Basic Maths Section [D] Calculus

BASIC MATHS & VECTOR SECTION [D] : CALCULUS Calculus is the study of how things change. In this we study the relationship between continuously varying functions. (A) DIFFERENTIAL CALCULUS The purpose of differential calculus to study the nature (i.e., increase or decrease) and the amount of variation in a quantity when another quantity (on which first quantity depends) varies independently. Average rate of change : Let a function y = f(x) be plotted as shown in figure. Average rate of change in y w.r.t. x in interval [x1, x2] is Average rate of change = change in y y y2  y1    slope of chord AB. change in x x x 2  x1 Instantaneous rate of change It is defined as the rate of change in y with x at a particular value of x. It is measured graphically by the slope of the tangent drawn to the y-x graph at the point (x,y) and algebraically by the first derivative of function y =f(x). Instantaneous rate of change = dy = slope of tangent = tan dx First Derivatives of Commonly used Functions (i) y = constant  (iii) y=ex  dy =0 dx dy = ex dx (v) y = sinx  dy = cosx dx (vii) y = tan x  dy = sec2 x dx (ii) y = xn  dy = nxn-1 dx (iv) y = ln x  1 dy = dx x (vi) y = cosx  dy = –sinx dx (viii) y = cotx  dy = –cosec2x dx Method of Differentiation or Rules of Differentiation (i) Function multiplied by a constant i.e., y = kf(x)  Ex. Find derivatives of the following functions : 4 (i) y = 2x3 (ii) y = (iii) y = 3ex x dy = kf   x  dx (iv) y = 6 ln x (v) y = 5 sin x Sol. 4 4 11 dy 44   4 4  11x x111  2 (ii) y y   4 x4 x  xx dxdx xx dy 3 3   2 23x33x131  6 x62x 2 (i) y y 2 x2 x  dx xx   (iii) yy33ee  dy 33eex x dx 5sinx x  (v) y y 5sin   (iv) y y 6 6lnlnx x  dy 5 5cos 5cosx x  cosx x 5cos dx (ii) Sum or Subtraction of Two functions i.e., y = f(x) ±g(x)  Ex. Find differentiation of y w.r.t x. (i) y = x2 – 6x Sol. (iii) dy = f   x  ± g  x  dx (ii) y = x5 + 2ex dy dy  2 x 21  6 1  2 x  6  5x (ii) dx dx Product of two functions : Product rule (i) y = f(x) ·g(x)   (iii) y = 4 ln x + cos x  2e x  5 x 4  2e x (iii) dy 4 1  4      sin x    sin x dx x x dy = f  x  × g  x  + f   x  × g  x  dx Ex. Find first derivative of y w.r.t. x. (i) y = x2 sin x Sol. (i) (iv) dy  4  e x  cos x   e x   sin x    4 e x  cos x  sin x  dx Division of two functions : Quotient rule f x dy f   x  g  x  - f  x  g   x  y=  = 2 g x dx g  x   Ex. Find differentiation of y w.r.t. x. Sol. (i) (ii) y =4(ex)cos x dy  x2  cos x   2 x  sin x   x2 cos x  2 x sin x dx   (ii) (ii) dy 1 6 6 6   dx x x   sin x 4 x3 (ii) y  x x e Here f(x) = sinx, g(x) = x So f'(x) = cos x,, g'(x) =1 dy  cos x  x    sin x 1 x cos x  sin x   Therefore dx x2 x2 (i) y  Here f(x) = 4x3, g(x) = ex So f   x   1 2 x2 , g   x   e x 2 x 3 x dy 12 x  e   4 x  e  12 x 2  4 x 3   Therefore, 2 dx ex ex  (v) Function of Functions : Chain rule Let f be a function of x, which in turn is a function of t. The first derivative of f w.r.t. t is equal to the product of df df df dx dx = × and Therefore dx dt dt dx dt Ex. Find first derivative of y w.r.t. x. (i) y = e–x (ii) y = 4 sin 3x (iii) 2 y  4e x 2 x BASIC MATHS & VECTOR Sol. (i) y  e  x  e z where z = –x So dy dy dz    e z  1   e z  e  x dx dz dx (ii) y = 4 sin 3x = 4 sinz where z = 3x So dy dy dz    4  cos z  3   12 cos 3 x dx dz dx 2 dy dy dz    4  z  2 x  2    8 x  8  e x  2 x dx dz dx The position of a particle moving along x-axis varies with time t as x=4t –t2 +1. Find the time interval(s) during which the particle is moving along positive x-direction. (iii) y  4 e x 2 Ex. Sol.   2 x  4 e z where z = x2 – 2x So If the particle moves along positive x-direction, its x-coordinate must increase with time t. dx 0. dt x-coordinate will increase with time t if dx  4  2t dt dx   444 222tt 00  000   tt  22 dt Hence, the particle moves in positive x-direction during time-interval 0 < t < 2. MAXIMUM AND MINIMUM VALUE OF A FUNCTION Higher order derivatives are used to find the maximum and minimum values of a function. At the points of maxima and minima, first derivative becomes zero. y B A Maxima minima x At point ‘A’ (minima) : As we see in figure, in the neighborhood of A, slope is increases so Condition for minima : dy d2 y  0 and 0 dx dx2 At point ‘B’ (maxima) : As we see in figure, in the neighborhood of B, slope is decreases so Condition for maxima : Ex. Sol. dy d2 y  0 and 0 dx dx2 The minimum value of y = 5x2 – 2x + 1 is (A) (C) 1 5 (B) 2 5 (C) 4 5 dy dy dy 11  00    55 22xx  2 1   0  0  x  dx dx dx 55 1 1 d2y Now at x= , =10 which is positive so minima at x = . 5 5 dx 2 For maximum/minimum value 2 d2 y > 0. dx2 4 1 1 Therefore ymin  5    2    1  5 5 5 (D) 3 5 d2 y <0 dx 2 Ex. The radius of a circular plate increase at the rate of 0.1 cm per second. At what rate does the area increase when the radius of plate is Sol. 5 cm ?  (A) 1 cm2/s (B) 0.1 cm2/s (A) Area of disk, A = r2 (where r = radius of disk) dA dr  dr     2r   2r dt dt  dt  Ex. Sol. (C) 0.5 cm2/s (D) 2 cm2/s dA 5  2   0.1  1 cm 2 / s dt  So A particle moves along the curve 12y = x3. Which coordinate changes at faster rate at x=10 ? (A) x–coordinate (B) y–coordinate (C) Both x and y–coordinate (D) Data insufficient (B) 2 dy  x   dx   12y = x  12dy = 3x dx  dt  2   dt  3 2 2  x Therefore for   >1 or x >2, y– coordinate changes at faster rate. 2 Ex. Water pours out at the rate of Q from a tap, into a cylindrical vessel of radius r. The rate at which the height of water level rises when the height is h, is (A) Sol. (B) Q rh (B) Q r 2  Ex. (C) Volume : V = r2h  w.r.t. x keeping y and z constant and so on. U = Differentiation of U x Column – I Column – II (A) For U = x2 yz, at (5,0,0) (P) (B) For U = x2 + yz at (5, 0,0) (Q) (C) For U = x (y+z) at (5, 0,0) (R) (D) For U = x y+z at (5,0,0) (S) 2 2 (D) Q r 2 h dh Q dV dh dV  2  r 2  Q so But dt r dt dt dt r U ˆ U ˆ U ˆ i j k For conservative force field, F   x y z where F  Force, U  Potential energy and Sol. Q 2 r 2 Fx = 0 Fy = 0 Fz = 0 U=0 (A) – P,Q,R,S ; (B) – Q,R ; (C) – P,S ; (D) – P,S r For (A) : F  2 xyz ˆi  x2 zjˆ  x2 ykˆ  Fx=0, Fy=0, Fz=0, U=0 r For (B) : F  2 xiˆ  zjˆ  ykˆ Fx  0, Fy=0, Fz=0, U 0 r For (C) : F  2 x  y  z  ˆi  x 2 ˆj  x 2 kˆ Fx=0, Fy 0, Fz 0, U=0 r For (D) : F  2 xy ˆi  x 2 ˆj  kˆ Fx =0, Fy 0, Fz 0, U=0 BASIC MATHS & VECTOR Ex. If surface area of a cube is changing at a rate of 5 m2/s, find the rate of change of body diagonal at the moment when side length is 1 m. (A) 5 m/s Sol. (B) 53 m/s (C) 5 3 m/s 2 (D) 5 4 3 m/s (D) Surface area of cube S=6a2 (where a = side of cube) Body diagonal l  3 a . Therefore S=2l2 Differentiating it w.r.t. time (B) dSdS d ld l d ld l 1 dSdS 5  2  2l       m/s dtdt dtdt dtdt 4 3a dtdt 4 3   INTEGRAL CALCULUS Integration is the reverse process of differentiation. By help of integration we can find a function whose derivative is known. Consider a function F(x) whose differentiation w.r.t. x is equal to f(x) then  f  x  dx  F  x   c here c is the constant of integration and this is called indefinite integration. Few basic formulae of integration are : n (i)  x dx  x n 1 c n 1 (ii) (iii)  sin xdx   cos x  c (iv)  cos xdx  sin x  c (v)  sec2 xdx  tan x  c (vi)  e x dx  e x  c n (vii)   ax  b  dx   ax  b n 1 a n  1  (ix)  sin  ax  b  dx  c  cos  ax  b  (xi)  sec 2  ax  b  dx  Ex. 1  x dx  ln x  c a tan  ax  b  a (viii) ln  ax  b  dx  ax  b  a c (x)  cos  ax  b  dx  c (xii)  e ax  b dx  c sin  ax  b  a c e ax  b c a Integrate the following w.r.t. x. (i) 4x3 (ii) x  1 x (iii) 1 2x  3 (iv) cos (4x+3) (v) cos2x Sol.  x31  4x4 3 4 x dx  4  c   c  x4  c   (i)  3  1 4   (iii) dx  2x  3  1 1 x2   ln x  c (ii)   x  dx   xdx   dx  x x 2  ln  2 x  3 c 2 2 (v)  cos xdx   (iv)  cos  4 x  3  dx  1  cos 2 x dx  1 dx  1 cos 2 xdx 2 cos 2 x dx   2 2 2 2 = sin  4 x  3 c 4 x 1 sin 2 x x 1   c   sin 2 x  c 2 2 2 2 4 Definite Integration When a function is integrated between a lower limit and an upper limit, it is called a definite integral. Consider a function F(x) whose differentiation w.r.t. x is equal to f(x), in an interval a  x  b then b  f  x  dx  F  b  F  a  a Area under a curve and definite integration y y=F(x) y x x=a x x=b dx Area of small shown element = ydx = f(x) dx b If we sum up all areas between x=a and x= b then 5 Ex. The integral (A) Sol. (B) 5 125 3  x dx 1 2  f  x  dx = shaded area between curve and x-axis. a is equal to (B) 124 3 (C) 1 3 (D) 45 5  x 3   53 13  125 1 124 x dx   3   3  3 3 3 3 1  1   Ex. 2 The following curve represent rate of change of a variable y w.r.t x. The change in the value of y when x changes from 0 to 11 is: dy/dx 20 9 0 (A) 60 (B) 25 3 11 6 (C) 35 x (D) 85 BASIC MATHS & VECTOR Sol.  dy  As dy =   dx So y   dy   dx  Area under the curve A1  11  dy    dx  dx y 0 20 1  6  20  60 2 A1 1 A 2    11  6 10   2 5 2 y  A1  A 2  60  25  35 0 3 6 9 A2 11 x -10 Average value of a continuous function in an interval : Average value of a function y = f(x), over an interval a  x  b is given by b yav   ydx a b  dx b   ydx a ba a Ex. Sol. Determine the average value of y = 2x +3 in the interval 0  x  1. (A) 1 (B) 5 (C) 3 (D) (D) 4 1 yav  Ex. Sol.  ydx 1 1   x2      2 x  3 dx   2    3 x   12  3 1  02  3  0   1  3  4 1 0 0   2  0 0   The average value of alternating current I=I0sint in time interval  0,  is   2I 4I (A) 0 (B) 2I0 (C) 0   (A) (D) I0  /  Ex.  Idt   /  /    I 0   cos t    I0 I 2I cos   cos 0   0  1  1  0          0 0  0  The velocity-time graph of a car moving along a straight road is shown in figure. The average velocity of the car is first 25 seconds is I av  0  I 0 sin tdt  Velocity(m/s) 20 0 Sol. (A) 20 m/s (B) (B) 14 m/s 10 20 25 Time(s) (C) 10 m/s (D) 17.5 m/s 25 Average velocity =  vdt 0 25  0  Area of v-t graph between t=0 to t = 25 s 1  25  10       20    14 m/s 25 25  2  