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01_Straight_Line 02_Circle 03_Permutation_and_combination 04_Complex_numbers 05_Application_of_Derivative 06_Binomial_theorem 07_Progression__Series.pdf 08_Definite_Integral.pdf 09_Indefinite_Integral.pdf 10_Probability.pdf

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  1. APPLICATION OF DERIVATIVE INTRODUCTION In this section we examine some applications in which derivatives are used to represent and interpret the rates at which things change in the world around us. Let S be the displacement described by a particle over the time interval t. Then S will be a function of t, we write as S = f(t). The velocity is the derivative of the position function S = f (t) with respect to time. At time t the velocity is dS dt Acceleration is the derivative of velocity with respect to time. If a body's position at time t is S = f(t), the body's acceleration at time t is d2S dv a(t) = — dt2 dt Geometrical Meaning of the Derivative y = f(x) The tangent to the curve y = f(x) at the point (x, y) makes an angle with the positive dy dy x-axis. Then— — tanw . Thus the derivative — or f'(x) represents the -x dx dx slope of the tangent to the curve at the point (x, y). Tangent and Normal to a Curve PT is the tangent to the curve y = f(x) at the point normal to the curve at P. dy The slope of the tangent at P(XI, yo is, dx 1 The slope of the normal at P(XI, yo is, dy dx Hence the equation of the tangent PT is, y —Yl o P(XI, yo. dy dx (XI,YI) PN is the Y = f(x) POI, YD x 904-11] x (x l), 1 dy dx dy (x —xo or, (x - + (y- YO dx (XI,YI) If 'PI' be the projection of the point P on the x-axis then T PI is called and the equation of the normal PN is, sub-tangent (projection of line o the segment PT on the x-axis) and NPI is called the sub normal (projection of line segment PN on the x-axis). PT is called the length of tangent and PN is called the length of normal from x-axis. 2 dx TPI = dy 2 dy dx dx dy dy dx
  2. Application of Derivative Angle between two Curves The angle between two curves (or the angle of Y intersection of two curves) is defined as the angle between the two tangents at their point of intersection. As the figure shows, the angle between the two curves, is given by = - tan+ = tan (WI - 11J2) tanklJl —tanklJ2 1 + tan tan where tamlJ1 = f'(X1) and tamlJ2 = g'(X1). 2 y=f(x) x Two curves are said to cut each other orthogonally if the angle between them is a right angle, that is, if (l) = 900 , in which case we will have, tanilJ1 tamlJ2 = -1. Two curves touch each other if the angle between the tangents to the curves at the point of intersection is 00 in which case we will have, tampi = tamlJ2. Example 1 : Solution: Tangent at point PI (other then (0, 0)) on the curve y = x 3 meets the curve again at P2. The tangent at P2 meets the curve at P3 and so on. Show that the abscissa of PI, P2, P3, area (APIP2P3) form a G.P. also, find the ratio area (AP2P3P4) Let a point PI on y = x 3 be (h, h3) tangent at PI is y - h3 = 3h2 (x- h), it meets y = x3 at P2 x3 - h3 = 3h2 (x- h) + Xh - 2h2 = 0 (x- h) (x + 2h) = O . x = -2h for P2 as x = h is for point PI . P2 is (-2h, -8h3) tangent at P2 is y + 8h it meets y = at P3, - 2hx - 8h2 = 0 (x- 4h) (x + 2h) = O . p 3 is (4h, 64h3) . x = 4h for P3 continuing like this, we get x = -8h for P4 etc, hence the abscissae of PI, P2 , P3 . are h, -2h, 4h, -8h, . which are in G.P. = API p2p3 Al A2 = AP2p3p4 h 1 2h -8h3 2 4h 64h3 Al 1 1 1 16 1 — 4h 8h3 64 h 3 512h3 1 1 On solving
  3. 3 Monotonocity Let y = f(x) be a given function with 'D' as its domain. Let DI CD. Application of Derivative (ii) (iii) (iv) Increasing Function f(x) is said to be increasing in DI if for every e DI, > f(X1) > f(X2). Non-decreasing Function f(x) is said to be non-decreasing in DI if for every e DI , > f(X1) > f(X2). Decreasing Function f(x) is said to be decreasing in DI if for every Xl, e DI, > f(X1) < f(X2). Non-increasing Function f(x) is said to be non-increasing in DI if for every e DI, > f(X1) < f(X2). Basic Theorems Let f(x) be a function that is continuous in [a, b] and differentiable in (a, b). Then (i) f(x) is a non-decreasing function in [a, b] if f '(x) > 0 in (a, b); (ii) f(x) is an increasing function in [a, b] if f '(x) > 0 in (a, b); (iii) f(x) is a non-increasing function in [a, b] if f '(x) < 0 in (a, b); (iv) f(x) is a decreasing function in [a, b] if f '(x) < 0 in (a, b). Remarks (iv) (v) (vi) (vii) If f '(x) > O V x e (a, b) and points which make f '(x) equal to zero (in between (a, b)) do not form an interval, then f (x) would be increasing in [a, b]. If f '(x) < O V x e (a, b) and points which make f '(x) equal to zero (in between (a, b)) do not form an interval, f (x) would be decreasing in [a, b]. Iff (0) = 0 and f '(x) > OV x e R, then f (x) < 0 V xe ( 00, 0) and f(x) > 0 V x e (0, 00). If f(0) = 0 and f '(x) < OV x e R, then f(x) > 0 V x e ( 00, 0) and f (x) < 0 V x e (0, 00). A function is said to be monotonic if its either increasing or decreasing. The points for which f '(x) is equal to zero or does not exist are called critical points. Here it should also be noted that critical points are the interior points of an interval. The stationary points are the points where f '(x) = 0 in the domain. Example 2 : Solution: Prove that the following functions are increasing for the given intervals, y = ex + sin x, xe R + y = sin x + tan e (0 , Tt/2) (iii) Y = x + sin x, (i) f (x) = ex + sin x, f = ex + COS X Clearly f '(x) > 0 V xeR+ (as ex xeR+ and —1 < cosx < 1, x e R+) Hence f (x) is increasing. (ii) f (x) = sin x + tan x —2 x, x e (0 , 11/2) f '(x) = cos x + sec2 x -2 X e (0 , Tt/2) as cos x > cos x, f '(x) > cos2 x + sec2 x -2 = (cos x — sec X e (0 , Tt/2) Hence f (x) is increasing in (0 , 11/2) (iii) f (x) = x + sin x, — + COS X f > O, as -1 < COS X < 1 Here f '(x) = 0 cos x = -1 —x = (2n + 1) n, nel Zeros of f '(x) don't form an interval. Hence f(x) would be increasing for all real values of x.
  4. Application of Derivative Maxima and Minima f(x) is said to have a local or relative maximum at x = c, if there exists a neighbourhood (c - h, c + h), (contained in the domain of f) , of c such that f(c) > f(x) 4 f(x) is said to have a local or relative minimum at x c if there exists a neighbourhood (c - h, c + h), (contained in the domain of f), of c such that f(c) < f(x) V x e (c — h, c) u (c, c + h). f(x) is said to have relative or local extremum at x if it has relative maximum or relative minimum at c If f(x) has a local maximum (minimum) at c, then f(c) is called a local maximum (minimum) value of f. Theorem If f(x) has local extremum at x = c then either f ' (c) = 0 or f '(c) does not exist. The converse of this theorem is not always true. That is, the fact that f ' (c) = 0 does not necessarily imply that f (x) has local extremum at x = a. For example, consider the function f(x) = x3. Its derivative f '(x) = 3x 2 vanishes at x 0. However, as the graph shows, x 3 of x . The number c in the domain of function f is called a critical point of f, if either f' (c) = 0 or f'(c) does not exist. 0 is not a local extremum 3 Example 3 : Solution: o Find the coordinate of the point on Y2 = 8x which is closest from + (y + 6) Let Point on parabola Y2 = 8x be (2u2, 4u), centre of circle = (0, -6) Distance between centre of circle and point on parabola (2012 ) 2 + (401 + D = 4014 + (401 +6) 2 where D = s 2 = 1 6013 = O = O (012-01 +3) + l) = + u -u -u +301+3 Now , 48u2 + 84 = . 48u2 +32 > Oat u = -1 du > D is minimum at u = -1. Point on parabola (2, -4) x Concept of Global Maximum/Minimum in [a, b] Let y = f(x) be a continuous function with domain D. Let [a, b] c D. Global maximum/minimum of f(x) in [a, b] is basically the greatest/least value of f(x) in [a, b]. Global maximum and minimum in [a, b] would always occur at critical points of f(x) within [a, b] or at the end points of the interval. In order to find the global maximum and minimum of f(x) in [a, b], find out all the critical points of f(x) in (a, b). Let Cl, C2, cn be the different critical points. Find the value of the function at these critical points. Let f(C1), f(C2), • .. be the values of the function at critical points. Say, Ml = max {f(a), f(C1), f(C2), , f(cn), f(b)} and W = min {f(a), f(C1), f(C2), , f(cn), f(b)}. Then Ml is the greatest value of f (x) in [a, b] and M2 is the least value of f (x) in [a, b].
  5. 5 Global Maximum / Minimum in (a, b) Application of Derivative Method for obtaining the greatest and least values of f(x) in (a, b) is almost same as the method used for obtaining the greatest and least values in [a, b], however with a caution. Let y = f (x) be a function and Cl, c2, , cn be different critical points of the function in (a, b). , f(cn)} Let Ml = max.{f(cl), f(C2), f(C3), and min.{f(cl), f(C2), f(C3), , f(cn)}• Now if Lim f(x) > Ml or < M2, f(x) would not have global maximum (or global minimum) in (a, b). (or This means that if the limiting values at the end points are greater than Ml or less than M2, then f(x) would not have global maximum/minimum in (a, b). On the other hand if Ml > Lim f (x) and Lim f(x), then Ml (andx.b—0) and M2 would respectively be the global maximum and global minimum of f (x) in (a, b). Tests for Maxima and Minima If f '(c) = 0, then we have three tests to decide whether f(x) have local maxima or local minima or neither at First Derivative Test Let f(x) be continuous in some neighbourhood (c f(x) has a local maximum at x = c if ; (a) x = c is a critical point of f(x); (b) f '(x) > 0 in (c - h, c), and (c) f'(x) < 0 in (c, c + h). h, c + h) of c. Then Here, in moving from left to right through the critical point c, f '(x) changes sign from plus to minus. f(x) has a local minimum at x = c if ; (a) x = c is a critical point of f(x); (b) f'(x) < 0 in (c - h, c), and (c) f'(x) > 0 in (c, c + h). Here, in moving from left to right through the critical point c, f '(x) changes sign from minus to plus. (iii) If f'(x) does not change sign in moving through c, then there is neither a maximum nor a minimum at x Second Derivative Test Let f be a function such that, (i) f(x) is continuous in (c - h, c + h) ; (ii) f'(c) = 0; (iii) f" (c) exists. Then, (a) f(x) has local maximum at x = c, if f"(c) < 0. (b) f(x) has local minimum at x = c, if f"(c) > 0. The second derivative test does not help when f'(c) depend upon the following test. nth Derivative Test Let f be a function such that = 0 and f"(c) = 0. In such a situation we shall have to f n(c) 0. Then,
  6. Application of Derivative (a) f(x) has a local maximum at x = c ; if n is even and f (n) (b) f(x) has a local minimum at x = c ; if n is even and f (n) (c) f(x) has no local extremum at x = c ; if n is odd. Remarks 6 If f'(c) does not exist or f(x) is discontinuous at x = c, then we should decide maximum / minimum by basic definition. (In the above case drawing the graph of the function becomes handy) Example 4 : Solution: Let f (x) = 2x3 — + 12 x + 6. Discuss the global maxima and minima of f (x) in [0, 2] and f (x) = - + 12 x +6 —f' (x) = + 12 = 6 (x2-3x + 2) = 6 (x-1) (x-2). First of all let us discuss [0, 2]. Clearly the critical point of f (x) in [0, 2] is x = 1. f(0) = 6, f (1) = 11, f (2) = 10. Thus x 0 is the point of global the point of global maximum. Now let us consider (1 , 3). minimum Clearly x = 2 is the only critical point in (1 , 3). f(2) = 10. Lim f (x) = 11 and Lim f (x) = 15. x-»3-0 Thus x - 2 is the point of global minimum in (1, 3) does not exist. Rolle's Theorem If a function f(x) is continuous in the closed interval [a , b]. differentiable in open interval (a , b ). (iii) f(a) = f (b) then there will be at least one point c e (a, b) such that f'(c) = 0. Geometrical Interpretation If f(x) satisfies the conditions of Rolle's theorem in [a, b] its derivative will vanish at least once in (a, b). If A (a, f (a)), B (b, f (b)) and f(a) = f(b) (third condition of Rolle's theorem) Slope of line AB = 0 We will have at least one point belonging to of (1, a f(x) in [0, 2] and x 1 3) and the global maximum is in o c b (a, b) so that tangent drawn to the curve at that point will be parallel to the line AB. Applications of Rolle's Theorem x If y = f (x) satisfies the Rolle's theorem in [a, b], then f'(x) = 0 for some xe (a, b). As any solution of f'(x) = 0 will give us a root of f'(x) = 0, we can say that at least one root of f'(x) = 0 will belong to (a, b) if f(x) satisfies all conditions of Rolle's Theorem.
  7. 7 (iii) Application of Derivative Let x = a and x = b be the roots of f(x) = 0 and y = f(x) satisfies the conditions of Rolle's theorem in [a, b]. Here f(a) = f(b) = 0. Hence we can say that between two roots of f(x) = 0 at least one root of f'(x) = 0 will lie. Let y = f(x) be a polynomial function of degree n. If f(x) = 0 has real roots only, then f'(x) = 0, f"(x) = 0, (x) = 0 will have real roots. It is in fact the general version of above mentioned application, because if f (x) = 0 have all real roots, then between two consecutive roots of f(x) = 0, exactly one root of f' (x) = 0 will lie. Lagrange's Mean Value Theorem If a function f(x) is continuous in the closed interval [a , b]. differentiable in open interval (a , b ). Then there will be at least one point c e (a, b) such that f'(c) = b-a — 45x + 1035, using Rolle's Theorem, prove that atleast one root Example 5 : Solution: 101 - If p (x) = 51x 1/100 lies between (45 , 46). Let g (x) = = 102 101 2 1 102 2 45 2 102 101 45 2 1 1 1/100 = (45)100 ——(45)100 +1 + c = c Now g (45 ) — 2 2 (46)102 g (46) = 2 23 (46) 101 45 ——(46)2 2 So g' (x) = p (x) will have atleast one root in given interval. Geometrical Interpretation Let A (a, f (a)) and B = (b, f (b)). Slope of the chord AB b—a As f' (x) gives us the slope of tangent at the point (x, y), this theorem simply says that there will be at least one point in (a, b) e.g. (points Cl, and Q) such that tangent drawn to the curve at this point will be parallel to the chord connecting points A and B. 2 0 a b x f (b)—f (a) We can have one more interpretation, i.e. f' (x) is the instantaneous rate of change of f(x) and gives b—a us the average rate of change of f(x) over [a, b]. So this theorem simply says that the average rate of change of the function over a given interval will be equal to instantaneous rate of change of function on at least one point of that interval. Another Version of Lagrange's Mean Value Theorem If we write b = a + h, then since, a < c < b, c = a + Oh where 0
  8. Application of Derivative 8 0 would have at least one root Example 6 : Solution: If 2a + 3b + 6c = 0 then prove that the equation ax2+ bx + c = in (0, 1); a, b, ce R. Let f '(x) ax2 +bx+c f (x) Also, f (1) = 6 ax3 bx2 3 2 Hence all the conditions of Rolle's theorem are satisfied in [O, 1]. So, f' (x) = 0 for atleast one value in (0, 1).

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