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

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1 APPLICATIN F DERIVATIVE INTRDUCTIN In this section we eamine 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 b a particle over the time interval t. Then S will be a function of t, we write as S = f(t). The velocit is the derivative of the position function S = f (t) with respect to time. At time t the velocit is v (t) = ds dt Acceleration is the derivative of velocit with respect to time. If a bod's position at time t is S = f(t), the bod's acceleration at time t is a(t) = dv dt = ds dt Geometrical Meaning of the Derivative The tangent to the curve = f() at the point (, ) makes an angle with the positive -ais. Then d tan d d. Thus the derivative or f'() represents the d slope of the tangent to the curve at the point (, ). Tangent and Normal to a Curve P (, ) = f() PT is the tangent to the curve = f() at the point P(, ). PN is the normal to the curve at P. = f() d The slope of the tangent at P(, ) is, d (, ). P(, ) The slope of the normal at P(, ) is, d d (, ). T P 90+ N Hence the equation of the tangent PT is, ( ) d d (, ) or, ( ) + ( ) d ( ), and the equation of the normal PN is, d (, ) d d = 0., If P be the projection of the point P on the -ais then TP is called the sub-tangent (projection of line segment PT on the -ais) and NP is called the sub normal (projection of line segment PN on the -ais). PT is called the length of tangent and PN is called the length of normal from -ais. d PT = d d PN = d TP = d d NP = d d

2 Application of Derivative Angle between two Curves The angle between two curves (or the angle of 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 b = tan = tan ( ) tan tan = tan tan, where tan = f'( ) and tan = g'( ). = g () P(, ) = f() T T ( ) Two curves are said to cut each other orthogonall if the angle between them is a right angle, that is, if = 90, in which case we will have, tan tan =. Two curves touch each other if the angle between the tangents to the curves at the point of intersection is 0 o, in which case we will have, tan = tan. Eample : Tangent at point P (other then (0, 0)) on the curve = 3 meets the curve again at P. The tangent at P meets the curve at P 3 and so on. Show that the abscissa of P, P, P 3,.P n, area PP P3 form a G.P. also, find the ratio. area P P P Solution: Let a point P on = 3 be (h, h 3 ) 3 4 tangent at P is h 3 = 3h ( h), it meets = 3 at P 3 h 3 = 3h ( h) + h + h 3h = 0 + h h = 0 ( h) ( + h) = 0 = h for P as = h is for point P P is ( h, 8h 3 ) tangent at P is + 8h 3 = 3(h) ( + h), it meets = 3 at P 3, 3 + 8h 3 =h ( + h) h 8h = 0 ( 4h) ( + h) = 0 = 4h for P 3 P 3 is (4h, 64h 3 ) continuing like this, we get = 8h for P 4 etc, hence the abscissae of P, P, P 3. are h, h, 4h, 8h, which are in G.P. = P P P 3 = P P 3 P h h h 8h h 8h 3 4h 64h 3. 4h 64h 3 8h 5h 3 n solving. 6

3 3 Application of Derivative Monotonocit Let = f() be a given function with D as its domain. Let D D. (i) Increasing Function f() is said to be increasing in D if for ever, D, > f( ) > f( ). (ii) (iii) (iv) Non-decreasing Function f() is said to be non-decreasing in D if for ever, D, > f( ) f( ). Decreasing Function f() is said to be decreasing in D if for ever, D, > f( ) < f( ). Non-increasing Function f() is said to be non-increasing in D if for ever, D, > f( ) f( ). Basic Theorems Let f() be a function that is continuous in [a, b] and differentiable in (a, b). Then (i) f() is a non-decreasing function in [a, b] if f '() 0 in (a, b); (ii) f() is an increasing function in [a, b] if f '() > 0 in (a, b); (iii) f() is a non-increasing function in [a, b] if f '() 0 in (a, b); (iv) f() is a decreasing function in [a, b] if f '() < 0 in (a, b). Remarks (i) If f '() 0 (a, b) and points which make f '() equal to zero (in between (a, b)) do not form an interval, then f () would be increasing in [a, b]. (ii) If f '() 0 (a, b) and points which make f '() equal to zero (in between (a, b)) do not form an interval, f () would be decreasing in [a, b]. (iii) If f (0) = 0 and f '() 0 R, then f () 0 (, 0) and f() 0 (0, ). (iv) If f(0) = 0 and f '() 0 R, then f() 0 (, 0) and f () 0 (0, ). (v) (vi) (vii) A function is said to be monotonic if its either increasing or decreasing. The points for which f '() is equal to zero or does not eist are called critical points. Here it should also be noted that critical points are the interior points of an interval. The stationar points are the points where f '() = 0 in the domain. Eample : Prove that the following functions are increasing for the given intervals, (i) = e + sin, R + (ii) = sin + tan, (0, /) (iii) = + sin, R Solution: (i) f () = e + sin, R + f '() = e + cos Clearl f '() > 0 R + (as e >, R + and cos, R + ) Hence f () is increasing. (ii) f () = sin + tan, f '() = cos + sec - (0, /) as cos > cos, (0, /) f '() > cos + sec - = (cos sec ) Hence f () is increasing in (0, /) (iii) f () = + sin, R f '() = + cos f '() 0, as - cos Here f '() = 0 cos = - = (n + ), n I > 0, (0, /) Zeros of f '() don t form an interval. Hence f() would be increasing for all real values of.

4 Application of Derivative 4 Maima and Minima f() is said to have a local or relative maimum at = c, if there eists a neighbourhood (c h, c + h), (contained in the domain of f), of c such that f(c) > f() (c h, c)(c, c + h). f() is said to have a local or relative minimum at = c if there eists a neighbourhood (c h, c + h), (contained in the domain of f), of c such that f(c) < f() (c h, c) (c, c + h). f() is said to have relative or local etremum at = c if it has relative maimum or relative minimum at = c. If f() has a local maimum (minimum) at c, then f(c) is called a local maimum (minimum) value of f. Theorem If f() has local etremum at = c then either f ' (c) = 0 or f '(c) does not eist. The converse of this theorem is not alwas true. That is, the fact that f ' (c) = 0 does not necessaril impl that f () has local etremum at = a. For eample, consider the function f() = 3. Its derivative f '() = 3 vanishes at = 0. However, as the graph shows, = 0 is not a local etremum of 3. 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 eist. = 3 Eample 3 : Find the coordinate of the point on = 8 which is closest from + ( + 6) =. Solution: Let Point on parabola = 8 be (, 4), centre of circle = (0, -6) Distance between centre of circle and point on parabola S = 4 6 D = 4 4 +(4 +6) where D = s dd d = 63 +(4 +6).4 = = = = 0 ( - +3) ( + ) = 0 = - Now, dd d = = > 0 at = - D is minimum at = -. Point on parabola (, -4) Concept of Global Maimum/Minimum in [a, b] Let = f() be a continuous function with domain D. Let [a, b] D. Global maimum/minimum of f() in [a, b] is basicall the greatest/least value of f() in [a, b]. Global maimum and minimum in [a, b] would alwas occur at critical points of f() within [a, b] or at the end points of the interval. In order to find the global maimum and minimum of f() in [a, b], find out all the critical points of f() in (a, b). Let c, c,, c n be the different critical points. Find the value of the function at these critical points. Let f(c ), f(c ),,f(c n ) be the values of the function at critical points. Sa, M = ma {f(a), f(c ), f(c ),, f(c n ), f(b)} and M = min {f(a), f(c ), f(c ),, f(c n ), f(b)}. Then M is the greatest value of f () in [a, b] and M is the least value of f () in [a, b].

5 5 Application of Derivative Global Maimum / Minimum in (a, b) Method for obtaining the greatest and least values of f() 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 = f () be a function and c, c,, c n be different critical points of the function in (a, b). Let M = ma.{f(c ), f(c ), f(c 3 ),, f(c n )} and M = min.{f(c ), f(c ), f(c 3 ),, f(c n )}. Now if Lim a0 (or b0) f() > M or < M, f() would not have global maimum (or global minimum) in (a, b). This means that if the limiting values at the end points are greater than M or less than M, then f() would not have global maimum/minimum in (a, b). n the other hand if M > Lim f () and M < Lim f(), then M a0 (andb0) and M would respectivel be the global maimum and global minimum of f () in (a, b). a0 (andb0) Tests for Maima and Minima If f (c) = 0, then we have three tests to decide whether f() have local maima or local minima or neither at = c. First Derivative Test Let f() be continuous in some neighbourhood (c h, c + h) of c. Then (i) f() has a local maimum at = c if ; (a) = c is a critical point of f(); (b) f () > 0 in (c h, c), and (c) f() < 0 in (c, c + h). Here, in moving from left to right through the critical point c, f () changes sign from plus to minus. (ii) f() has a local minimum at = c if ; (a) = c is a critical point of f(); (b) f() < 0 in (c h, c), and (c) f() > 0 in (c, c + h). Here, in moving from left to right through the critical point c, f () changes sign from minus to plus. (iii) If f() does not change sign in moving through c, then there is neither a maimum nor a minimum at = c. Second Derivative Test Let f be a function such that, (i) f() is continuous in (c h, c + h) ; (ii) f(c) = 0; (iii) f(c) eists. Then, (a) f() has local maimum at = c, if f(c) < 0. (b) f() has local minimum at = c, if f(c) > 0. The second derivative test does not help when f(c) = 0 and f(c) = 0. In such a situation we shall have to depend upon the following test. n th Derivative Test Let f be a function such that (i) f (c) = f (c) =... = f (n ) (c) = 0; (ii) f n (c) 0. Then,

6 Application of Derivative 6 (a) f() has a local maimum at = c ; if n is even and f (n) (c) < 0. (b) f() has a local minimum at = c ; if n is even and f (n) (c) > 0. (c) f() has no local etremum at = c ; if n is odd. Remarks If f(c) does not eist or f() is discontinuous at = c, then we should decide maimum / minimum b basic definition. (In the above case drawing the graph of the function becomes hand) Eample 4 : Let f () = Discuss the global maima and minima of f () in [0, ] and (, 3). Solution: f () = f' () = = 6 ( 3 + ) = 6 (-) (-). First of all let us discuss [0, ]. Clearl the critical point of f () in [0, ] is =. f(0) = 6, f () =, f () = 0. Thus = 0 is the point of global minimum of f() in [0, ] and = is the point of global maimum. Now let us consider (, 3). Clearl = is the onl critical point in (, 3). f() = 0. Lim f () = and 0 Lim f () = Thus = is the point of global minimum in (, 3) and the global maimum in (, 3) does not eist. Rolle S Theorem If a function f() is (i) continuous in the closed interval [a, b]. (ii) differentiable in open interval (a, b ). (iii) f(a) = f (b) then there will be at least one point c (a, b) such that f(c) = 0. Geometrical Interpretation If f() satisfies the conditions of Rolle s theorem in [a, b] its derivative will vanish at least once in (a, b). c 3 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 A c c B We will have at least one point belonging to (a, b) so that tangent drawn to the curve at that point will be parallel to the line AB. a b Applications of Rolle s Theorem (i) If = f () satisfies the Rolle s theorem in [a, b], then f() = 0 for some (a, b). As an solution of f() = 0 will give us a root of f() = 0, we can sa that at least one root of f() = 0 will belong to (a, b) if f() satisfies all conditions of Rolle s Theorem.

7 7 Application of Derivative (ii) Let = a and = b be the roots of f() = 0 and = f() satisfies the conditions of Rolle s theorem in [a, b]. Here f(a) = f(b) = 0. Hence we can sa that between two roots of f() = 0 at least one root of f() = 0 will lie. (iii) Let = f() be a polnomial function of degree n. If f() = 0 has real roots onl, then f() = 0, f() = 0,..., f n () = 0 will have real roots. It is in fact the general version of above mentioned application, because if f () = 0 have all real roots, then between two consecutive roots of f() = 0, eactl one root of f() = 0 will lie. Lagrange s Mean Value Theorem If a function f() is (i) continuous in the closed interval [a, b]. (ii) differentiable in open interval (a, b ). Then there will be at least one point c (a, b) such that f(c) = f b f a b a. Eample 5 : If p () = , using Rolle s Theorem, prove that atleast one root lies between (45 /00, 46). Solution: Let g () = p()d = c = c. Now g (45 / ) = c = c g (46) = c c. So g () = p () will have atleast one root in given interval. Geometrical Interpretation Let A (a, f (a)) and B (b, f (b)). f (b) f (a) Slope of the chord AB. ba As f() gives us the slope of tangent at the point (, ), this theorem simpl sas that there will be at least one point in (a, b) e.g. (points c, c and c 3 ) such that tangent drawn to the curve at this point will be parallel to the chord connecting points A and B. We can have one more interpretation, i.e. f() is the instantaneous rate of change of f() and f (b) f (a) gives ba us the average rate of change of f() over [a, b]. So this theorem simpl sas 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 + h where 0 < <. Thus mean value theorem can be stated as follows: (i) If f () is continuous in [a, a + h] (ii) f() eists in (a, a + h), then there eists at least one number (0 < < ) such that f(a + h) = f (a) + h f (a + h). A a C C C 3 b B

8 Application of Derivative 8 Eample 6 : Solution: If a + 3b + 6c = 0 then prove that the equation a + b + c = 0 would have at least one root in (0, ); a, b, c R. Let f '() a b c f (0) = d Also, f () = a 3b 6c d = d 6 3 a b f () c d 3 Hence all the conditions of Rolle s theorem are satisfied in [0, ]. So, f () = 0 for atleast one value in (0, ).