Y

8s

A S P

V+ôv
X
Fig. 49.
 . Cartesian Formula-f find the radius of
of t ure fn
cartesian curve y = [x).

Let the equation of the curve in cartesian lorm

orm be y=
y= fx
fx)
We know that tanyy =
dy
dx

Differentiating this w.r.t. x, we get

dy dy ds dy
secy d d dx2 sec2 ds dx d2
1 sery y
secy.
P cos dx2 P dr
2 3/2

(1 tany)32= "y
dx2
= .
(dy
1+|dx) dy
dr
2 3/2

P>

dy
where
dr2
y0.
0.
d2

In short, we write this fomula as follows:

(1+y3/2
2
IILParametric Forula
e tthe equation of the curve be x ={), y>g0.
dy
Then dy d _ &)
dx dx
dt

and .

dx
dt
1
'(2
Ifwe putf'() x, g() =y
=

then ddx and dy*y'-y*

dx2 (x
3/2
2 3/2
1+ *) .
From the formula p =

y X-yx
d2 (r)3
3
(x2+y2
3

(2+ya2
xy" - y ' '
EXAMPLE 8. Find the radius of curnvalure al anypoint of the curve

x =
a(0 + sin0), y =
a(1 -

cos0).
(P.U. 61A, 72A; B.U. 595, 62A; R.U. 66A; M.U. 78H, 89
sOLUTION: Given that x =al®+ sin6), y =q1 -

cos0).
dx
a(1 + cos6), dy a sin6,

x = a sin6, a2 a cost.
02
From the formula,
( 2 + y23/2
p

a(1 +cos9)2+ asin2013/2

a cos(1 + cos9)+ a^sin20
al1 + 2cos0 + cos29 + sin20]3/2 a(2 + 2 cos0)5/2
alcose+ cos20 +sin20} 1 cos0

a.23/2(1 + Cos6 23/2 a1

1 Cos6
= .
+
cos6)"
s20)1/2
=
23/2a 2 cos =
23/2. a21/2cos 2
2

= 4a cos
2
Note: At 0 = 0, p 4a. (B.U. 55A)

2 2
equation.
Folium
4.42. Find the radius of curvature at the point (3a/2, 3a/2) of the
Bxample (Mangalore, 1999)
3+y3axy.
Differentiating with respect to x, we get
32+3y = 3a|y +x dy
dr dx
or o- ax)=ay -x* .() dy at (3a/2, 3a/2) = -

1.
dx
Differentiating (i),
+(-ax) =a dy a t (3a/2, 3a/2) = -

32/3a
d2
d4

Hence p 3a/2)=l+dy/dr)°* [1+(-1)j3/2 3a

3a nn . magnitude)
at (3a/2, 32/3a 82
dy/d2
 D
Tage No:

dallanla
T indA
Cwhve
eh dáVatae bedal
khe bed
he

1 d
dP
Tis tL AquiAd.amlaudeh tl N
eh in H beclad
(4 Radius of curvature for polar curve r = f(0), is given by

(+r23/2
P2+2r rr2
EXAMPLE 1. Find the radius of curvature at any point ofthe curner
ve T=
sOLUTION:Given The equation of the curve is r= a
Differentiating w.r.t. 6, we get
dr
m.mmo.m = mr

d m
dr
m.mr= m'r.

From the formula, (+r232

P2+ 272 rr,2
(Pmr232 P(1 + m)3/2
22m22- m2r2 2+m2R

P1+ m3/2
N1+M+m
(1+m
 the
PLE2 i n d t 6
ind
radius of vature at any point of the curve
nE2 r= a(1 + cos8) (P.U. 45S, S65; B.U. 565, 76; R.U.
62A)
N:Given
NUUTION.
n The
1h equation of the curve is
r = a1 + cos0)

fierentiatung
w.r.t. 8, we get
Dt

dr
a - a sin6, acose

(+2
p =

Fromthe formula, 2+ 27 -

2
a(1 +cos6) + asin015/2
(1 +cos0)+ 2asin0 +a(1 + cos6).a cose
a(1+ 2cose +cos20+ sin20)]3/2
al(1 + 2cos6 +cos20) + 2sin20 + (cos6 + cos20)}

a2+2cos0a1+cos@32
a(3+3cos6) 3(1 + cose)
23/2 a ( 1 +cos6)1/2232 a (2cos20)2
3

23/2.a-24cos 4a
3

1642 16a2 1+ cos6

Cor. pcosp?-
3 9 2

9 8a 2ar
(B.U. 93H)
P V2ar.