Note 1437658785 - Unit PDF

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Unit 2...

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ENVELOPE A curve which touches each member of a given family of curves is called envelope of that family. Procedure to find envelope for the given family of curves: Case 1: Envelope of one parameter family of curves Let us consider y = f(x,α) to be the given family of curves with ‘α’ as the parameter. Step 1: Differentiate w.r.t to the parameter α partially, and find the value of the parameter Step 2: By Substituting the value of parameter α in the given family of curves, we get the required envelope. Special Case: If the given equation of curve is quadratic in terms of parameter,i.e. Aα2+Bα+c=0, then envelope is given by discriminant = 0 i.e. B2- 4AC=0 Case 2: Envelope of two parameter family of curves. Let us consider y = f(x,α, β) to be the given family of curves, and a relation connecting the two parameters α and β, g(α, β) = 0 Step 1: Consider α as independent variable and β depends α . Differentiate y = f(x,α, β) and g(α, β) = 0, w.r. to the parameter α partially. Step 2: Eliminating the parameters α, β from the equations resulting from step 1 and g(α, β) = 0, we get the required envelope. Problems on envelope of one parameter family of curves : 1. Find the envelope of Solution : Differentiate

y  mx  am p where m is the parameter and a, p are constants y  mx  am p

(1)

with respect to the parameter m, we get,

0  x  pam p1 1   x  m    p  1  pa 

(2)

Using (2) eliminate m from (1) 1

p

 x  x  y    p  1 x  a   p  1  pa   pa 

  x  p 1  x x  y p  1    a p 1     pa   pa 

i.e.

p

p p 1 p p 1 p ap y  x p  ( x )

which is the required equation of envelope of (1) 2. Determine the envelope of x sin  y cos  a , where θ being the parameter. Solution : Differentiate , x sin  y cos  a

(1)

with respect to θ , we get, x cos  y sin  a

(2)

As θ cannot be eliminated between (1) and (2) ,we solve (1) and (2) for x and y in terms of θ. For this, multiply (2) by sinθ and (1) by cosθ and then subtracting, we get, y  a (sin    cos  ) . Using similar simplification, we get, x  a ( sin  cos )

.

3. (Leibnitz’s problem) Calculate the envelope of family of circles whose centres lie on the x-axis and radii are proportional to the abscissa of the centre. Solution : Let (a,0) be the centre of any one of the member of family of curves with a as the parameter. Then the equation of family of circles with centres on x-axis and radius proportional to the abscissa of the centre is

( x  a)2  y2  ka2

(1)

where k is the proportionality constant. Differentiating (1) with respect to a, we get,  2( x  a)  2ka

i.e. a 

x . 1 k

2

From (1) ,

k x   x2   y2  x  2 1k   (1  k)

2 2  2  2 i.e. k  k x  1  k  y  0,

k 1

4. Find the envelope of xsec2   y cos ec2  a , where θ is the parameter. 2  2    Solution : The given equation is rewritten as , x1  tan    y  1  cot    a

4 2 i.e. x tan   ( x  y  a) tan   y  0 , which is a quadratic equation in t  tan 2  . Therefore the required envelope is given by the 2

discriminant equation : B -4AC = 0

2 i.e. ( x  y  a)  4 xy  0 2 2 2 i.e. x  y  2 xy  2 ax  2 ay  a  0

.

Envelope of Two parameter family of curves : 1. Find the envelope of family of straight lines ax+by=1, where a and b are parametersconnected by the relation ab = 1 Solution : ax by  1

(1)

ab  1

(2)

Differentiating (1) with respect to a ( considering ‘a’ as independent variable and ‘b’ depends on a ). x

i.e.

db y 0 da

db  x  da y

(3)

Differentiating (2) with respect to a b a

db 0 da

db  b i.e. da  a

(4)

From (3) and (4), we have

x b  y a

i.e.



ax by ax by 1    1 1 2 2

a

1 and 2x

b

1 2y

(5)

Using (5) in (2), we get the envelope as 4xy = 1

2. Find the envelope of family of straight lines connected by the relation Solution :

a

b 1

x  a

y  1 , where a and b are parameters b

x  a a

y 1 b

(1)

b1

(2)

Differentiating (1) with respect to a x y db  0  2 a 3 2  2 b3 2 da

i.e.

32 db  x b  da y a3 2

(3)

Differentiating (2) with respect to a 1 2 a

i.e.



1

db 0 2 b da

db  b  da a

(4)

From (3) and (4), we have

x b 1 y a

i.e.



x a  a a

y b  b x and

x y  a b 1 a  b 1

b

y

(5)

14 14 1 Using (5) in (2), we get the envelope as x  y

x

y

3. Find the envelope of family of straight lines a  b  1 , where a and b are parameters 2 3

connected by the relation a b = c

5

x y  1 a b 2 3

ab =c

(1)

5

(2)

Differentiating (1) with respect to a, y db x  0 2 a b2 da

i.e.

db  b 2 x  da a2 y

(3)

Differentiating (2) with respect to a 2 ab3  3a2 b2

i.e.

db 0 da

db  2 b  da 3a

(4)

From (3) and (4), we have

3x 2y  a b

i.e.



x y x y  a  b a b 1 3 2 5 5 a

5x 3 and

b

5y 2

Using (5) in (2), we get the envelope as

(5)

x 2 y3 

72 5 c 3125 x2

y2

4. Find the envelope of the family of circles whose centres lie on the ellipse a 2  b 2  1 and which pass through its centre.

Solution: Let (α,β) be the centre of arbitrary member of family of circles which lie on the x2 y2 ellipse a 2  b 2  1 , whose centre is (0,0). Therefore, equation of the circles passing through

origin and having centreat (α,β) is x 2  y 2  2x  2  y  0

(1)

with 2 2  2 1 a2 b

(2)

Differentiating (1) with respect to α ( ‘α’ as independent variable and ‘β’ depends on α ), x

d y0 d

d  x i.e. d   y

(3)

Differentiating (2) with respect to α 2 2 d   0 2 a b2 d  d   b2  i.e. d a2  From (3) and (4), we have

x b 2  y a2

x y xy k    2 2 2 2 1    i.e.   , where k = αx+βy a2 b2 a2 b2

(4)





a2 x and k

 

b2 y k

(5)

2 2 From (1), we have , x  y  2k

(6)

2

Using (5) and (6) in (2), we get the envelope as

 x 2  y2     

 4  a 2 x2  b2 y2  



x2

5. Determine the equation of the envelope of family of ellipses parameters a and b are connected by the relation

a

2



2

(1)

2

a b  2 1 2 l m

(2)

Differentiating (1) with respect to a,  2 x 2 2 y 2 db  0 a3 b 3 da

i.e.

db  b 3x 2  da a3 y2

(3)

Differentiating (2) with respect to a 2 a 2b db  0 l 2 m 2 da

i.e.

db  m 2 a  da l 2b

From (3) and (4), we have

b

2

 1where the

a 2 b2  2  1, l and m are non-zero l2 m

constants. x2 y2  1 a2 b2

y2

(4)

b 4 x 2 m2  a4 y2 l2 x2 y2 x2 y2  a 2  b2  a 2 b 2  1 a2 b2 a 2 b2 1  l2 m2 l2 m2

i.e.

a 4  l 2 x 2 and



i.e.

a 2  lx and

b 4  m2 y2

b 2  my

(5) x y  1 l m

Using (5) in (2), we get the envelope as

Problems on Evolute as envelope of its normals : x

2

y2

1. Determine the evolute of hyperbola a 2  b 2  1 by considering it as an envelope of its normal Solution : Let P (a cosht, b sinht) be any point on the given hyperbola. Then dy b cosh t b dy  dt   coth t dx dx a sinh t a dt

Equation of normal line to the hyperbola is

( y  b sinh t ) 



a x  a cosh t  b cosh t

by ax   a 2  b2 sinh t cosh t

Differentiating (2) partially with respect to t, we have,

(1)

(2)

 by (sinh t ) 2

cosh t 

ax (cosh t ) 2

sinh t  0

13

 by    ax 



tanh t   

13

 by    h

 sinh t  ∓

andcosh t   ax   h 

13

(3)

h  (ax) 2 3  (by) 2 3

Where

Using (3) in (2) , we get,

ax by h  a 2  b2 h 1 3 1 3  (by ) (ax )

i.e.

(ax )

i.e. ( ax)

23

23

 (by ) 2 3

(ax )

23

 (by )2 3

2 2  ( by) 2 3   a  b   



12

 a 2  b2

23

2. By considering the evolute of a curve as the envelope of its normal, find the evolute of x  cos    sin  , y  sin    cos 

Solution : dy dy  sin   d   tan  dx  cos  dx . d

Equation of normal line to the hyperbola is ( y  (sin    cos )) 

1  x  cos   sin   tan 



i.e.

y sin  sin 2   sin cos   x cos  cos 2    sin  cos 

y sin   x cos  1

(1)

Differentiating (1) with respect to the parameter θ, we have

y cos  x sin  0

(2)

Multiplying (1) by cosθ and (2) by sinθ and then subtracting, we have,

x  cos 

(3)

Similarly we get,

y  sin 

Eliminating θ between (3) and (4) we get the required evolute as

(4) x 2  y 2 1...