22.11 - Change In Quantity

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.
• It's volume will be 5³ = 125 cm³.
4. Suppose that, the side is increased by 2%.
• Then the new length of side = 1.02(l) = 1.02(5) = 5.1 cm
• 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
• Thus we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{dV}    & {~=~}    &{f'(l_1) . dl}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{ (3 {l_1}^2) . (l_2 - l_1)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{ 3({5}^2) . (5.1 - 5)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{3 (25) (0.1)}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{7.5~\rm{cm^3}}    \\
\end{array}$

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Let us cross check by calculating the actual change in volume ΔV.
$\rm{\Delta V \,=\, V_2 - V_1 \,=\, 5.1^3 - 5^3 \,=\,132.651 \, - \, 125 \,=\,7.651 ~ {cm}^3}$

• When we use differential approximation, we get 7.5.
• 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)

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The link below gives a few more solved examples:

Exercise 22.4

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In the next section, we will see Maxima and Minima.

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22.10 - Calculating The Amount of Error

In the previous section, we completed a discussion on differential approximation. In this section, we will see how it can be used to find amount of error.

The basic details can be demonstrated using an example. It can be written in 6 steps:
1. We know that, volume of a sphere can be obtained using the formula: $\rm{V = \frac{4}{3} \pi r^3}$
   ♦ V is the volume of the sphere.
   ♦ r is the radius of the sphere.
2. Based on the above formula, we can say that, volume is a function of radius. That is.,
$\rm{V = f(r) = \frac{4}{3} \pi r^3}$
3. Suppose that, the radius of a sphere is measured to be 5 cm with an error of 0.1 cm.
• Then the actual radius r will be such that:
4.9 ≤ r ≤ 5.1
4. Consider the situation:
   ♦ The actual radius is 5.1 cm.
   ♦ We use 5 cm for calculating the volume.
• In such a situation, there will be error in the calculated volume.
• How much error will be present in the calculated volume?
Answer can be written in 4 steps:
(i) Calculated volume =
$\rm{V_1 \,=\, f(r_1) \,=\, \frac{4}{3} \pi {r_1}^3 \,=\, \frac{4}{3} \pi (5^3)}$
(ii) Actual volume =
$\rm{V_2 \,=\, f(r_2) \,=\, \frac{4}{3} \pi {r_2}^3 \,=\, \frac{4}{3} \pi ({5.1}^3)}$
(iii) So the error in the calculated volume =
$\rm{V_2 - V_1 \,=\, \Delta V \,=\, \frac{4}{3} \pi ({5.1}^3) ~-~\frac{4}{3} \pi ({5}^3)}$
(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, r₂ is close to r₁)
   ♦ dV = f'(r₁).dr
• Thus we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{dV}    & {~=~}    &{f'(r_1) . dr}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{4}{3} \pi (3 {r_1}^2) . (r_2 - r_1)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{4 \pi  (5^2) . (5.1 - 5)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{4 \pi  (25) . (0.1)}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{10 \pi~\rm{cm^3}}    \\
\end{array}$
5. Consider the situation:
   ♦ The actual radius is 4.9 cm.
   ♦ We use 5 cm for calculating the volume.
• In such a situation also, there will be error in the calculated volume.
• How much error will be present in the calculated volume?
Answer can be written in 4 steps:
(i) Calculated volume =
$\rm{V_1 \,=\, f(r_1) \,=\, \frac{4}{3} \pi {r_1}^3 \,=\, \frac{4}{3} \pi (5^3)}$
(ii) Actual volume =
$\rm{V_2 \,=\, f(r_2) \,=\, \frac{4}{3} \pi {r_2}^3 \,=\, \frac{4}{3} \pi ({4.9}^3)}$
(iii) So the error in the calculated volume =
$\rm{V_2 - V_1 \,=\, \Delta V \,=\, \frac{4}{3} \pi ({4.9}^3) ~-~\frac{4}{3} \pi ({5}^3)}$
(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, r₂ is close to r₁)
   ♦ dV = f'(r₁).dr
• Thus we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{dV}    & {~=~}    &{f'(r_1) . dr}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{4}{3} \pi (3 {r_1}^2) . (r_2 - r_1)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{4 \pi  (5^2) . (4.9 - 5)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{4 \pi  (25) . (-0.1)}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{-10 \pi~\rm{cm^3}}    \\
\end{array}$
6. So we can write:
The error (dV) in calculated volume will be such that:
−10π ≤ dV ≤ 10π.

---

Let us cross check by calculating ΔV
• Consider the situation:
   ♦ The actual radius is 5.1 cm.
   ♦ We use 5 cm for calculating the volume.
Then we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\Delta V}    & {~=~}    &{V_2 - V_1}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{4}{3} \pi ({5.1}^3)~-~\frac{4}{3} \pi ({5}^3)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{10.2013333333333 \pi ~\rm{cm^3}}    \\
\end{array}$

• Consider the situation:
   ♦ The actual radius is 5.1 cm.
   ♦ We use 5 cm for calculating the volume.
Then we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\Delta V}    & {~=~}    &{V_2 - V_1}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{4}{3} \pi ({4.9}^3)~-~\frac{4}{3} \pi ({5}^3)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{-9.80133333333329 \pi ~\rm{cm^3}}    \\
\end{array}$

• When we use differential approximation, we get 10π and −10π.
• 10π is approximately equal to 10.2013 π.
• −10π is approximately equal to −9.8013333 π.

Solved example 22.37
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. 
Solution:
1. The actual radius r will be such that:
8.97 ≤ r ≤ 9.03
2. Consider the situation:
   ♦ The actual radius is 9.03 cm.
   ♦ We use 9 cm for calculating the volume.
3. First we fix r₁ and r₂:
r₂ must be the value which causes the difficulty. So we put:
   ♦ r₂ = 9.03 cm
   ♦ r₁ = 9.0 cm
4. Then we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{dV}    & {~=~}    &{f'(r_1) . dr}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{4}{3} \pi (3 {r_1}^2) . (r_2 - r_1)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{4 \pi  (9^2) . (9.03 - 9)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{4 \pi  (81) . (0.03)}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{9.72 \pi~\rm{cm^3}}    \\
\end{array}$
5. Consider the situation:
   ♦ The actual radius is 8.97 cm.
   ♦ We use 9 cm for calculating the volume.
• Here we get the same result with opposite sign because dx is the same value but with opposite sign.
• That means, dV in this case is −9.72π cm³.
6. We can write:
The error (dV) in calculated volume will be such that:
−9.72π ≤ dV ≤ 9.72π.

Solved example 22.38
If the length of a cube is measured as 6 cm with an error of 0.2 cm, then find the approximate error in calculating its volume. 
Solution:
1. The actual length (l) will be such that:
5.8 ≤ l ≤ 6.2
2. Consider the situation:
   ♦ The actual length is 6.2 cm.
   ♦ We use 6 cm for calculating the volume.
3. First we fix l₁ and l₂:
l₂ must be the value which causes the difficulty. So we put:
   ♦ l₂ = 6.2 cm
   ♦ l₁ = 6.0 cm
4. Then we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{dV}    & {~=~}    &{f'(l_1) . dl}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{ (3 {l_1}^2) . (l_2 - l_1)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{3  (6^2) . (6.2 - 6)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{3 (36) . (0.2)}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{21.6~\rm{cm^3}}    \\
\end{array}$
5. Consider the situation:
   ♦ The actual length is 5.8 cm.
   ♦ We use 6 cm for calculating the volume.
• Here we get the same result with opposite sign because dx is the same value but with opposite sign.
• That means, dV in this case is −21.6 cm³.
6. We can write:
The error (dV) in calculated volume will be such that:
−21.6 ≤ dV ≤ 21.6.

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In the next section, we will see change in quantity.

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22.9 - Solved Examples on Differentials

In the previous section, we saw the details about differentials. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 22.34
Using differentials, find the approximate value of $\rm{\sqrt[3]{25}}$
Solution:
1. Let $\rm{f(x) = \sqrt[3]{x}}$
• Then we want $\rm{f(25) = \sqrt[3]{25}}$
2. First we must fix x₁ and x₂
• x₂ should be taken as the value which causes difficulty. So we can put x₂ = 25
• x₁ should be selected in such a way that:
   ♦ It is a convenient number
   ♦ It is close to x₂.
• We can take x₁ = 27
The cube root of 27 is already known. So it is a convenient number. Also, 27 is close to 25
3. Thus we have: x₁ = 27 and x₂ = 25
Then dx = (25 − 27) = −2
4. We have:
Δy = [f(x₂) − f(x₁)] = [f(25) − f(27)]
⇒ f(25) = Δy + f(27)
5. Instead of finding Δy, we can find dy. This is because, x₂ is close to x₁ and so Δy ≈ dy.
• Then we can write: f(25) ≈ dy + f(27)
6. We can easily calculate dy as:
dy = f'(27).dx
= [(1/3)(x)^-2/3]_x=27 (−2)
= [(1/3)(27)^-2/3] (−2)
= [(1/3)(3)^-2] (−2)
= [1/27] (−2)
= −0.07407407407
7. So from (5), we get:
⇒ f(25) ≈ dy + f(27)
≈ −0.07407407407 + [\rm{\sqrt[3]{27}]
≈ −0.07407407407 + [3]
≈ 2.92592592593

This is the same result that we obtained using linear approximation method. See solved example 22.24 of section 22.7.

Solved example 22.35
Using differentials, find the approximate value of cos 89^o.
Solution:
• 89^o = $\frac{89 \pi}{180}$ radians
• 90^o = $\frac{90 \pi}{180}~=~\frac{\pi}{2}$ radians
1. Let $\rm{f(x) = \cos x}$
• Then we want $\rm{f(\frac{89 \pi}{180}) = \cos \frac{89 \pi}{180}}$
2. First we must fix x₁ and x₂
• x₂ should be taken as the value which causes difficulty. So we can put x₂ = $\frac{89 \pi}{180}$
• x₁ should be selected in such a way that:
   ♦ It is a convenient number
   ♦ It is close to x₂.
• We can take x₁ = $\frac{90 \pi}{180}$
The cosine of this angle is already known. So it is a convenient number. Also, 90 is close to 89
3. Thus we have: x₁ = $\frac{90 \pi}{180}$ and x₂ = $\frac{89 \pi}{180}$
Then dx = x₂ − x₁ = $\frac{- \pi}{180}$
4. We have:
Δy = [f(x₂) − f(x₁)] = $\rm{\left[f(\frac{89 \pi}{180}) - f(\frac{90 \pi}{180})\right]}$
⇒ $\rm{f(\frac{89 \pi}{180})~=~\Delta y ~+~f(\frac{90 \pi}{180})}$
5. Instead of finding Δy, we can find dy. This is because, x₂ is close to x₁ and so Δy ≈ dy.
• Then we can write:
$\rm{f(\frac{89 \pi}{180})~≈~ dy ~+~f(\frac{90 \pi}{180})}$
⇒ $\rm{f(\frac{89 \pi}{180})~≈~ dy ~+~\cos (\frac{ 90 \pi}{180})}$
⇒ $\rm{f(\frac{89 \pi}{180})~≈~ dy ~+~\cos (\frac{\pi}{2})}$
⇒ $\rm{f(\frac{89 \pi}{180})~≈~ dy ~+~ 0}$
6. We can easily calculate dy as:
$\rm{dy ~=~\left[f'(\frac{90 \pi}{180})\right] dx ~=~\left[f'(\frac{\pi}{2})\right] dx}$
$\rm{=~\left[(- \sin x)_{x = \pi/2} \right] dx}$
$\rm{=~\left[- \sin (\pi/2) \right] dx}$
$\rm{=~\left[- \sin (\pi/2) \right] (-\pi / 180)}$
$\rm{=~\left[- 1 \right] (-3.14 / 180)}$
= 0.01745329
7. So from (5), we get:
$\rm{f(\frac{89 \pi}{180})~≈~ dy ~+~ 0}$
⇒ $\rm{f(\frac{89 \pi}{180})~≈~ 0.01745329}$

This is the same result that we obtained using linear approximation method. See solved example 22.27 of section 22.7.

Solved example 22.36
Using differentials, find the approximate value of (0.999)^1/10.
Solution:
1. Let $\rm{f(x) = (1-x)^{1/10}}$
• Then we want $\rm{f(0.001)}$
• This is because, $\rm{f(0.001) = (1-x)^{1/10} = (0.999)^{1/10}}$
2. First we must fix x₁ and x₂
• x₂ should be taken as the value which causes difficulty. So we can put x₂ = 0.001
• x₁ should be selected in such a way that:
   ♦ It is a convenient number
   ♦ It is close to x₂.
• We can take x₁ = 0
zero makes the calculations easier. So it is a convenient number. Also, 0 is close to 0.001
3. Thus we have: x₁ = 0 and x₂ = 0.001
Then dx = (0.001 − 0) = 0.001
4. We have:
Δy = [f(x₂) − f(x₁)] = [f(0.001) − f(0)]
⇒ f(0.001) = Δy + f(0)
5. Instead of finding Δy, we can find dy. This is because, x₂ is close to x₁ and so Δy ≈ dy.
• Then we can write: f(0.001) ≈ dy + f(0)
6. We can easily calculate dy as:
dy = f'(0).dx
= $\rm{\left[\frac{1}{10} (1-x)^{-9/10} (-1) \right]_{x = 0} dx}$
= $\rm{\left[\frac{-1}{10} (1-x)^{-9/10} \right]_{x = 0} (0.001)}$
= $\rm{\left[\frac{-1}{10} (1)^{-9/10} \right] (0.001)}$
= $\rm{\left[\frac{-1}{10} (1) \right] (0.001)}$
= $\rm{\left[\frac{-1}{10} \right] (0.001)}$
= $\rm{-0.0001}$
7. So from (5), we get:
f(0.001) ≈ dy + f(0)
≈ −0.0001 + (1 − 0)^1/10.
≈ −0.0001 + (1)^1/10.
≈ −0.0001 + 1.
≈ 0.9999.

This is the same result that we obtained using linear approximation method. See solved example 22.29 of section 22.7.

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22.8 - Differentials

In the previous section, we completed a discussion on linear approximation. In this section, we will see differentials.

The basic details about differentials can be written in 14 steps:
1. We are given a function f.
• Assume that:
   ♦ When the input for this function is x₁, the output is y₁
   ♦ When the input for this function is x₂, the output is y₂
2. Let us write the changes in input and output:
• When the input changes from x₁ to x₂,
   ♦ change in input = Δx = x₂ − x₁
• When the output changes from y₁ to y₂,
   ♦ change in output = Δy = y₂ − y₁
3. Often in scientific and engineering problems, we are more interested in Δy, rather than the individual outputs y₁ and y₂.
• So we need to develop a method which will help us to easily find Δy.
4. We know that:
y₂ = f(x₂) and y₁ = f(x₁)
• So Δy = (y₂ − y₁) = f(x₂) − f(x₁)
5. But from (2), we have: x₂ = x₁ + Δx
• So the result in (4) can be written as:
Δy = f(x₂) − f(x₁) = f(x₁ + Δx) − f(x₁)
6. If x₂ is close to x₁, the quantity Δx will be small. Then we can calculate f(x₁ + Δx) using L(x₁+Δx).
This is possible because, f(x₁ + Δx) ≈ L(x₁+Δx)
• Recall the solved example 22.25 of the previous section, in which we calculated f(3.02) using L(3.02)_a=3. In that example,
   ♦ x₁ is 3
   ♦ x₂ is 3.02
   ♦ Δx is 0.02
7. So our next task is to find L(x)_x1. It can be done in 4 steps:
(i) The slope at x₁ is f'(x₁)
(ii) The y-coordinate at x₁ is f(x₁)
(iii) So the tangent at x₁ can be obtained as:
y − f(x₁) = f'(x₁) (x − x₁)
⇒ y = f'(x₁) (x − x₁) + f(x₁)
(iv) Thus we get:
L(x)_x1 = f'(x₁) (x − x₁) + f(x₁)
8. Since (x₁ + Δx) is close to x₁, we get:
f(x₁ + Δx) ≈ [L(x₁ +  Δx)_x1]
≈ f'(x₁) (x₁ +  Δx − x₁) + f(x₁)
≈ f'(x₁) (Δx) + f(x₁)
• That is., f(x₁ + Δx) ≈ f'(x₁) (Δx) + f(x₁)
• Also, since (x₁ + Δx) is close to x₁, the quantity Δx is small, and so Δx can be replaced by dx.
• So we get:
f(x₁ + dx) ≈ f'(x₁) (dx) + f(x₁)
9. Now the result in (5) becomes:
Δy = [f(x₁ + dx)] − f(x₁)
≈ [f'(x₁) (dx) + f(x₁)] − f(x₁)
≈ f'(x₁) (dx)
• That is., Δy ≈ f'(x₁) (dx)
10. Now consider the familiar equation:
$\rm{\frac{dy}{dx}~=~f'(x)}$
• Multiplying both sides by dx, we get:
dy = f'(x) dx
11. So the result in 9 becomes:
Δy ≈ [f'(x₁) (dx) = dy]
• That.,
   ♦ Δy ≈ f'(x₁) (dx)
   ♦ Δy ≈ dy
• We can write:
The change in output (denoted by Δy), is approximately equal to dy. The quantity dy, can be calculated by multiplying the 'derivative at x₁' by dx.
12. The fig.22.24 below clearly shows the difference between Δy and dy.

Fig.22.25

   ♦ P and Q are points on f
   ♦ S is a point on L
• The actual change in output is QR. But instead of QR, we find SR.
• This is because, Q and S are very close to each other. Also, SR (which is dy), can be calculated easily by multiplying the 'derivative at x₁' by dx.
13. For calculating dy, we use the equation:
dy = f'(x₁) (dx)
• In this equation, dy and dx are called differentials.
14. The method which involves the use of dy instead of Δy, is known as differential approximation.

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Let us see some solved examples

Solved example 22.30
Using differentials, find the approximate value of f(3.02) where f(x) = 3x² + 5x + 3
Solution:
1. First we must fix x₁ and x₂
• x₂ should be taken as the value which causes difficulty. So we can put x₂ = 3.02
• x₁ should be selected in such a way that:
   ♦ It is a convenient number
   ♦ It is close to x₂.
• We can take x₁ = 3
For the given function, calculations with '3' is easy. So it is a convenient number. Also, 3 is close to 3.02
2. Thus we have: x₁ = 3 and x₂ = 3.02
Then dx = (3.02 − 3) = 0.02
3. We have:
Δy = [f(x₂) − f(x₁)] = [f(3.02) − f(3)]
⇒ f(3.02) = Δy + f(3)
4. Instead of finding Δy, we can find dy. This is because, x₂ is close to x₁ and so Δy ≈ dy.
• Then we can write: f(3.02) ≈ dy + f(3)
5. We can easily calculate dy as:
dy = f'(3).dx
= (6x + 5)_x=3 (0.02)
= (23)(0.02) = 0.46
6. So from (4), we get:
f(3.02) ≈ dy + f(3)
≈ 0.46 + [3(3)² + 5(3) + 3]
≈ 0.46 + [27 + 15 + 3]
≈ 0.46 + [45]
≈ 45.46

This is the same result that we obtained using linear approximation method. See solved example 22.25 of the previous section. 

Solved example 22.31
Using differentials, find the approximate value of $\rm{\sqrt{36.6}}$
Solution:
1. Let $\rm{f(x) = \sqrt{x}}$
• Then we want $\rm{f(36.6) = \sqrt{36.6}}$
2. First we must fix x₁ and x₂
• x₂ should be taken as the value which causes difficulty. So we can put x₂ = 36.6
• x₁ should be selected in such a way that:
   ♦ It is a convenient number
   ♦ It is close to x₂.
• We can take x₁ = 36
The square root of 36 is already known. So it is a convenient number. Also, 36 is close to 36.6
3. Thus we have: x₁ = 36 and x₂ = 36.6
Then dx = (36.6 − 36) = 0.6
4. We have:
Δy = [f(x₂) − f(x₁)] = [f(36.6) − f(36)]
⇒ f(36.6) = Δy + f(36)
5. Instead of finding Δy, we can find dy. This is because, x₂ is close to x₁ and so Δy ≈ dy.
• Then we can write: f(36.6) ≈ dy + f(36)
6. We can easily calculate dy as:
dy = f'(36).dx
= [(1/2)(x)^-1/2]_x=36 (0.6)
= [(1/2)(36)^-1/2] (0.6)
= [(1/2)(6)^-1] (0.6)
= [1/12] (0.6)
= 0.05
7. So from (5), we get:
f(36.6) ≈ dy + f(36)
≈ 0.05 + [√36]
≈ 0.05 + [6]
≈ 6.05

This is the same result that we obtained using linear approximation method. See solved example 22.21 of section 22.6. 

Solved example 22.32
Using differentials, find the approximate value of $\rm{\sqrt{9.1}}$
Solution:
1. Let $\rm{f(x) = \sqrt{x}}$
• Then we want $\rm{f(9.1) = \sqrt{9.1}}$
2. First we must fix x₁ and x₂
• x₂ should be taken as the value which causes difficulty. So we can put x₂ = 9.1
• x₁ should be selected in such a way that:
   ♦ It is a convenient number
   ♦ It is close to x₂.
• We can take x₁ = 9
The square root of 9 is already known. So it is a convenient number. Also, 9 is close to 9.1
3. Thus we have: x₁ = 9 and x₂ = 9.1
Then dx = (9.1 − 9) = 0.1
4. We have:
Δy = [f(x₂) − f(x₁)] = [f(9.1) − f(9)]
⇒ f(9.1) = Δy + f(9)
5. Instead of finding Δy, we can find dy. This is because, x₂ is close to x₁ and so Δy ≈ dy.
• Then we can write: f(9.1) ≈ dy + f(9)
6. We can easily calculate dy as:
dy = f'(9).dx
= [(1/2)(x)^-1/2]_x=9 (0.1)
= [(1/2)(9)^-1/2] (0.1)
= [(1/2)(3)^-1] (0.1)
= [1/6] (0.1)
= 0.016667
7. So from (5), we get:
f(9.1) ≈ dy + f(9)
≈ 0.01667 + [√9]
≈ 0.01667 + [3]
≈ 3.01667

This is the same result that we obtained using linear approximation method. See solved example 22.22 of section 22.6. 

Solved example 22.33
Using differentials, find the approximate value of $\rm{\sqrt[3]{8.1}}$
Solution:
1. Let $\rm{f(x) = \sqrt[3]{x}}$
• Then we want $\rm{f(8.1) = \sqrt[3]{8.1}}$
2. First we must fix x₁ and x₂
• x₂ should be taken as the value which causes difficulty. So we can put x₂ = 8.1
• x₁ should be selected in such a way that:
   ♦ It is a convenient number
   ♦ It is close to x₂.
• We can take x₁ = 8
The cube root of 8 is already known. So it is a convenient number. Also, 8 is close to 8.1
3. Thus we have: x₁ = 8 and x₂ = 8.1
Then dx = (8.1 − 8) = 0.1
4. We have:
Δy = [f(x₂) − f(x₁)] = [f(8.1) − f(8)]
⇒ f(8.1) = Δy + f(8)
5. Instead of finding Δy, we can find dy. This is because, x₂ is close to x₁ and so Δy ≈ dy.
• Then we can write: f(8.1) ≈ dy + f(8)
6. We can easily calculate dy as:
dy = f'(8).dx
= [(1/3)(x)^-2/3]_x=8 (0.1)
= [(1/3)(8)^-2/3] (0.1)
= [(1/3)(2)^-2] (0.1)
= [1/12] (0.1)
= 0.0083333
7. So from (5), we get:
f(8.1) ≈ dy + f(8)
≈ 0.008333 + [$\rm{\sqrt[3]{8}}$]
≈ 0.008333 + [2]
≈ 2.008333

This is the same result that we obtained using linear approximation method. See solved example 22.23 of section 22.6.

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In the next section, we will see a few more solved examples.

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