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

If an error of $1^o$ is made in measuring the angle of a sector of radius $30 \ cm$, then the approximate error in its area is

  1. $450 cm^2$
  2. $25\pi cm^2$
  3. $2.5\pi cm^2$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Area of sector $\displaystyle A=\dfrac{\pi r^{2}\theta}{360}$
Given $r=30 cm, d{\theta}=1^{0}$
Approximate error in A is $\displaystyle=dA=(\dfrac{dA}{d\theta})\Delta \theta$
                             $\displaystyle = \dfrac{900\pi}{360} $
$\displaystyle \Rightarrow dA =2.5 \pi cm^{2}$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If error in  measuring the edge of a cube is $k$% then the percentage error in estimating its volume is

  1. $k$
  2. $3k$
  3. $\displaystyle \frac{k}{3}$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the actual length of the cube be a.

Therefore the measured length of the cube will be 
$=a(1\pm0.0k)$
$=a(1\pm\dfrac{k}{100})$
Considering positive error, 
$a'=a(1+\dfrac{k}{100})$
$V'=a^{3}(1+\dfrac{k}{100})^{3}$

$=a^{3}(1+3(\dfrac{k}{100})+3(\dfrac{k}{100})^{2}+(\dfrac{k}{100})^{3})$

Since $\dfrac{k}{100}<<1$, hence we neglect the higher order terms.
Thus 
$V'=a^{3}(1+3(\dfrac{k}{100}))$

Actual volume V
$V=a^{3}$
Therefore 
$V'-V=a^{3}(1+\dfrac{3k}{100})-a^{3}$

$=a^{3}(\dfrac{3k}{100})$

$\dfrac{V'-V}{V}=\dfrac{a^{3}\dfrac{3k}{100}}{a^{3}}$

$=\dfrac{3k}{100}$

$=\dfrac{3k}{100}$

$\dfrac{V'-V}{V}\times 100=3k$
Therefore percentage error in volume is $3k$.

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

If the radius of a sphere is measured as $9 \ cm$ with an error of $ 0.03 \ cm$ then, find the approximate error in calculating its volume.

  1. $\displaystyle 9.72\pi\:\: cm^{3}$
  2. $\displaystyle 7.92\pi\:\: cm^{3}$
  3. $\displaystyle 8.72\pi\:\: cm^{3}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $r=9 cm, \Delta r=0.03cm$

We know Volume of sphere with radius 'r' is $V=\cfrac{4}{3}\pi r^3$
$\therefore \Delta V=4\pi r^2\Delta r$
$\Rightarrow \Delta V=4\pi\times 81\times .03=9.72\pi  cm^3 $(using given values)

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

The percentage error in the $11^{th}$ root of the number $28$ is approximately ____________ times the percentage error in $28$

  1. $\dfrac { 1 }{ 28 } $
  2. $\dfrac { 1 }{ 11 } $
  3. $11$
  4. $28$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Suppose $y = x^{11}$

$dy = 11x^{10}dx$
Dividing by y on both sides, we have $ \cfrac{\Delta y}{y} = \cfrac{11x^{10}\Delta x}{y} = \cfrac{11x^{10}\Delta x}{x^{11}}$
$\therefore \cfrac{\Delta y}{y} = \cfrac{11 \Delta x}{x}$

Here, when $y = 28$, $x$ will be the $11^{th}$ root of $28$.
$\therefore \cfrac{\Delta (28)}{28} = \cfrac{11 \Delta (\sqrt[11]{28})}{\sqrt[11]{28}}$
Hence, the percentage error in the $11$th root of $28$ would approximately be $\cfrac{1}{11}$ times the error in $28$

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

State true or false:
By the method of Newton-Raphson, the cube root of $10$ after the first iteration is $2.167$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $f\left( x \right) ={ x }^{ 3 }-10$
$f\left( x \right) =3{ x }^{ 2 }$
Letting initial guess be ${ x } _{ 0 }=2$ (${ 2 }^{ 3 }=8$ which is close to $10$)
We have
${ x } _{ 1 }=2-\left[ \cfrac { { (2) }^{ 3 }-10 }{ 3\times { (2) }^{ 2 } }  \right] \quad \left[ \because { x } _{ n+1 }={ x } _{ n }-\cfrac { f\left( { x } _{ n } \right)  }{ f'\left( { x } _{ n } \right)  }  \right] $
${ x } _{ 1 }=2-\left[ \cfrac { -2 }{ 12 }  \right] $
${ x } _{ 1 }=2+\cfrac { 1 }{ 6 } $
${ x } _{ 1 }=2+0.167$
${ x } _{ 1 }=2.167$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

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

  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. $x _{n + 1} = \dfrac {1}{3}\left [2x _{0} + \dfrac {y}{x _{0}^{2}}\right ]$
Reveal answer Fill a bubble to check yourself
A Correct answer
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]  } $ 
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

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

  1. $\cdot8203825\ E\ 09$
  2. $\cdot8303645\ E\ 06$
  3. $\cdot1559662\ E\ 06$
  4. $\cdot8305645\ E\ 09$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $1.23 \times 10^{10}$

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

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

                                                 $=0.1559662 \times 10^6$

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

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

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

  1. $\cdot12057\ E\ 09$
  2. $\cdot0033\ E-12$
  3. $\cdot0033\ E\ 09$
  4. $\cdot12057\ E\ 05$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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