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

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

Errors and approximations: These solutions were designed by the ‘Basics in Maths‘ team. These notes to do help 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) = x^{2} + x at x = 10 when ∆x = 0.1.

**Sol: **

Given y = f(x) = x^{2} + 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 – (10^{2} + 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 = 60^{0} with ∆x = 1^{0} (1^{0} = 0.0174 radians)

**Sol: **

Given y = cos x, x = 60^{0} and ∆x = 1^{0}

∆y = f (x + ∆x) – f (x)

= cos (60^{0} + 1^{0}) – cos 60^{0}

= cos (61^{0}) – 0.5

= 0.4848 – 0.5

= – 0.0152

dy = f’ (x) ∆x

= – sin x (1^{0})

= – sin 60^{0} × 0.0174

=– 0.8660 × 0.0174

= – 0.0150

**(iii) **y = x^{2} + 3x + 6, x = 10 with ∆x = 0.01

**Sol: **

y = x^{2} + 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 – (10^{2} + 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

**Sol: **

**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 the area be A

Area of the square A = x^{2}

** **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

A = x^{2}

∆A = 2x × ∆x

The approximate percentage error in area A

= 2 × 2 =4

**Question 4**

From the following. Find the approximations

**Sol: **

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

Approximate value is

f (x + ∆x) = f(x) + f’ (x) ∆x

= 10 – 0. 0033

= 9.9967

**Sol:**

**Sol:**

**(iv) **Sin 62^{0}

**Sol: **

Let f(x) = sin x, where x = 60^{0} and ∆x =2^{0}

Approximate value is

f (x + ∆x) = f(x) + f’ (x) ∆x

= sin 60^{0} + cos x (2^{0})

= sin 60^{0}+ cos 60^{0} (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 r^{2}

∆A = 8π r ∆r

= 8 ×3.14× 14 × 0.02

= 7.0336