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.
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.
In the next section, we will see a few more solved examples.
Previous
Contents
Next
Copyright©2024 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment