Tag: approximation

Questions Related to approximation

By Newton - Raphson's method the formula for finding the square root of any number $y$ is:

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $x _{n + 1} = \dfrac {1}{2}\left [x _{n} + \dfrac {y}{x _{n}}\right ]$

  2. $x _{n + 1} = \dfrac {1}{2}\left [x _{0} + \dfrac {y}{x _{0}}\right ]$

  3. $x _{n + 1} = \dfrac {1}{3}\left [2x _{n} + \dfrac {y}{x _{n}^{2}}\right ]$

  4. $x _{n + 1} = \dfrac {1}{3}\left [2x _{0} + \dfrac {y}{x _{0}^{2}}\right ]$

Show answer
Correct Option: A
Explanation:
Let $x = \sqrt { y } $
$\implies { x }^{ 2 } = y$
$\implies { x }^{ 2 }-y = 0$
Iterative eqn. for Newton Raphson method is

${ x } _{ n+1 } = { { x } _{ n } }-\dfrac { f({ { x } _{ n } }) }{ f\prime ({ { x } _{ n } }) } $ 

Substitute $f(x) = { x }^{ 2 }-y$
$\implies f'(x) = 2x$ ........... $[\because\ y$ is any number $\therefore  f'(y) =0]$

     ${ x } _{ n+1 } = { { x } _{ n } }-\dfrac { { x } _{ n }^{ 2 }-y }{ 2{ x } _{ n } } $ 
              $= { x } _{ n }-\dfrac { { x } _{ n }^{ 2 } }{ 2{ x } _{ n } } +\dfrac { y }{ 2{ x } _{ n } } $ 

               $= { x } _{ n }-\dfrac { { x } _{ n } }{ 2 } +\dfrac { y }{ 2{ x } _{ n } } $
 
    ${ x } _{ n+1 } = \dfrac { { x } _{ n } }{ 2 } +\dfrac { y }{ 2{ x } _{ n } } $ 

    $\boxed { { x } _{ n+1 } = \dfrac { 1 }{ 2 } \left[ { x } _{ n }+\dfrac { y }{ { x } _{ n } }  \right]  } $ 

The value of $\cdot4125\ E\ 05 \times \cdot3781\ E01.$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\cdot8203825\ E\ 09$

  2. $\cdot8303645\ E\ 06$

  3. $\cdot1559662\ E\ 06$

  4. $\cdot8305645\ E\ 09$

Show answer
Correct Option: C
Explanation:

$0.4125:E:05 \times 0.3781:E:01=?$

The Scientific format displays a number in exponential notation, replacing part of the number with $E+n$, where $E$ (stands for exponent) multiplies the preceding number by $10$ to the $n^{th}$ power.

That is $1.23E+10$ can be written as $1.23 \times 10^{10}$

$0.4125:E:05 \times 0.3781:E:01=(0.4125 \cdot 10^5) \times (0.3781 \cdot 10^1)$

                                                 $=(0.4125 \times 0.3781)10^6$

                                                 $=0.1559662 \times 10^6$

$0.4125:E:05 \times 0.3781:E:01=0.1559662:E:06$

The value of $\cdot7378\ E\ 05 -\cdot2347\ E05.$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\cdot12057\ E\ 09$

  2. $\cdot12057\ E\ 05$

  3. $\cdot64045\ E\ 09$

  4. $\cdot5031\ E\ 05$

Show answer
Correct Option: D
Explanation:

We need to find value of $\cdot7378\ E\ 05 -\cdot2347\ E05.$
Take out $E05$ common to get 
$ (.7378-.2347)E05$ $= \cdot5031\ E\ 05 $

The value of $\cdot4365\ E\ 05 +\cdot2735\ E06$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\cdot12057\ E\ 09$

  2. $\cdot31715\ E\ 06$

  3. $\cdot64045\ E\ 09$

  4. $\cdot517336\ E\ 09$

Show answer
Correct Option: B
Explanation:

$⋅4365 E 05+⋅2735 E06 = (.04365+.2735)E06$4

$=(.31715)E06$

The value of $\cdot4657\ E\ - 12 -\cdot4624$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\cdot12057\ E\ 09$

  2. $\cdot0033\ E-12$

  3. $\cdot0033\ E\ 09$

  4. $\cdot12057\ E\ 05$

Show answer
Correct Option: B
Explanation:

$0.4657E-12-0.4624E=(0.4657E-0.4624E)-12$
                                              $=0.0033E-12$
Therefore the value of $0.4657E-12-0.4624E$ is $0.0033E-12$


The value of $\cdot3214\ E\ - 02 \times \cdot3781\ E\ 05.$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\cdot699045\ E\ 05$

  2. $\cdot699045\ E\ 02$

  3. $\cdot699055\ E\ 02$

  4. $\cdot698045\ E\ 02$

Show answer
Correct Option: B

The percentage error in the surface area of a cube with edge x cm, when the edge is increased by $11\%$ is _________.

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $11$

  2. $22$

  3. $10$

  4. $44$

Show answer
Correct Option: B
Explanation:

Surface area of cube $=$ S $=6x^2$
$\therefore \dfrac{dS}{dt}=12x\dfrac{dx}{dt}$
Side of cube increase $11\%$.
$\therefore \dfrac{dx}{dt}=11\%$ increment in side
$=\dfrac{11x}{100}$
$\therefore \dfrac{dx}{dt}=12\left(\dfrac{11x}{100}\right)$
$=6\left(\dfrac{22}{100}\right)x^2$
$=6x^2\left(\dfrac{22}{100}\right)$
$=22\%$ in surface area
$\therefore$ Surface area increase $22\%$.

The focal length of a mirror is given by $\dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$. If equal errors ($\alpha$) are made in measuring $u$ and $v$, then the relative error in $f$ is

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\dfrac {2}{\alpha}$

  2. $\alpha \left (\dfrac {1}{u}+\dfrac {1}{v}\right )$

  3. $\alpha \left (\dfrac {1}{u}-\dfrac {1}{v}\right )$

  4. none of these

Show answer
Correct Option: B
Explanation:

Given, $\displaystyle \dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$
$\Rightarrow \displaystyle -\dfrac {\Delta v}{v^2}+\dfrac {\Delta u}{u^2}=-\dfrac {2\Delta f}{f^2}$
Since, given equal errors in measuring u and v i.e.$\Delta u=\Delta v=\alpha$
$\Rightarrow {\alpha}\left(\dfrac{1}{u}-\dfrac{1}{v}\right)\left(\dfrac{1}{u}+\dfrac{1}{v}\right)=-\dfrac {2\Delta f}{f^2}$
But, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{2}{f}$
$\Rightarrow\displaystyle \frac{ \Delta f}{f}={\alpha}\left(\dfrac{1}{u}+\dfrac{1}{v}\right)$
Hence, correct option is B

The period of oscillation $T$ of a pendulum of length $l$ at a place of acceleration due to gravity $g$ is given by $T=2\pi \sqrt {\dfrac {l}{g}}$. If the calculated length is $0.992$ times the actual length and if the value assumed for $g$ is $1.002$ times its actual value, the relative error in the computed value of $T$ is

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $0.005$

  2. $-0.005$

  3. $0.003$

  4. $-0.003$

Show answer
Correct Option: B
Explanation:

Relative error will calculate by
$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{\Delta L}{L}-\dfrac{\Delta g}{g}\right]$

$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{0.992L}{L}-\dfrac{1.002 g}{g}\right]$

$\Delta T=-0.005T$

The area of a triangle is computed using the formula $S=\dfrac {1}{2}$ bc sin A. If the relative errors made in measuring b, c and calculating S are respectively $0.02$, $0.01$ and $0.13$ the approximate error in A when $A=\pi /6$ is

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $0.05$ radians

  2. $0.01$ radians

  3. $0.05$ degree

  4. $0.01$ degree

Show answer
Correct Option: A
Explanation:

Error formula for given equation is
$\dfrac{\Delta s}{s}=\dfrac{\Delta b}{b}+\dfrac{\Delta c}{c}+\dfrac{\Delta sinx}{sinx}$
$0.13=0.02+0.01+\dfrac{\Delta sinx}{1/2}$
$\Delta sinx=0.05$