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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