Tag: errors and approximations

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Questions Related to errors and approximations

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

If the percentage error in the edge of a cube is 1, then error in its volume is

  1. $1 \%$
  2. $2 \%$
  3. $3 \%$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Volume of cube $V=x^{3}$
$\Rightarrow \displaystyle \dfrac{dV}{dx}=3x^{2}$
Percentage error in x is 1%.
$\Rightarrow \displaystyle \dfrac{\Delta x}{x}=\dfrac{1}{100}$
$\Rightarrow \displaystyle \Delta x=\dfrac{x}{100}$
Approximate error in V $\displaystyle=dV=(\dfrac{dV}{dx}) \Delta x$
                                  $\displaystyle = \dfrac{3{x}^{3}}{100}$                              
Percentage error in V $\displaystyle= \dfrac{dV}{V} $
                                   $\displaystyle=\dfrac{3}{100}=3\%$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

In a $\Delta ABC$ if sides a and b remain constant such that $\alpha$ is the error in C, then relative error in its area is

  1. $\alpha \cot C$
  2. $\alpha \sin C$
  3. $\alpha\tan C$
  4. $\alpha\cos C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Area of triangle $S\displaystyle =\dfrac {1}{2}ab \sin C$
$\displaystyle \Rightarrow \dfrac {dS}{dC}=\dfrac {1}{2}ab \cos C$
Now, approximate error in S is $\Delta S=\dfrac {dS}{dC}\Delta C$
$\displaystyle\Rightarrow \Delta S=\dfrac {1}{2}ab \cos C \alpha              [\because \Delta C=\alpha]$
$\displaystyle\Rightarrow \dfrac {\Delta S}{S}=\dfrac {\dfrac {1}{2}ab \cos C}{\dfrac {1}{2}ab \sin C}\alpha=\alpha \cot C$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

In a $\Delta ABC$ the sides b and c are given. If there is an error $\Delta A$ in measuring angle A, then the error $\Delta a$ in side a is given by

  1. $\dfrac {S}{2a}\Delta A$
  2. $\dfrac {2S}{a}\Delta A$
  3. bc sin A $\Delta A$

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

In $\triangle ABC$ we have
$\Rightarrow d\left( 2bc\cos { A }  \right) =d\left( { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } \right) \ \Rightarrow -2bc\sin { A } dA=-2ada\ \Rightarrow bc\sin { A } dA=ada$
$\displaystyle \Rightarrow \frac { 2 }{ a } \left( \frac { 1 }{ 2 } bc\sin { B }  \right) dA=da$
$\displaystyle \Rightarrow da=\frac { 2S }{ a } dA$
$\displaystyle \Rightarrow \triangle a=\frac { 2S }{ a } dA\ \left[ \because dx\equiv \triangle a\quad and\quad dA=BA \right] $

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

If errors of $1\%$ each are made in the base radius and height of a cylinder, then the percentage error in its volume is

  1. $1\%$
  2. $2\%$
  3. $3\%$

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

Given, percentage error in r is 1%
$\Rightarrow \displaystyle \frac{\Delta r}{r}=\frac{1}{100}$

$\Rightarrow \displaystyle \Delta r=\frac{r}{100}$
Also given, percentage error in h is 1%
$\Rightarrow \displaystyle \frac{\Delta h}{h}=\frac{1}{100}$

$\Rightarrow \displaystyle \Delta h=\frac{h}{100}$
Now, volume of cylinder $V=\pi r^{2}h$
$\Delta V=\pi [r^{2}\Delta h+2rh\Delta r]$
$\displaystyle \Delta V=\pi[r^{2}\frac{h}{100}+2rh\frac{r}{100}]$

$\displaystyle\Delta V=\pi r^{2}h[\frac{3}{100}]$
$\Rightarrow\displaystyle\frac{\Delta V}{V}=\frac{3}{100}$
Percentage error in V is 3%
 

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

The circumference of a circle is measured as $56$ cm with an error $0.02$ cm. The percentage error in its area is

  1. $\dfrac {1}{7}$
  2. $\dfrac {1}{28}$
  3. $\dfrac {1}{14}$
  4. $\dfrac {1}{56}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Circumference of circle $C=2\pi r=56cm$


$\Rightarrow \displaystyle r=\frac{28}{\pi}$

Also, $\displaystyle \frac{dC}{dr}=2\pi$

Area of circle $A=\pi r^{2}$

$\displaystyle \frac{dA}{dr}=2\pi r$

$\Rightarrow\displaystyle \frac{dA}{dC}=r =\frac{28}{\pi}$

Approximate error in A $=\displaystyle dA=(\frac{dA}{dC})\Delta C$

                                      $= r (0.02)$

$\Rightarrow\displaystyle \frac{dA}{A}= \frac{0.02}{\pi r}=\frac{1}{1400}$

Percentage error in A is $\displaystyle\frac{1}{14}$%

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 } $
${ x } _{ 1 }=2+0.167$
${ x } _{ 1 }=2.167$