Errors and Approximations || V.S.A.Q’S||

Errors and Approximations || V.S.A.Q’S||

These solutions designed by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students.

 

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.

 

Errors and Approximations

 

 

Question 1

Find dy and ∆y for the following functions for the values of x and ∆x which are shown against each of the functions

(i) y = f(x) = x2 + x at x = 10 when ∆x = 0.1.

Sol:

Given y = f(x) = x2 + x at x = 10, ∆x = 0.1

∆y = f (x + ∆x) – f (x)

     = f (10 + 0.1) – f (10)

    = f (10.1) – f (10)

    = (10. 1)2 + 10.1 – (102 + 10)

   = 102.01 + 10.1 – 100 – 10

    = 112.11 – 110

    = 2.11

dy = f’ (x) ∆x

     = (2x + 1) (0.1)

     = [2(10) + 1] (0.1)

    = 21 × 0.1

    = 2.1

(ii)  y = cos x at x = 600 with ∆x = 10 (10 = 0.0174 radians)

Sol:

Given y = cos x, x = 600 and ∆x = 10

 ∆y = f (x + ∆x) – f (x)

      = cos (600 + 10) – cos 600

      = cos (610) – 0.5

      = 0.4848 – 0.5

      = – 0.0152

dy = f’ (x) ∆x

      = – sin x (10)

      = – sin 600 × 0.0174

      =– 0.8660 × 0.0174

      = – 0.0150

(iii)  y = x2 + 3x + 6, x = 10 with ∆x = 0.01 

Sol:

   y = x2 + 3x + 6

  ∆y = f (x + ∆x) – f (x)

       = f (10 + 0.01) – f (10)

       = f (10.01) – f (10)

      = (10.01)2 + 3 (10.01) + 6 – (102 + 3 (10) + 6)

       = 100. 2001 + 30.03 + 6 – 100 – 30 – 6

       =130. 2301 – 130

       = 0.2301

dy = f’ (x) ∆x

     = (2x + 3 + 0) (0.01)

     = (2× 10 + 3) (0.01)

      = 23 × 0.01

      = 0.23

(iv)  y =Differentiation 42 , x = 8 and ∆x =0.02

Sol:

Differentiation 42

 

Ts Inter Maths IA Concept

1/  3

 

Leave a Comment

Your email address will not be published.