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:
$\sqrt[3]{8.1}~=~2.00829$
In the next section, we will see a few more solved examples.
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