|

APPLICATION OF DERIVATIVES
Tags: Engineering Entrance  |  JEE Main  |  CBSE Board  |  JEE Main & Advanced  |  Differential Calculus  |  Class 11  |  Class 12
« Back to Content

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.
|

Name

Mobile

E-mail

City

Class