Tiêu đề APPLICATIONS OF DIFFERENTIAL CALCULUS (Q).pdf ✅

\\ Chapter – 1 Consider a curve y = f(x) and two points P(x, y) and Q(x + x, y + y) on it. Then x y dx dy x     0 lim =               x x x y y y x 0 lim = 0 lim x (slope of the chord PQ) = Slope of the tangent PT at P(x, y) = tan , where  is the angle which the tangent at P makes with the positive direction of the x-axis. Illustration 1 Question: Find the slope of tangent at the point having ordinate 3 on the curve x 3 = 3y 2 . Solution: Differentiating the given equation of curve w.r.t x, we get dx dy 3x 3.2y 2   y x dx dy 2 2  So, we require abscissa also to obtain this value. Substituting y = 3 in the equation of curve, we have x 3 = 3 (3)2 = 27  x = 3  The point of interest is (3, 3) Hence, slope of tangent at this point = 2 3 2( 3) 3 2 (3, 3)      dx dy GEOMETRICAL INTERPRETATION OF DERIVATIVE 1 THEORY CONTENT OF APPLICATIONS OF DIFFERENTIAL CALCULUS  y P(x , y) x O Q(x + x, y + y) T TANGENT AND NORMAL 2
Tangent at a point is a line which touches the curve at that point and normal at a point is a line which is perpendicular to the tangent at that point. Given the equation of a curve y = f(x) and a point A (x1, y1) on it, the equation of the tangent at A is y – y1 = ( ) x x1 dx dy atA        and the equation of the normal at A is ( ) 1 1 x x1 dx dy y y atA           Illustration 2 Question: Find the equation of tangent and normal in the previous illustration. Solution: Equation of tangent ( 3) 2 3 y  (3)   x   2y + 6 = 3x + 9  3x + 2y  3 = 0 Equation of normal ( 3) 3 2 y  (3)   x   3y + 9 = 2x  6  2x  3y  15 = 0  When the curve is given in parametric form i.e., x = g(t) and y = h(t) Equation of tangent at the point t = t1 is ( ( )) '( ) '( ) ( ) 1 1 1 1 x g t g t h t y  h t   and equation of normal is ( ( )) '( ) '( ) ( ) 1 1 1 1 x g t h t g t y  h t    Illustration 3 Question: Find the points on the curve y = x 3 – x 2 – x + 3 where the tangent is parallel to the x-axis. Solution: Given curve y = x 3 – x 2 – x + 3 dx dy = 3x 2 – 2x –1 Since the tangent is parallel to the x-axis, slope = tan 0 = 0 i.e., dx dy = 0 Hence 3x 2 – 2x –1 = 0 or (3x + 1) (x – 1) = 0  x = 3 1  or 1 For the first point, x = 3 1  and y = 27 88 3 3 1 3 1 3 1 3 2                          For the second point, x = 1 and y = 13 – 1 2 – 1 + 3 = 2 Hence the points are        27 88 , 3 1 and (1, 2)  Length of tangent, normal, subtangent and subnormal
Let the tangent and normal at P(x, y) meet the x-axis at T and N respectively. PT is called the length of the tangent and is equal to PM cosec  = sin 1 y (from  TMP) Hence length of tangent (PT) = m y y m 2 1 1 1 tan sec     where        dx dy m Similarly, the following results can easily be concluded.  PN is called the length of the normal = PM sec  = 2 y1 1m (from MNP)  TM is called the subtangent = PM. cot m y1 y1 tan     (from TMP)  MN is called the subnormal = PM tan  = | y1m | (from MNP) Illustration 4 Question: Find the length of tangent, normal, subtangent and subnormal to the curve 2 1 x x y   at the point having abscissa 2 . Solution: At x  2 , y   2  Point is P( 2 ,  2) Now, 2 2 2 2 2 2 (1 ) 1 (1 ) (1 ) ( 2 ) x x x x x x dx dy          3 (1 2) 1 2 2 at     p dx dy = m (let) Equation of tangent : y  2  3(x  2) It intersects x-axis at         , 0 3 4 2 T Length of tangent =   3 2 5 2 0 2 3 4 2 2 2            PT   Perpendicular dropped from P on x-axis is M ( 2, 0)  Subtangent = MT = 3 2 2 3 4 2   Equation of normal : ( 2) 3 1 y  2   x  It intersects x-axis at N (2 2, 0) Length of normal = PN = 20 and subnormal = MN = 3 2 . Alternative: y P(x1 , y1 ) x O T M N  
Length of tangent =   3 2 5 3 1 2 1 9 2      m y m Length of normal = 1  2 10 2 5 2 y  m     Length of subtangent = m y = 3 2 3 2   Length of subnormal = y  m   2  3  3 2 Given two curves C1 : y = f(x) and C2 : y = g(x) intersecting at some point P (x1, y1). Let PT1 be the tangent at P to curve C1 and let PT1 make an angle 1 with OX. Let PT2 be the tangent at P to curve C2 and let PT2 make an angle 2 with OX. The angle between two curves is defined to be the angle between the two tangents at the point of intersection.   (the angle between the curves) APB = T1PT2 = 2 – 1 from ABP tan  = tan (2 – 1) = 1 2 2 1 2 1 2 1 1 tan tan 1 tan tan m m m m          If  is the acute angle between the two curves 1 2 2 1 1 tan m m m m     where m1 = f (x) at P and m2 = g(x) at P. Remark: (i) curves intersect orthogonally if m1 m2 = –1; (ii) curves touch each other if m1 = m2. Illustration 5 Question: Find the angle of intersection of the curves y = x 3 and 6y = 7 – x 2 . Solution: The point of intersection is found by solving the equations simultaneously y = x 3 and y = 6 6 7 2 x   6x 3 = 7 – x 2 or 6x 3 + x 2 – 7 = 0 (x – 1) (6x 2 + 7x + 7) = 0 This gives x = 1 only, the other factor gives complex roots. If x = 1 y = 1 (by using equation 3 y  x ) Now, from C1: 2 3x dx dy  = 3 at x = 1 1 2  T2 T1 C1 C2 P(x1 , y1 ) A B x y O ANGLES BETWEEN TWO CURVES 3