study material-mathematics-differential calculus

# application-of-derivatives-2

__ Dumb Question__: How this derived ?

Ans: Let m

_{1}= tan

_{1}=

m

_{2}= tan

_{2}=

From fiq, = +

_{1}

=

_{2}-

_{1}

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

**Orthogonal curves**If curves are orthogonal, =

__: Find angle of intersection of curves y = x__

**Illustration**^{2}& y = 4 - x

^{2}

For intersection points of given curves,

(x

^{2}) = 2x

(4 - x

^{2}) = - 2x

At x = - ,

= 2 x = 2 & = - 2

Both angles are equal.

__:__

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

**Length of Tangent**PT =

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

**Subtangent**__: Length of segment PN intercapted b/w point on curve r x-axis.__

**Length of Normal**PN =

__: Projection of segment PN along x-axis i.e. c/d subnormal.__

**Subnormal**SN = |y,

__: How these relation derived ?__

**Dumb Question**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 =

__: Show that curve y = be__

**Illustration**^{x/a}. subnormal varies as square of ordinate ?

Ans:

**[**

__: What is ordinate ?__

**Dumb Question**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.