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

The pressure P and volume V of a gas are connected by the relation $PV^{1/4}=constant$. The percentage increase in the pressure corresponding to a deminition of $\dfrac12 \%$ in the volume is

  1. $\dfrac {1}{2}$ %
  2. $\dfrac {1}{4}$ %
  3. $\dfrac {1}{8}$ %

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$PV^{1/4}=constant$
$\displaystyle \Rightarrow P=\dfrac{k}{V^{{1}/{4}}}$
$\displaystyle \Rightarrow \dfrac{dP}{dV}=-\dfrac{k}{4}V^{-{5}/{4}}$
Percentage error in V $\displaystyle= -\dfrac{1}{2}\%$
$\Rightarrow\displaystyle \dfrac{\Delta V}{V} =-\dfrac{1}{200}$
$\Rightarrow\displaystyle {\Delta V}=-\dfrac{V}{200}$
Approximate change in $P\displaystyle=dP=(\dfrac{dP}{dV}){\Delta V}$
                                           $\displaystyle  =\dfrac{1}{800} {kV^{-1/4}}=\dfrac{1}{8}\%$ of P
Percentage increase in  $V \ \displaystyle =\dfrac{1}{8}\%$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If the ratio of base radius and height of a cone is 1:2 and percentage error in radius is $\lambda$ %, then the error in its volume is

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

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Volume of cone $V=\dfrac { 1 }{ 3 } \pi { r }^{ 2 }h $
Given, $\displaystyle \dfrac{r}{h}=\dfrac{1}{2}$
$\Rightarrow V=\dfrac { 2 }{ 3 } \pi { r }^{ 3 } $
$\displaystyle \dfrac{dV}{dr}=2\pi r^{2}$
Percentage error in measuring r $=\lambda$%
$\Rightarrow \displaystyle \dfrac{\Delta r}{r}=\dfrac{\lambda}{100}$
$\Rightarrow \displaystyle \Delta r =\dfrac{\lambda r}{100}$
Approximate error in V $\displaystyle=dV=(\dfrac{dV}{dr}) \Delta r $
                                      $\displaystyle=\dfrac{\lambda}{100} (2\pi r^{3})=\lambda\%$ of V
Percentage error in $V =\lambda\%$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If $y=x^n$, then the ratio of relative errors in $y$ and $x$ is

  1. $1:1$
  2. $2:1$
  3. $1:n$
  4. $n:1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$y=x^{n}$
$\Rightarrow \displaystyle \dfrac{dy}{dx}=nx^{n-1}$
Approximate error in y is $\displaystyle dy=\left (\dfrac{dy}{dx}\right) \Delta x$
                                      $=nx^{n-1} \Delta x$
Relative error in y is $\displaystyle \dfrac{dy}{y}=\dfrac{n}{x}\Delta x$
Approximate error in x is $\displaystyle dx=\left (\dfrac{dx}{dy}\right) \Delta y$
                                     $\displaystyle=\dfrac{1}{nx^{n-1}} \Delta y$
Relative error in x is $\displaystyle \dfrac{dx}{x}=\dfrac{1}{nx^{n}}\Delta y$
Required ratio $\displaystyle = \dfrac{\dfrac{n}{x}\Delta x}{\dfrac{1}{nx^{n}}\Delta y}$
                               $\displaystyle =n^{2}x^{n-1} \dfrac{\Delta x}{\Delta y}$
                               $\displaystyle =\dfrac{n}{1}$
So, the ratio of relative errors in y and x is $ n:1$.
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The circumference of a circle is measured as $28 cm$ with an error of $0.01 cm$. The percentage error in the area is

  1. $\dfrac {1}{14}$
  2. $0.01$
  3. $\dfrac {1}{7}$

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

Circumference $C=2\pi r$
$\Rightarrow\displaystyle r=\frac{14}{\pi}$
Also, $\displaystyle \frac{dC}{dr}=2\pi$
Area of circle $A=\pi r^{2} $
$\Rightarrow\displaystyle A=\frac{{14}^{2}}{\pi}$
Also, $\displaystyle \frac{dA}{dr}=2\pi r$
$\displaystyle \Rightarrow \frac{dA}{dC}=r=\frac{14}{pi}$
Approximate error in $A$ is $\displaystyle dA=( \frac{dA}{dC}) \Delta C$
                           $\displaystyle=\frac{14}{\pi}\frac{1}{100}$
                            $\displaystyle=\frac{1}{1400}$ of A
Percentage error in $A \ \displaystyle =\frac{1}{14}\%$

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

If there is an error of $0.01 cm$ in the diameter of a sphere then percentage error in surface area when the radius $= 5 cm$, is

  1. $0.005\%$
  2. $0.05\%$
  3. $0.1\%$
  4. $0.2\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Surface area of sphere $S=4\pi r^{2}$
$\displaystyle S=\pi D^{2}$
$\Rightarrow S=100\pi$
Also, $ \displaystyle \frac{dS}{dD}=2\pi D=20\pi$
Approximate error in S is $\displaystyle dS=(\frac{dS}{dD})\Delta D$
                                         $ =20\pi (0.01)$
                                          $=\dfrac{1}{500} S$
                                           $=0.2$% of S
Percentage error in $S=0.2%$

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 $\displaystyle\frac{1}{14}$%