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) = x1/3 and use it to estimate 251/3.
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 |
♦ 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 62o.
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 = 62o.
•
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 = 60o
cosine and sine of 60 is already known. So it is a convenient number. Also, 60 is close to 62.
•
Converting 60o 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 62o, 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 89o.
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 = 89o.
•
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 = 90o
cosine and sine of 90 is already known. So it is a convenient number. Also, 90 is close to 89.
•
Converting 90o 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 89o, 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 |
♦ 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.
In the next section, we will see Differentials.
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