Question

Find the approximate change in the volume \[\text{V}\] of a cube of side ‘x’ meters caused by increasing the side by 2%.

Answer 1

Hint: Find the new side by increasing the given side by 2%. Then calculate the volume of the cube for the new side and old side. And hence use the formula $\dfrac{{{\text{V}}_{2}}-{{\text{V}}_{1}}}{{{\text{V}}_{2}}}\times 100$ to calculate the percentage change in volume, where ${{\text{V}}_{\text{1}}}$ is old and ${{\text{V}}_{2}}$ is new volume of solution : the cube.

Complete step-by-step solution -
Here, it is given that the size of the cube is ‘x’ meters with volume ‘V’ and we need to determine the approximate change in volume ‘V’ if the side of the cube is increased by 2% .
Now we know the volume of the cube in terms of side of length ‘x’ meters can be given by \[{{\left( \text{side} \right)}^{3}}\ =\ {{x}^{3}}\]. Hence, we can equate volume V and ${{x}^{3}}$ as both are representing the volumes of the cube. So, we get
$\text{V}\ =\ {{x}^{3}}$………………………………………….…………(i)
Now, we are increasing the side ‘x’ by 2%, we can write the new side of the cube as
New side $=$ previous side $+$$2%$ of previous side
$\Rightarrow $New side of the cube
$=\ x\ +\ 2% o\text{f }x$
$=\ x\ +\ \dfrac{2}{100}\times x$
$=\ \dfrac{x}{1}+\dfrac{2x}{100}$
$=\ \dfrac{100x+2x}{100}\ $…………………………………………………(ii)
Now, we can calculate the volume of the cube with side $\dfrac{102x}{100}\ $. So, let us represent its volume by ${{\text{V}}_{\text{1}}}$. Hence, we get
${{\text{V}}_{\text{1}}}\ =\ {{\left( \text{side} \right)}^{3}}\ =\ {{\left( \dfrac{102x}{100} \right)}^{3}}$………………………………………...(iii)
Now we know the change in volume can be given by taking difference between new volume and old volume of the cube i.e. $\left( {{\text{V}}_{^{\text{1}}}}\ \text{-}\ \text{V} \right)$ . And we can get percentage change in volume by dividing the exact change in volume by old volume (V) and hence multiply it by 100 to get in percentage.
Hence, change in volume can be given by relation: -
\[=\ \dfrac{\text{change in volume}}{\text{old volume}}\ \times \ 100\]
\[=\ \dfrac{{{\text{V}}_{^{\text{1}}}}\ \text{-}\ \text{V}}{\text{V}}\ \times \ 100\]
Now, we can put values of \[{{\text{V}}_{^{\text{1}}}}\]and \[\text{V}\] in the above relation by substituting the values of \[{{\text{V}}_{^{\text{1}}}}\]and \[\text{V}\] from equations (iii) and (i) respectively, we get percentage change in volume as
\[=\ \dfrac{\left( \dfrac{102x}{100} \right){{\ }^{3}}-\ {{x}^{3}}}{{{x}^{3}}}\ \times \ 100\]
\[=\ \dfrac{{{\left( \dfrac{102}{100} \right)}^{3}}x{{\ }^{3}}-\ {{x}^{3}}}{{{x}^{3}}}\ \times \ 100\]
Now taking ‘\[{{x}^{3}}\]’ common from numerator and cancelling out by denominator, we get\[=\ \left( \dfrac{{{\left( 102 \right)}^{3}}}{{{\left( 100 \right)}^{3}}}-\dfrac{1}{1} \right)\times \ 100\]
\[=\ \left( \dfrac{{{\left( 102 \right)}^{3}}-\ {{\left( 100 \right)}^{3}}}{100\ \times 100\ \times 100} \right)\times \ 100\]
\[=\dfrac{{{\left( 100\ +\ 2 \right)}^{3}}-\ {{\left( 100 \right)}^{3}}}{100\ \times 100\ }\]………………………………………. (iv)
Now, use the identity of \[{{\left( a+\ b \right)}^{3}}\] to simplify the term \[{{\left( 100\ +\ 2 \right)}^{3}}\]which can be given as \[{{\left( a+\ b \right)}^{3}}\ =\ {{a}^{3}}\ +\ {{b}^{3}}+\ 3ab\left( a+b \right)\]
Hence, we can get
\[{{\left( 100+2 \right)}^{3}}\ =\ {{100}^{3}}\ +\ {{2}^{3}}+\ 3\times 100\times 2\left( 100+2 \right)\]
Now, we can put the above value of \[{{\left( 102 \right)}^{3}}\] in the expression (iv). Hence, we get
% change in volume
\[=\dfrac{{{100}^{3}}\ +\ {{2}^{3}}+\ 3\times 100\times 2\left( 100+2 \right)\ -\ {{\left( 100 \right)}^{3}}}{100\ \times 100\ }\]
\[=\dfrac{8\ +\ 600\times 102}{100\ \times 100\ }\]
\[=\dfrac{8\ +\ 61200}{100\ \times 100\ }\]
\[=\]\[\dfrac{61208}{100\times 100}\]
\[=\ 6.1208%\]
Now, the above value of \[%\]change in volume is the exact value, so, we can approximate \[6.1208%\]to \[6%\], as we need to determine exact change in volume. Hence,
Change in volume \[=\ 6.1208%\ \cong \ 6%\]
So, the approximation change in volume is \[6%\].

Note: As the change in volume is very less. So, we can use following approach for these kind of question as
We have volume of cube as
$\text{V}\ =\ {{x}^{3}}$
Take log to both sides
$\log \text{V}\ \text{=}\ \text{log }{{x}^{3}}=\ 3\log x$
Now, differentiate with respect to $x$
$\dfrac{1}{\text{V}}\dfrac{d\text{V}}{dx}\ =\ \dfrac{3}{x}$
$\Rightarrow \ \dfrac{d\text{V}}{\text{V}}\ =\ 3\dfrac{dx}{x}$
Now, we know that $d\text{V}$and $dx$ are very less values and represent the changes in volume and side. Hence, we can multiply by $100$ to both sides and get
$\dfrac{d\text{V}}{\text{V}}\times 100\ =\ 3\times \dfrac{dx}{x}\times 100$
Now, we know change in side is $2%$,
So, we can replace $\dfrac{dx}{x}\times 100$ by $2$and hence, get
$\dfrac{d\text{V}}{\text{V}}\times 100\ =\ 3\times 2$
$=\ 6%$
Using ${{\left( a+b \right)}^{3}}$ for expanding ${{102}^{3}}$ is the key point of the question. One may calculate ${{102}^{3}}$ by solving $102\times 102\times 102$ but that would be more complex. We can always use the above-mentioned approach for (less change) these kinds of questions and can get answers easily.