22.12 - Maxima And Minima

In the previous section, we saw change in quantity. In this section, we will see maxima and minima.

Some basics can be written in 4 steps:
1. Consider the function f(x) = sin x.
• We can give any real number as the input. The maximum possible output is 1.
    ♦ We will not get an output greater than 1
• We can give any real number as the input. The minimum possible output is −1.
    ♦ We will not get an output lesser than −1
• We have seen this fact in our trigonometry classes.
• Also this fact can be visually seen in the graph of the sine function.
2. Consider another function f(x) = 3x² + 5x + 7.
• It’s graph is shown in fig.22.26 below:

Fig.22.26

• We see that:
    ♦ when x increases, the graph goes up to +∞.
    ♦ when x decreases, then also the graph goes up to +∞.
• So there is no maximum value for this function.
3. Let us try to find the minimum value.
• From the graph, we get a feeling that, −4 is the minimum value.
• But when a horizontal line (shown in yellow color) is drawn through (0,−4), it cuts the graph at two points A and B.
• A portion of the graph is below the yellow line. That means, −4 is not the minimum value.
4. Often in science, engineering and business problems, we will want to find the exact maximum and minimum values of various functions. Derivatives can help us to achieve this goal. In this section, we will first become familiar with absolute extrema.

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Some basics about absolute extrema can be written in 5 steps:
1. Consider the function f(x) = x² + 2.
The graph is shown in fig.22.27 below:

Fig.22.27

2. We see that:
    ♦ when x increases, the graph goes up to +∞.
    ♦ when x decreases, then also the graph goes up to +∞.
• So there is no maximum value for this function.
3. Let us try to find the minimum value.
From the graph, we get a feeling that, 2 is the minimum value. When a horizontal line (shown in yellow color) is drawn through (0,2), it just touches the graph at a point. No portion of the graph is below the yellow line. That means, 2 is indeed the minimum value.
4. Analytically also, we can prove that 2 is the minimum value. It can be done in 3 steps:
(i) In the given function, there are two terms: x² and 2
    ♦ x² will be always +ve
    ♦ 2 is +ve
    ♦ The two terms are being added together
(ii) So the value of the function will be always greater than or equal to 2.
(iii) Therefore, the minimum value is 2, which occurs when x = 0
5. In this situation we say two points:
(i) The function f(x) = x² + 2 has an absolute minimum.
    ♦ The absolute minimum is 2
    ♦ The absolute minimum occurs at x = 0
(ii) The function f(x) = x² + 2 does not have an absolute maximum.

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Now we can define absolute extrema. It can be written in 7 steps:
1. f is a function defined over an interval I.
2. c is an element in I. That is., c∈I
3. We take each element of I and use it as the input for f.
And we find that each output is less than f(c).
Symbolically, we write this as: f(c) ≥ f(x)  for all x∈I
4. We take each element of I and use it as the input for f.
And we find that each output is greater than f(c).
Symbolically, we write this as: f(c) ≤ f(x)  for all x∈I
5. If the situation in (3) occur, we say that:
f has an absolute maximum on I at c.
6. If the situation in (4) occur, we say that:
f has an absolute minimum on I at c.
7. If either of (3) or (4) occur, we say that:
f has an absolute extremum on I at c.
(extremum is the singular of extrema)

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Extreme Value Theorem
This can be explained in 3 steps:
1. When we are given a function f, it may fall into any one of the following four categories:
(i) f has an absolute maximum but no absolute minimum.
(ii) f has an absolute minimum but no absolute maximum.
(iii) f has an absolute maximum and an absolute minimum.
(iv) f has neither absolute maximum nor absolute minimum.
2. But if two conditions are satisfied, we can guarantee that, there will be an absolute maximum and there will be an absolute minimum.
Condition 1:
• The interval under consideration, must be closed and bounded.
Explanation for "closed and bounded: We know that, a closed interval has square brackets on both ends. That is, []. For an interval to be bounded, neither of the end points should be infinity. So for condition 1, the interval must be in the form [a,b].
Condition 2:
• f must be continuous over [a,b]
3. The guarantee given by the two conditions is known as extreme value theorem.

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Local extrema
We have seen absolute extrema. Now we will see local extrema. It can be explained in 7 steps:
1. In fig.22.28 below, the function f is defined on (−∞,∞).

Fig.22.28

• The function f has:
   ♦ A peak point at x = a
   ♦ A valley point at x = b
   ♦ A peak point at x = c
2. It is clear that, f does not have an absolute minimum. This is because:
   ♦ As x decreases, the value of f approaches −∞
   ♦ As x increases, then also, the value of f approaches −∞
3. It is clear that, the peak at x = c, represents the absolute maximum.
4. The peak at x = a, also has significance.
• The value f(a) is larger than the other values in the immediate vicinity.
• This fact can be expressed mathematically in the form of three conditions:
(i) There is a positive number 'h' such that, we can form the interval (a−h,a+h) on the x-axis.
(ii) Take each input x value from this interval. Write the corresponding f(x) value for each of those inputs.
(iii) All those f(x) values will be less than f(a).
5. If we can find a number 'h' which satisfies the three conditions above, then we say that, x =a is a local maximum.
• So the function has an absolute maximum and a local maximum.
• But the absolute maximum at x = c satisfies the three conditions. So it is a local maximum also.
• We can write:
The function has two local maxima. The local maxima at x = c happens to be the absolute maximum also.
6. Similarly, x = b is a local minimum.
• So the function does not have an absolute minimum. But it has a local minimum.
7. Local maximum is also called relative maximum.
• Similarly, local minimum is also called relative minimum.

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Critical points
This can be explained in 6 steps:
1. On many occasions, graph of the function may not be available. Even if we have the graph, it may not be easy to point out the absolute and local extrema. For example, in the fig.22.26 that we saw at the beginning of this section, there are several points below the horizontal yellow line. We will be able to point out the absolute minimum, only if we zoom in.
2. So it is clear that, we need an analytical method to find the extrema. In this regard, we must first become familiar with critical points.
3. Consider the graph in fig.22.28 above.
• Local extrema occur at x=a, x=b and x=c. We can write two important facts about these three points.
(i) At x=a and x=b, the derivative f'(x) is zero. (Note that, the tangents at these points are horizontal. So slope of tangents at these points is zero)
(ii) At x=c, the derivative does not exist. (At corner points, derivative does not exist)
4. Consider all possible input values of a function (ie., all elements of the domain). Based on the two facts written in (3), we can form two categories:
Category I:
The input values at which derivative is zero, will fall in this category.
Category II:
The input values at which derivative does not exist, will fall in this category.
5. For a given function, if an input x value, falls in category I OR category II, then that x value is a critical point.
• So for the graph in fig.22.28 above, x=a, a=b and x=c are critical points.
6. The above step (5) gives us an effective way to find critical points. But we must keep in mind that, all critical points do not represent extrema. Let us see an example:
• Fig.22.29 below shows the graph of f(x) = x³.
The derivative is given by: f'(x) = 3x²
So at x = 0, the derivative is zero. (tangent is horizontal)
Since the derivative is zero, it is a critical point. But it is not an extremum.

Fig.22.29

• Let us see another example.
Fig.22.30 below shows the graph of $\rm{f(x) \,=\,2(x)^{5/3} \,-\,x^{2/3}}$.

Fig.22.30

At x = 0, we cannot draw the tangent because it is a cusp point. Remember that, we cannot draw tangents at corner points and cusp points.
So at x = 0, the derivative does not exist.
Since the derivative does not exist, it is a critical point. But it is not an extremum.

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Now we will see a solved example on critical points
Solved Example 22.42
Find all critical points for the function f(x) = x³ − 3x + 3
Solution:
1. We have f'(x) = 3x² − 3 = 3(x² − 1)
2. Equating f'(x) to zero, we get:
3(x² − 1) = 0
⇒ (x² − 1) = 0
⇒ x = +1 or x = −1
4. So the points in category I are: x = 1 and x = −1
5. We obtained f'(x) = 3(x² − 1)
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
6. So the critical points are: x = 1 and x = −1
7. Fig.22.31 shows the graph:
 

Fig.22.31

• From the graph, it is clear that:
   ♦ x=1 is a critical point. And it is the absolute minimum.
   ♦ x=−1 is a critical point. And it is the absolute maximum.

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

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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.

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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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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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In the next section, we will see amount of error.

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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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22.7 - Solved Examples on Linear Approximation

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

Solved example 22.24
Write the linear approximation of f(x) = x^1/3 and use it to estimate 25^1/3.
Solution:
1. Given $\rm{f(x) = x^{1/3}}$
• We are asked to write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is $f'(x) ~=~\rm{\frac{1}{3} (x)^{-2/3}}$  
• So the slope at 'a' is $\rm{f'(a) ~=~\frac{1}{3} (a)^{-2/3}}$
3. At the point where x = a, the y coordinate can be calculated as follows:
$\rm{y = x^{1/3}= a^{1/3}}$
4. So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - a^{1/3}}    & {~=~}    &{\frac{1}{3} a^{-2/3} (x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{1}{3} a^{-2/3} (x - a)~+~a^{1/3}}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = \frac{1}{3} a^{-2/3} (x - a)~+~a^{1/3}$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 25
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 27
Cube root of 27 is already known. So it is a convenient number. Also, 27 is close to 25
7. Based on the result in (5), we can write:
Linear approximation of f at x = 27 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{\frac{1}{3} a^{-2/3} (x - a)~+~a^{1/3}}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=27}}    & {~=~}    &{\frac{1}{3} (27^{-2/3}) (x - 27)~+~27^{1/3}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{3} (3^{-2}) (x - 27)~+~3}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{(3^{-3}) (x - 27)~+~3}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(25)}    & {~≈~}    &{L(25)}    \\
{~\color{magenta}    3    }    &{\implies}    &{25^{1/3}}    & {~≈~}    &{(3^{-3}) (25 - 27)~+~3}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~≈~}    &{\frac{-2}{27}~+~3}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~≈~}    &{2.9259259}    \\
\end{array}$
9. It is interesting to note that, if we use a calculator or computer to calculate the cube root of 25, we will get:
$25^{1/3}~=~2.9240177$

Solved example 22.25
Find the approximate value of f(3.02) where
f(x) = 3x² + 5x + 3.
Solution:
1. Given f(x) = 3x² + 5x + 3
• We will first write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is f'(x) = 6x + 5
• So the slope at 'a' is f'(a) = 6a + 5
3. At the point where x = a, the y coordinate can be calculated as follows:
y = 3x² + 5x + 3 = 3a² + 5a + 3
4. So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - (3a^2 + 5a + 3)}    & {~=~}    &{(6a+5) (x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{(6a+5) (x - a)~+~(3a^2 + 5a + 3)}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = (6a+5) (x - a)~+~(3a^2 + 5a + 3)$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 3.02
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 3
'3' is a whole number which makes the calculations easy. So it is a convenient number. Also, 3 is close to 3.02
7. Based on the result in (5), we can write:
Linear approximation of f at x = 3 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{(6a+5) (x - a)~+~(3a^2 + 5a + 3)}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=3}}    & {~=~}    &{(18+5) (x - 3)~+~(27 + 15 + 3)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{(23) (x - 3)~+~(45)}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(3.02)}    & {~≈~}    &{L(3.02)}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~≈~}    &{(23) (3.02 - 3)~+~(45)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~≈~}    &{(23) (0.02)~+~(45)}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~≈~}    &{45.46}    \\
\end{array}$
9. It is interesting to note that, if we use a calculator or computer to calculate the given function at 3.02, we will get:
f(x) = 3(3.02)² + 5(3.02) + 3 = 45.4612
10. Fig.22.23 below shows the graph.
   ♦ f is drawn in red color
   ♦ L is drawn in green color

Fig.22.23

   ♦ A is a point on f
   ♦ B is a point on f
   ♦ B' is a point on L
• We want the y-coordinate at B. But, for ease of calculation, we use linear approximation and obtain the y-coordinate at B'.
• In the graph, A, B and B' are so close to each other that, they are not distinctly visible from each other.

 

Solved example 22.26
Write the linear approximation of f(x) = sin x and use it to estimate sin 62^o.
Solution:
1. Given $\rm{f(x) = \sin x}$
• We are asked to write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is $\rm{f'(x) ~=~\cos x}$  
• So the slope at 'a' is $\rm{f'(a) ~=~\cos a}$
3. At the point where x = a, the y coordinate can be calculated as follows:
y = sin x = sin a
4.  So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - \sin a}    & {~=~}    &{\cos a \,(x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\cos a \, (x - a) + \sin a}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{x \cos a ~-~ a \cos a ~+~\sin a}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = x \cos a ~-~ a \cos a ~+~\sin a$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 62^o.
• Converting this to radians, we get:
$\rm{62^o ~=~62 \times \frac{\pi}{180}~=~\frac{62 \pi}{180}}$
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 60^o
cosine and sine of 60 is already known. So it is a convenient number. Also, 60 is close to 62.
• Converting 60^o to radians, we have:
$\rm{60^o ~=~\frac{\pi}{3}}$
7. Based on the result in (5), we can write:
Linear approximation of f at x = π/3 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{x \cos a ~-~ a \cos a ~+~\sin a}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=\pi / 3}}    & {~=~}    &{x \cos \frac{\pi}{3} ~-~ a \cos \frac{\pi}{3} ~+~\sin \frac{\pi}{3}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2} ~-~ \frac{\pi}{6} ~+~ \frac{\sqrt3}{2}}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(\frac{62 \pi}{180})}    & {~≈~}    &{L(\frac{62 \pi}{180})}    \\
{~\color{magenta}    3    }    &{\implies}    &{\sin \frac{62 \pi}{180}}    & {~≈~}    &{\frac{62 \pi}{180} ~-~ \frac{\pi}{6} ~+~ \frac{\sqrt3}{2}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~≈~}    &{0.883478
}    \\
\end{array}$
9. It is interesting to note that, if we use a calculator or computer to calculate the sine of 62^o, we will get:
$\sin 62^o~=~0.882947$

Solved example 22.27
Write the linear approximation of f(x) = cos x and use it to estimate cos 89^o.
Solution:
1. Given $\rm{f(x) = \cos x}$
• We are asked to write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is $\rm{f'(x) ~=~-\sin x}$  
• So the slope at 'a' is $\rm{f'(a) ~=~-\sin a}$
3. At the point where x = a, the y coordinate can be calculated as follows:
y = cos x = cos a
4.  So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - \cos a}    & {~=~}    &{-\sin a \,(x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{-\sin a \, (x - a) + \cos a}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{-x \sin a ~+~ a \sin a ~+~\cos a}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = -x \sin a ~+~ a \sin a ~+~\cos a$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 89^o.
• Converting this to radians, we get:
$\rm{89^o ~=~89 \times \frac{\pi}{180}~=~\frac{89 \pi}{180}}$
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 90^o
cosine and sine of 90 is already known. So it is a convenient number. Also, 90 is close to 89.
• Converting 90^o to radians, we have:
$\rm{90^o ~=~\frac{\pi}{2}}$
7. Based on the result in (5), we can write:
Linear approximation of f at x = π/2 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{-x \sin a ~+~ a \sin a ~+~\cos a}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=\pi / 2}}    & {~=~}    &{-x \sin \frac{\pi}{2} ~+~ a \sin \frac{\pi}{2} ~+~\cos \frac{\pi}{2}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{-x ~+~ \frac{\pi}{2}}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(\frac{89 \pi}{180})}    & {~≈~}    &{L(\frac{89 \pi}{180})}    \\
{~\color{magenta}    3    }    &{\implies}    &{\cos \frac{89 \pi}{180}}    & {~≈~}    &{-\frac{89 \pi}{180} ~+~ \frac{\pi}{2}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~≈~}    &{0.01745329}    \\
\end{array}$
9. It is interesting to note that, if we use a calculator or computer to calculate the cosine of 89^o, we will get:
$\cos 89^o~=~0.017452406$

Solved example 22.28
Write the linear approximation of f(x) = (1+x)ⁿ and use it to estimate (1.01)³.
Solution:
1. Given $\rm{f(x) = (1+x)^n}$
• We are asked to write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is $\rm{f'(x) ~=~n(1+x)^{n-1}}$  
• So the slope at 'a' is $\rm{f'(a) ~=~n(1+a)^{n-1}}$
3. At the point where x = a, the y coordinate can be calculated as follows:
$\rm{y = (1+a)^n}$
4.  So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - (1+a)^n}    & {~=~}    &{n(1+a)^{n-1} \,(x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{nx(1+a)^{n-1}~-~na(1+a)^{n-1}~+~(1+a)^{n}}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = nx(1+a)^{n-1}~-~na(1+a)^{n-1}~+~(1+a)^{n}$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 0.01.
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 0
Then (1+a) will become 1. So it is a convenient number. Also, 0 is close to 0.01.
7. Based on the result in (5), we can write:
Linear approximation of f at x = 0 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{nx(1+a)^{n-1}~-~na(1+a)^{n-1}~+~(1+a)^{n}}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=0}}    & {~=~}    &{nx(1) ~-~n(0)(1)~+~1}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{nx ~+~ 1}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(0.01)}    & {~≈~}    &{L(0.01)}    \\
{~\color{magenta}    3    }    &{\implies}    &{(1+0.01)^n}    & {~≈~}    &{n(0.01) + 1}    \\
{~\color{magenta}    4    }    &{{\implies}}    &{{(1.01)^3}}    & {~≈~}    &{3(0.01) + 1}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~≈~}    &{1.03}    \\
\end{array}$
9. It is interesting to note that, if we use a calculator or computer to calculate (1.01)³, we will get:
$(1.01)^3~=~1.030301$

Solved example 22.29
Write the linear approximation of f(x) = (1−x)^1/10 and use it to estimate (0.999)^1/10.
Solution:
1. Given $\rm{f(x) = (1-x)^{1/10}}$
• We are asked to write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is:
$\rm{f'(x) ~=~\frac{1}{10}(1- x)^{-9/10} (-1)~=~\frac{-1}{10}(1- x)^{-9/10}}$  
• So the slope at 'a' is $\rm{f'(a) ~=~\frac{-1}{10}(1- a)^{-9/10}}$
3. At the point where x = a, the y coordinate can be calculated as follows:
$\rm{y = (1-a)^{1/10}}$
4.  So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - (1-a)^{1/10}}    & {~=~}    &{\frac{-1}{10}(1- a)^{-9/10} \,(x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{-1}{10}(1- a)^{-9/10} \,(x - a)~+~(1-a)^{1/10}}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = \frac{-1}{10}(1- a)^{-9/10} \,(x - a)~+~(1-a)^{1/10}$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 0.001.
This is because, 0.999 = (1 − 0.001)
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 0
Then (1−a) will become 1. So it is a convenient number. Also, 0 is close to 0.001.
7. Based on the result in (5), we can write:
Linear approximation of f at x = 0 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{\frac{-1}{10}(1- a)^{-9/10} \,(x - a)~+~(1-a)^{1/10}}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=0}}    & {~=~}    &{\frac{-1}{10}(1) \,(x)~+~1}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{-x}{10} ~+~ 1}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(0.001)}    & {~≈~}    &{L(0.001)}    \\
{~\color{magenta}    3    }    &{\implies}    &{(1-0.001)^{1/10}}    & {~≈~}    &{\frac{-0.001}{10} + 1}    \\
{~\color{magenta}    4    }    &{{\implies}}    &{{(0.999)^{1/10}}}    & {~≈~}    &{-0.0001 + 1}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~≈~}    &{0.9999}    \\
\end{array}$
9. It is interesting to note that, if we use a calculator or computer to calculate (0.999)^1/10, we will get:
$(0.999)^{1/10}~=~0.9998999$
10. Fig.22.24 below shows the graph.
   ♦ f is drawn in red color
   ♦ L is drawn in green color

Fig.22.24

   ♦ A is a point on f
   ♦ B is a point on f
   ♦ B' is a point on L
• We want the y-coordinate at B. But, for ease of calculation, we use linear approximation and obtain the y-coordinate at B'.
• In the graph, A, B and B' are so close to each other that, they are not distinctly visible from each other.

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In the next section, we will see Differentials.

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22.6 - Linear Approximation

In the previous section, we completed a discussion on Tangents and Normals. In this section, we will see Approximations.

Some basic details can be written in 12 steps:
1. Fig.22.21 below shows the graph of f(x) = x³ − 6x + 2
It is drawn in red color.

Fig.22.21

2. Consider point A where x = 2
We get: f(2) = (2)³ − 6(2) + 2 = −2
3. The green line is the tangent at x = 2
• Let us find the equation of this tangent.
f'(x) = 3x2` − 6
⇒ f'(2) = 3(2)2` − 6 = 6
y − (−2) = 6(x − 2)
⇒ y + 2 = 6 x − 12
⇒ y = 6x − 14
4. This tangent can be considered as a function of x. We will name this function as L. So we can write:
L(x) = 6x − 14
5. Let us calculate the value of L at x = 2
L(2) = 6(2) − 14 = −2
6. From (2) and (5), we can write: f(2) = L(2)
• In general, we can write: f(a) = L(a)
This is possible because, the tangent L is drawn at x = a
7. Consider another point B where x = 1.9.
    ♦ At A, x = 2
    ♦ At B, x = 1.9
• So we can say that, B is close to A
    ♦ f(1.9) = (1.9)³ − 6(1.9) + 2 = −2.541
    ♦ L(1.9) = 6(1.9) − 14 = −2.6
• We see that, f(1.9) is approximately equal to L(1.9)
• When we calculate L(1.9), we get the coordinates of a new point as (1.9, − 2.6). We will name this point as B'.
In the graph, we see that, B is very close to B'.
8. From the above discussion, we can write 3 points:
(i) We wanted the value of f at B. For that, we used the tangent at A
(ii) If we directly use f, we get: f(1.9) = −2.541
(iii) If we use the tangent at A, we get: L(1.9) = −2.6
(iv) We can write: f(1.9) ≈ L(1.9)
9. We accept the fact that, there is considerable difference between −2.541 and −2.6
• This difference is due to the fact that, B is not very close to A.
    ♦ x value at B is 1.9
    ♦ x value at A is 2
• If B is very close to A, we will get better approximations. Let us try such values for B:
(i) At B, if x = 1.95, we get:
f(1.95) = −2.2851 and L(1.95) = −2.3
(ii) At B, if x = 1.98, we get:
f(1.98) = −2.1176 and L(1.98) = −2.12
(iii) At B, if x = 1.99, we get:
f(1.99) = −2.059401 and L(1.99) = −2.06
(iv) At B, if x = 1.999, we get:
f(1.999) = −2.005994001 and L(1.999) = −2.006
(iv) At B, if x = 1.9999, we get:
f(1.9999) = −2.000599940001 and L(1.999) = −2.0006
10. It is clear that, if B is very close to A, then the tangent at A can be used to find the approximate value of f at B.
• The method can be summarized in 6 steps:
(i) We want the approximate value of f at B. Assume that, at B, x = b.
• So we want the approximate value of f(b).
(ii) For that, we choose a convenient point A, which is very close to B.
• Assume that, at A, x = a
(iii) We write the function L at A.
(iv) Then the approximate value of f(b) will be L(b)
(v) In the function f, the power of x may be one, two or above. But in the function L, the power of x will be always one. So it is easier to find L(b) than to find f(b).
(vi) The function L is known by three different names. We can use any on of them
   ♦ L is called linear approximation of f at x = a.
   ♦ L is called tangent line approximation of f at x = a.
   ♦ L is called linearization of f at x = a.
11. This method will work also when B (very close to A), is to the right of A. Let us see such points.
(i) At B, if x = 2.05, we get:
f(2.05) = −1.6848 and L(2.05) = −1.7
(ii) At B, if x = 2.02, we get:
f(2.02) = −1.877592 and L(2.02) = −1.88
(iii) At B, if x = 2.01, we get:
f(2.01) = −-1.939399 and L(2.01) = −1.94
(iv) At B, if x = 2.001, we get:
f(2.001) = −-1.993993999 and L(2.001) = −1.994
(iv) At B, if x = 2.0001, we get:
f(2.0001) = −1.999399939999 and L(2.0001) = −1.9994
12. This method will work only if B is very close to A. In the fig.22.21 above, a point C is marked. This point C is not close to A.
    ♦ x value at C is 2.4
    ♦ x value at A is 2
• We see that:
    ♦ f(2.4) is 1.424
    ♦ L(2.4) = 0.4
• There is a very large difference between 1.424 and 0.4
• This is because, there is a large difference between 2 and 2.4.

---

Let us see some solved examples:

Solved example 22.21
Write the linear approximation of f(x) = √x and use it to estimate $\sqrt{36.6}$.
Solution:
1. Given $\rm{f(x) = \sqrt{x}}$
• We are asked to write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is $f'(x) ~=~\rm{\frac{1}{2} (x)^{-1/2}}$  
• So the slope at 'a' is $\rm{f'(a) ~=~\frac{1}{2} (a)^{-1/2}~=~\frac{1}{2 \sqrt{a}}}$
3. At the point where x = a, the y coordinate can be calculated as follows:
y = √x = √a
4. So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - \sqrt{a}}    & {~=~}    &{\frac{1}{2 \sqrt{a}} (x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{1}{2 \sqrt{a}} (x - a) + \sqrt{a}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~-~\frac{a}{2 \sqrt{a}} ~+~ \sqrt{a}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~-~\frac{\sqrt{a}}{2} ~+~ \sqrt{a}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = \frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 36.6
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 36
Square root of 36 is already known. So it is a convenient number. Also, 36 is close to 36.6
7. Based on the result in (5), we can write:
Linear approximation of f at x = 36 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=36}}    & {~=~}    &{\frac{x}{2 \sqrt{36}} ~+~\frac{\sqrt{36}}{2}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 (6)} ~+~\frac{6}{2}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{12} ~+~3}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(36.6)}    & {~≈~}    &{L(36.6)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\sqrt{36.6}}    & {~≈~}    &{\frac{36.6}{12} ~+~3}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~≈~}    &{3.05 ~+~ 3}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~≈~}    &{6.05}    \\
\end{array}$
9. It is interesting to note that, if we use a calculator or computer to calculate the square root of 36.6, we will get:
$\sqrt{36.6}~=~6.0497$
10. Fig.22.22 below shows the graph.
   ♦ f is drawn in red color
   ♦ L is drawn in green color

Fig.22.22

   ♦ A is a point on f
   ♦ B is a point on f
   ♦ B' is a point on L
• We want the y-coordinate at B. But, for ease of calculation, we use linear approximation and obtain the y-coordinate at B'.
• In the graph, B and B' are so close to each other that, B' is not distinctly visible from B.  

Solved example 22.22
Write the linear approximation of f(x) = √x and use it to estimate √(9.1).
Solution:
1. Given $\rm{f(x) = \sqrt{x}}$
• We are asked to write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is $f'(x) ~=~\rm{\frac{1}{2} (x)^{-1/2}}$  
• So the slope at 'a' is $\rm{f'(a) ~=~\frac{1}{2} (a)^{-1/2}~=~\frac{1}{2 \sqrt{a}}}$
3. At the point where x = a, the y coordinate can be calculated as follows:
y = √x = √a
4. So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - \sqrt{a}}    & {~=~}    &{\frac{1}{2 \sqrt{a}} (x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{1}{2 \sqrt{a}} (x - a) + \sqrt{a}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~-~\frac{a}{2 \sqrt{a}} ~+~ \sqrt{a}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~-~\frac{\sqrt{a}}{2} ~+~ \sqrt{a}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = \frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 9.1
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 9
Square root of 9 is already known. So it is a convenient number. Also, 9 is close to 9.1
7. Based on the result in (5), we can write:
Linear approximation of f at x = 9 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=9}}    & {~=~}    &{\frac{x}{2 \sqrt{9}} ~+~\frac{\sqrt{9}}{2}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 (3)} ~+~\frac{3}{2}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{6} ~+~\frac{3}{2}}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(9.1)}    & {~≈~}    &{L(9.1)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\sqrt{9.1}}    & {~≈~}    &{\frac{9.1}{6} ~+~\frac{3}{2}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~≈~}    &{1.51667 ~+~ 1.5}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{3.01667}    \\
\end{array}$
9. It is interesting to note that, if we use a calculator or computer to calculate the square root of 9.1, we will get:
$\sqrt{9.1}~=~3.0166$

Solved example 22.23
Write the linear approximation of $\rm{f(x)\,=\,\sqrt[3]{x}}$ and use it to estimate $\rm{\sqrt[3]{8.1}}$.
Solution:
1. Given $\rm{f(x)\,=\,\sqrt[3]{x}}$
• We are asked to write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is $f'(x) ~=~\rm{\frac{1}{3} (x)^{-2/3}}$  
• So the slope at 'a' is $\rm{f'(a) ~=~\frac{1}{3} (a)^{-2/3}}$
3. At the point where x = a, the y coordinate can be calculated as follows:
$\rm{y\,=\,\sqrt[3]{a}\,=\,a^{1/3}}$
4. So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - a^{1/3}}    & {~=~}    &{\frac{1}{3} (a)^{-2/3} (x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{1}{3} (a)^{-2/3} (x) ~-~ \frac{1}{3} (a)^{-2/3}(a)~+~a^{1/3}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{3} (a)^{-2/3} (x) ~-~ \frac{1}{3} a^{1/3}~+~a^{1/3}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{3} (a)^{-2/3} (x) ~+~ \frac{2}{3} (a)^{1/3}}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = \frac{1}{3} (a)^{-2/3} (x) ~+~ \frac{2}{3} (a)^{1/3}$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 8.1
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 8
Cube root of 8 is already known. So it is a convenient number. Also, 8 is close to 8.1
7. Based on the result in (5), we can write:
Linear approximation of f at x = 8 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{\frac{1}{3} (a)^{-2/3} (x) ~+~ \frac{2}{3} (a)^{1/3}}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=8}}    & {~=~}    &{\frac{1}{3} (8)^{-2/3} (x) ~+~ \frac{2}{3} (8)^{1/3}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{3} (2)^{-2} (x) ~+~ \frac{2}{3} (2)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{12} ~+~\frac{4}{3}}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(8.1)}    & {~≈~}    &{L(8.1)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\sqrt[3]{8.1}}    & {~≈~}    &{\frac{8.1}{12} ~+~\frac{4}{3}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~≈~}    &{0.675 ~+~ 1.33333}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~≈~}    &{2.00833}    \\
\end{array}$                           
9. It is interesting to note that, if we use a calculator or computer to calculate the cube root of 8.1, we will get:
$\sqrt[3]{8.1}~=~2.00829$

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

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