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

Inter maths 1b 2 marks Questions

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

Question 2

The side of a square is increased from 3cm to 3.01cm find the approximate increase in the area of the square.

Sol:

Let x be the side of the square and area be A

Area of the square A = x2

 x = 3 and ∆x = 0.01

∆A = 2x × ∆x

      = 2(3) (0.01)

      = 6 × 0.01

      = 0.06

Question 3

If an increase in the side of a square is 2% then find the approximate percentage of increase in its area.

Sol:

Let x be the side of the square and A be its area

Differentiation 36  = 2

 A = x2

∆A = 2x × ∆x

The approximate percentage error in area A

Differentiation 37

= 2 × 2 =4

Question 4

From the following. Find the approximations 

(i) Differentiation 38

Sol:

Let f(x) =  , where x = 1000 and ∆x =– 1

f’ (x) =Differentiation 39

Approximate value is

f (x + ∆x) = f(x) + f’ (x) ∆x

Differentiation 40 

 = 10 – 0. 0033

 = 9.9967

(ii) Differentiation 43

Sol:

Differentiation 44

(iii) Differentiation 46

Sol:

Differentiation 45

(iv) Sin 620

Sol:

Let f(x) = sin x, where x = 600 and ∆x =20

Approximate value is

 f (x + ∆x) = f(x) + f’ (x) ∆x

= sin 600 + cos x (20)

 = sin 600+ cos 600 (0.0348)

= 0.8660 + 0.5 × 0.0348

= 0.8660 + 0.0174

=0.8834

Question 5

The radius of a sphere is measured as 14cm. Later it was found that there is an error of 0.02cm in measuring the radius. Find the approximate error in the surface area of the sphere.

Sol:

Given r = 14 cm and ∆r =0.02cm

Surface area of sphere =A = 4π r2

∆A = 8π r ∆r

      = 8 ×3.14× 14 × 0.02

      = 7.0336

 


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