Errors and approximations - class-XII

Using binomial approximation, (1 - x)^(1/n) is approximately equal to 1 - x/n for small x. Here, let x = 0.001 and n = 10, so (0.999)^(1/10) = (1 - 0.001)^(1/10) is approximately 1 - 0.001 / 10 = 1 - 0.0001 = 0.9999.

Here ${x} _{0}=14, f(x)={x}^{2}-78.8$
and $f'(x)=2x$
$\therefore$ ${x} _{1}={x} _{0}-\cfrac{f({x} _{0}}{f'({x} _{0})}$
$=14-\cfrac { { \left( 14 \right)  }^{ 2 }-\left( 78.8 \right)  }{ 2\times 14 } =9.814$

$E05$ can be taken out common. 

$E08$ can be taken common 

in a mutiplication while multiplying we have to add the percentage errors 

Let $r$ be the radius of the sphere.

Volume of cube$V=x^{3}$
$\displaystyle \frac{dV}{dx}=3x^{2}$
Percentage error in measuring side $= k%$
$\displaystyle \Rightarrow \frac{\delta x}{x}=\frac{k}{100}$
$\displaystyle {\delta{x}}=\frac{xk}{100}$
Approximate error in estimating volume $=dV=(\frac{dV}{dx}){\delta x}=3x^{2}\frac{xk}{100}$
Percentage error in estimating volume$=\frac{dV}{V}\times 100=3k$

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

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 $S=0.2%$

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] $

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%
 

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

Let the actual length of the cube be a.

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

Suppose $y = x^{11}$

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

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 $1.23 \times 10^{10}$

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

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

                                                 $=0.1559662 \times 10^6$

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

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

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

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 $22%$.

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

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

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

$\Delta T=-0.005T$

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

To find the cube root of $24$ using Newton - Raphson method,
we need to solve $f(x)=x^3-24$.

$ \Rightarrow f'(x)=3x^2$

Notice $3^3=27$

Therefore the cube root of $24$ is slightly less than $3$.

We have $f(x)=x^3-24, f'(x)=3x^2$

Let us start estimating the root $x$

Let the first estimation be $a=2.9$ (slightly less than 3)

Hence the subsequent estimates will be $b=a-\dfrac{f(a)}{f'(a)},c=b-\dfrac{f(b)}{f'(b)}$.

$f(a)=f(2.9)=(2.9)^3-24=0.389$ and $f'(a)=f'(2.9)=3(2.9)^2=25.23$

Therefore $b=2.9-\dfrac{0.389}{25.23}\approx 2.88458$

Now $c=2.88458-\dfrac{f(2.88458)}{f'(2.88458)}=2.88449$

Hence the cube root of $24$ is $2.884$

We have to find the second,third and fourth approximation of root of the equation $x^3-3x-5=0$ in the interval $(2,2.5)$ using successive Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{2+2.5}{2}=2.25$

Since $f(c _0)f(a _0)=f(2.25)f(2)>0$

Therefore set $a _1=2.25,b _1=b _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{2.25+2.5}{2}=2.375$

Since $f(c _1)f(a _1)=f(2.375)f(2.25)<0$

Therefore set $a _2=a _1,b _2=c _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{2.25+2.375}{2}=2.3125$

Since $f(c _2)f(a _2)=f(2.3125)f(2.25)<0$

Therefore set $a _3=a _2,b _3=c _2$

$\textbf{Iteration 4: k=3}$

$c _3=\dfrac{a _3+b _3}{2}=\dfrac{2.25+2.3125}{2}=2.28125$

Thus the second,third and fourth approximations are $2.375,2.3125,2.28125$ respectively.

Here, $x^3-2x-5=0$

Here, $x^4-x-10=0$

We have to find the second,third and fourth approximation of root of the
equation $x^3-x-4=0$ in the interval $(1,2)$ using successive
Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{1+2}{2}=1.5$

Since $f(c _0)f(a _0)=f(1.5)f(1)>0$

Therefore set $a _1=1.5,b _1=b _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{1.5+2}{2}=1.75$

Since $f(c _1)f(a _1)=f(1.75)f(1.5)>0$

Therefore set $a _2=c _1,b _2=b _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{1.75+2}{2}=1.875$

Since $f(c _2)f(a _2)=f(1.875)f(1.75)<0$

Therefore set $a _3=a _2,b _3=c _2$

$\textbf{Iteration 4: k=3}$

$c _3=\dfrac{a _3+b _3}{2}=\dfrac{1.75+1.875}{2}=1.8125$

Thus the second,third and fourth approximations are $1.75,1.875,1.8125$ respectively.

Here, $f(x)=x^3-x-4=0$

We have to find the second,third and fourth approximation of root of the
equation $x^3+x^2-1=0$ in the interval $(0,1)$ using successive
Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{0+1}{2}=0.5$

Since $f(c _0)f(a _0)=f(0.5)f(0)>0$

Therefore set $a _1=0.5,b _1=b _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{0.5+1}{2}=0.75$

Since $f(c _1)f(a _1)=f(0.75)f(0.5)>0$

Therefore set $a _2=c _1,b _2=b _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{0.75+1}{2}=0.875$

Since $f(c _2)f(a _2)=f(0.875)f(0.75)<0$

Therefore set $a _3=a _2,b _3=c _2$

$\textbf{Iteration 4: k=3}$

$c _3=\dfrac{a _3+b _3}{2}=\dfrac{0.75+0.875}{2}=0.8125$

Thus the second,third and fourth approximations are $0.75,0.875,0.8125$ respectively.

Here, $x^3-x^2-1=0$

Here, $x^3-x-1=0$

$0.8642:E:02 \div 0.2562 : E: 02 = ?$

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 $1.23 \times 10^{10}$

$0.8642:E:02 \div 0.2562 : E: 02 = \dfrac{0.8642 \times 10^2}{0.2562 \times 10^2} $

                                                 $=\dfrac{0.8642}{0.2562}$

                                                $=3.371459$

                                                $=0.3371459 \times 10^1$

                                                $=0.33715 : E : 01$

$0.8642:E:02 \div 0.2562 : E: 02 =0.33715 : E : 01$

Here, $x^3-5x-7=0$

We have to find the third approximation of root of the
equation $x^3-x^2-1=0$ in the interval $(1,2)$ using successive
Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{1+2}{2}=1.5$

Since $f(c _0)f(a _0)=f(1.5)f(1)<0$

Therefore set $a _1=a _0,b _1=c _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{1+1.5}{2}=1.25$

Since $f(c _1)f(a _1)=f(1.25)f(1)>0$

Therefore set $a _2=c _1,b _2=b _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{1.25+1.5}{2}=1.375$

Thus the third approximation of the root is $1.375$ respectively.

Function can be written as $f(x)=x^3-2$

$0.4267:E:10\div0.2437 : E :-02=?$

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 $1.23 \times 10^{10}$

$0.4267:E:10\div0.2437 : E :-02=\dfrac{0.4267 \times 10^{10}}{0.2437 \times 10^{-2}}$

                                                       $= 1.75092 \times 10^{12}$

                                                       $= 1.751 \times 10^{12}$

                                                       $= .1751 \times 10^{13}$

                                                       $= 0.1751 : E :13$

Hence $0.4267:E:10\div0.2437 : E :-02=0.1751 : E :13$

Here, $x^3-9x+1=0$

$\dfrac { dr }{ r } =0.05\Rightarrow dr=(0.05)r$