APPLICATION OF DERIVATIVES |
Introduction
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 = y1 & cot = [see in fig.] = cot(90 - ) PS tan Subnormal = SN = Length of tangent = PT = Length normal = PN = Illustration: Show that curve y = bex/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(x1, y1)y1 is ordinate of point P. y = bex/a ............................................... (i) differentiating curve y = bex/a w.r.t. x. & Let P(x1, y1) lie on curve Length of subnormal = So, subnormal varies as square of ordinate. |




1 =
(x2) = 2x
(4 - x2) = - 2x
,
[see in fig.]
= cot(90 -


