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APPLICATION OF DERIVATIVES
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Dumb Question:
Dumb Question: How this derived ?
Ans: Let   m1 = tan 1 =

              m2 = tan 2 =

From fiq,   = + 1

           = 2 - 1

          

          

Orthogonal curves: If angle of intersection of two curves is right angle, two curves are c/d orthogonal curves.

If curves are orthogonal,  =



Illustration: Find angle of intersection of curves   y = x2 & y = 4 - x2

For intersection points of given curves,

(x2) = 2x

(4 - x2) = - 2x

At  x = - ,

= 2 x = 2   &   = - 2







  

Both angles are equal.

Length of Tangent, Sub-Tangent, Normal & sub-Normal:
Length of Tangent: Length of segment PT of tangent b/w point of tangent & x-axis is c/d length of tangent.

 

PT =

Subtangent
: Projection of segment PT along x-axis i.e. St c/d subtangent.

Length of Normal
: Length of segment PN intercapted b/w point on curve r x-axis.
PN =

Subnormal
: Projection of segment PN along x-axis i.e. c/d subnormal.
SN = |y,


Dumb Question
: How these relation derived ?

Ans: Since PT makes angle

with x-axis, then
tan

=




Subtangent = ST = PS cot


But PS = y
1
  &   cot

=

[see in fig.]



= cot(90 -



PS tan



Subnormal = SN =


Length of tangent = PT =


Length normal = PN =


Illustration
: Show that curve y = be
x/a
. subnormal varies as square of ordinate ?

Ans:
[Dumb Question
: What is ordinate ?

Ans: y-axis component is c/d ordi i.e. in P(x
1
, y
1
)y
1
is ordinate of point P.
y = be
x/a
............................................... (i)
differentiating curve   y = be
x/a
  w.r.t. x.


& Let P(x
1
, y
1
) lie on curve


Length of subnormal =


                             


So, subnormal varies as square of ordinate.
     |     

 
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