ1| = Π
2
or tanϕ1tanϕ2 = −1, then we say that
the two curves intersect orthogonally.
2. tan(Π
4
+ θ) = 1+tanθ
1−tanθ
and cot(Π
4
+ θ) = 1−tanθ
1+tanθ
Problems:
I) Find the angle between the radius vector and the tangent
for the following curves
1. r = a(1 − cosθ)
Solution:
Taking logarithms on both sides, logr = loga + log(1 − cosθ)
Differentiating with respect to θ we get
1
r
dr
dθ
= 0 + sinθ
1−cosθ
cotϕ =
2sin( θ
2
)cos(θ
2
)
2sin2(θ
2
)
= cot(θ
2
) ⇒ ϕ = θ
2

2. rm
= am
(cosmθ + sinmθ)
Solution:
Taking logarithms on both sides,
mlogr = mloga + log(cosmθ + sinmθ)
m
r
dr
dθ
= 0 + −msinmθ+mcosmθ
cosmθ+sinmθ
1
r
dr
dθ
= cosmθ−sinmθ
cosmθ+sinmθ
cotϕ = cosmθ(1−tanmθ)
cosmθ(1+tanmθ)
= (1−tanmθ)
(1+tanmθ)
= cot(Π
4
+ mθ)
Thus we have ϕ = Π
4
+ mθ

3. l
r
= 1 + ecosθ
Solution:
Taking logarithms on both sides, logl − logr = log(1 + ecosθ)
0 − 1
r
dr
dθ
= −esinθ
1+ecosθ
1
r
dr
dθ
= esinθ
1+ecosθ
cotϕ = esinθ
1+ecosθ
tanϕ = 1+ecosθ
esinθ
Thus we have ϕ = tan−
1 1+ecosθ
esinθ


II. Find the angle between the radius vector and tangent as indicated
1. r = a(1 + cosθ) at θ = π
3
Solution:
Taking logarithms on both sides, logr = loga + log(1 + cosθ)
1
r
dr
dθ
= 0 + −sinθ
1+cosθ
cotϕ =
−2sin( θ
2
)cos( θ
2
)
2cos2( θ
2
)
= −tan(θ
2
) = cot(π
2
+ θ
2
)
⇒ ϕ = π
2
+ θ
2
Thus ϕ(θ=π
3
) = π
2
+ π
6
= 2π
3

2. r = a(1 + sinθ) at θ = π
2
Solution:
Taking logarithms on both sides, logr = loga + log(1 + sinθ)
1
r
dr
dθ
= 0 + cosθ
1+sinθ
cotϕ = cosθ
1+sinθ
cotϕ(θ=π
2
) = 0
1+1
= 0
⇒ ϕ(θ=π
2
) = cot−1
(0) = π
2
Thus ϕ(θ=π
2
) = π
2

EXERCISES:
Find the angle between the radius vector and the tangent for
the following curves
1. r2
cos2θ = a2
2. rsec2
(θ
2
) = 2a
3. r2
= a2
(cos2θ + sin2θ)
4. r = acosec2
(θ
2
)
5. 2a
r
= 1 − cos(θ) at θ = 2π
3
6.rcos2
(θ
2
) = a at θ = 2π
3
ANSWERS:
1. ϕ = π
2
− 2θ, 2. ϕ = π
2
+ θ
2
, 3.ϕ = π
4
+ 2θ, 4.ϕ = −θ
2
, 5. ϕ = π
6
,
6.ϕ = π
6

II) Show that the following pairs of curves
intersect each other orthogonally
1. r = a(1 + cosθ) and r = b(1 − cosθ) (Jan-2020)
Solution:
Taking logarithms on both sides, logr = loga + log(1 + cosθ)
1
r
dr
dθ
= 0 + −sinθ
1+cosθ
cotϕ1 =
−2sin( θ
2
)cos( θ
2
)
2cos2( θ
2
)
cotϕ1 = −tan(θ
2
) = cot(Π
2
+ θ
2
)
Thus ϕ1 = Π
2
+ θ
2

Now from second curve we get logr = logb + log(1 − cosθ)
1
r
dr
dθ
= 0 + sinθ
1−cosθ
cotϕ2 =
2sin( θ
2
)cos( θ
2
)
2sin2( θ
2
)
= cot(θ
2
)
Thus ϕ2 = θ
2
Therefore angle of intersection ϕ = |ϕ2 − ϕ1| = |θ
2
− Π
2
− θ
2
| = Π
2
Thus the curves intersect each other orthogonally.

2. r = a(1 + sinθ) and r = a(1 − sinθ)
Solution:
Taking logarithms on both sides, logr = loga + log(1 + sinθ)
1
r
dr
dθ
= 0 + cosθ
1+sinθ
cotϕ1 = cosθ
1+sinθ
⇒ tanϕ1 = 1+sinθ
cosθ
Now from second curve we get logr = loga + log(1 − sinθ)
1
r
dr
dθ
= 0 + −cosθ
1−sinθ
cotϕ2 = −cosθ
1−sinθ
⇒ tanϕ2 = −1−sinθ
cosθ
We have tanϕ1 · tanϕ2 = 1+sinθ
cosθ
· −1−sinθ
cosθ
= −1−sin2θ
cos2θ
= −1

3. rn
= an
cosnθ and rn
= bn
sinnθ (Jan-2019, July-2018)
Solution:
Taking logarithms on both sides, nlogr = nloga + log(cosnθ)
n
r
dr
dθ
= 0 + −nsinnθ
cosnθ
cotϕ1 = −tannθ = cot(Π
2
+ nθ) ⇒ ϕ1 = (Π
2
+ nθ)
Now from second curve we get nlogr = nlogb + log(sinnθ)
n
r
dr
dθ
= 0 + ncosnθ
sinnθ
cotϕ2 = cotnθ ⇒ ϕ2 = nθ
ϕ = |ϕ2 − ϕ1| = |nθ − Π
2
− nθ| = Π
2

4. r = 4sec2
(θ
2
) and r = 9cosec2
(θ
2
)
Solution:
Taking logarithms on both sides, logr = log4 + 2logsec(θ
2
)
1
r
dr
dθ
= 0 + 2
sec( θ
2
)·tan( θ
2
)· 1
2
sec(θ
2
)
cotϕ1 = tan(θ
2
) = cot(Π
2
− θ
2
) ⇒ ϕ1 = Π
2
− θ
2
Now from second curve we get logr = log9 + 2logcosec(θ
2
)
1
r
dr
dθ
= 0 − 2
cosec( θ
2
)·cot(θ
2
)· 1
2
cosec( θ
2
)
cotϕ2 = −cot(θ
2
) = cot(−θ
2
) ⇒ ϕ2 = −θ
2
ϕ = |ϕ1 − ϕ2| = |Π
2
− θ
2
+ θ
2
)| = Π
2

5. r = aeθ
and reθ
= b
Solution:
Taking logarithms on both sides, logr = loga + θloge
1
r
dr
dθ
= 0 + 1
cotϕ1 = 1 ⇒ ϕ1 = Π
4
Now from second curve we get logr + θloge = logb
1
r
dr
dθ
+ 1 = 0
cotϕ2 = −1 ⇒ ϕ2 = −Π
4
ϕ = |ϕ1 − ϕ2| = |Π
4
+ Π
4
| = Π
2

EXERCISES:
Show that the following pairs of curves intersect each other
orthogonally
1. r2
sin2θ = a2
and r2
cos2θ = b2
2. rsec2
(θ
2
) = a and rcosec2
(θ
2
) = b
3. rn
cosnθ = an
and rn
sinnθ = bn
4. 2a
r
= 1 + cosθ and 2a
r
= 1 − cosθ
5. r2
= a2
cos(2θ) and r2
= a2
sin(2θ)
ANSWERS:
ϕ = |ϕ1 − ϕ2| = π
2
OR tanϕ1 · tanϕ2 = −1 for all the problems.

MODULE 1_Calculus_part 1_Presentation.pdf

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