In the previous section, we completed a discussion on calculation of error. In this section, we will see how it can be used to find change in quantity.
The basic details can be demonstrated using an example. It can be written in 4 steps:
1. We know that, volume of a cube can be obtained using the formula: $\rm{V = l^3}$
♦ V is the volume of the cube.
♦ l is the length of side of the cube.
2. Based on the above formula, we can say that, volume is a function of length. That is.,
$\rm{V = f(l) = l^3}$
3. Consider a cube with side 5 cm.
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It's volume will be 53 = 125 cm3.
4. Suppose that, the side is increased by 2%.
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Then the new length of side = 1.02(l) = 1.02(5) = 5.1 cm
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In such a situation, there will be an increase in the volume of the cube.
•
How much increase will be taking place?
Answer can be written in 4 steps:
(i) Original volume =
$\rm{V_1 \,=\, f(l_1) \,=\, {l_1}^3 \,=\, 5^3}$
(ii) New volume =
$\rm{V_2 \,=\, f(l_2) \,=\, {l_2}^3 \,=\, 5.1^3}$
(iii) So the change in volume =
$\rm{V_2 - V_1 \,=\, \Delta V \,=\, 5.1^3 ~-~5^3}$
(iv) But based on differential approximation that we learned in the previous section, we need not find the exact ΔV. we can write:
♦ ΔV ≈ dV (This is possible because, l2 is close to l1)
♦ dV = f'(l1).dl
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Thus we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{dV} & {~=~} &{f'(l_1) . dl} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{ (3 {l_1}^2) . (l_2 - l_1)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{ 3({5}^2) . (5.1 - 5)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{3 (25) (0.1)} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{7.5~\rm{cm^3}} \\
\end{array}$
Let us cross check by calculating the actual change in volume ΔV.
$\rm{\Delta V \,=\, V_2 - V_1 \,=\, 5.1^3 - 5^3 \,=\,132.651 \, - \, 125 \,=\,7.651 ~ {cm}^3}$
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When we use differential approximation, we get 7.5.
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7.5 is approximately equal to 7.651.
Solved example 22.39
Find the approximate change in the volume V of a cube of side x meters caused by increasing the side by 1%.
Solution:
1. We have:
♦ V = f(l) = l3
♦ ΔV ≈ dV (This is possible when l2 is close to l1)
♦ dV = f'(l1).dl
2. Original length = x m
Increased length = 1.01x m
3. Fixing l1 and l2:
l2 must be the value which causes the difficulty. So we put:
♦ l2 = 1.01x cm
♦ l1 = x cm
4. Then we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{dV} & {~=~} &{f'(l_1) . dl} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{ (3 {l_1}^2) . (l_2 - l_1)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{ 3({x}^2) . (1.01x - x)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{3 x^2 (0.01 x)} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{0.03 x^3~\rm{m^3}} \\
\end{array}$
Solved example 22.40
Find the approximate change in the surface area of a cube of side x meters caused by decreasing the side by 1%.
Solution:
1. We have:
♦ S = f(l) = 6l2
♦ ΔS ≈ dS (This is possible when l2 is close to l1)
♦ dS = f'(l1).dl
2. Original length = x m
Decreased length = 0.99x m
3. Fixing l1 and l2:
l2 must be the value which causes the difficulty. So we put:
♦ l2 = 0.99x cm
♦ l1 = x cm
4. Then we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{dV} & {~=~} &{f'(l_1) . dl} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{ 6( 2 l_1) . (l_2 - l_1)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{ 12{x} . (0.99x - x)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{12 x (0.01 x)} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{0.12 x^2~\rm{m^2}} \\
\end{array}$
Solved example 22.41
The approximate change in the volume of a cube of side x meters caused by increasing the side by 3% is
(A) 0.06 x3 m3 (B) 0.6 x3 m3 (C) 0.09 x3 m3 (B) 0.9 x3 m3
Solution:
1. We have:
♦ V = f(l) = l3
♦ ΔV ≈ dV (This is possible when l2 is close to l1)
♦ dV = f'(l1).dl
2. Original length = x m
Increased length = 1.03x m
3. Fixing l1 and l2:
l2 must be the value which causes the difficulty. So we put:
♦ l2 = 1.03x cm
♦ l1 = x cm
4. Then we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{dV} & {~=~} &{f'(l_1) . dl} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{ (3 {l_1}^2) . (l_2 - l_1)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{ 3({x}^2) . (1.03x - x)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{3 x^2 (0.03 x)} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{0.09 x^3~\rm{m^3}} \\
\end{array}$
5. So the correct option is (C)
The link below gives a few more solved examples:
In the next section, we will see Maxima and Minima.
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