7 Appllied of Differentials
Differentials are not merely mathematical concepts found in textbooks—they are powerful analytical tools for understanding real-world change. Through the fundamental ideas of rates of change and local approximations of functions, differentials help us predict, measure, and optimize various phenomena in everyday life as well as in professional fields [1], [2].
The video below takes us through real-world applications of differential concepts—showing how ideas about change and approximation appear in object motion, temperature variation, fluid flow, population dynamics, and various technical processes in engineering and science. By exploring these examples, we can appreciate that differentials are not merely formulas, but the foundation for precise analysis and informed decision-making..
7.1 Derivatives in Metallurgy
7.1.1 Carbon Diffusion Function
We want to model the carbon concentration \(C(x)\) as a function of depth \(x\). Let us derive this systematically.
1. Observing the Data Pattern
Given data:
| Depth \(x\) (mm) | Carbon (%) |
|---|---|
| 0.5 | 0.2 |
| 1.0 | 0.8 |
| 1.5 | 1.8 |
To identify the type of function, examine ratios:
- Ratio of carbon increase for doubling depth:
\[\frac{C(1.0)}{C(0.5)} = \frac{0.8}{0.2} = 4\] - Ratio of carbon increase for tripling depth:
\[\frac{C(1.5)}{C(0.5)} = \frac{1.8}{0.2} = 9\]
These ratios suggest a power-law relationship:
\[C(x) \propto x^n\]
where \(n\) is the exponent to be determined.
2. Determining the Exponent
Assume the general form:
\[C(x) = k x^n\]
Using the data points \((x_1, C_1) = (0.5, 0.2)\) and \((x_2, C_2) = (1.0, 0.8)\):
\[C_2 / C_1 = (x_2 / x_1)^n\]
Substitute the values:
\[ \frac{0.8}{0.2} = \left( \frac{1.0}{0.5} \right)^n \] \[4 = 2^n\]
Taking logarithm base 2:
\[\log_2 4 = \log_2 2^n \implies 2 = n\]
Thus, the exponent is \(n = 2\), confirming a quadratic relationship:
\[C(x) = k x^2\]
3. Determining the Coefficient
Use any data point to solve for \(k\), e.g., \(x = 1.0\) mm, \(C = 0.8\):
\[0.8 = k (1.0)^2 \implies k = 0.8\]
4. Final Model
The carbon concentration as a function of depth is:
\[\boxed{C(x) = 0.8 x^2}\]
This mathematically derived function reproduces the observed quadratic increase of carbon with depth and can be used to predict concentration at other depths.
5. Derivative and Rate of Change
The rate of change is given by the derivative using the power rule:
\[C'(x) = \frac{d}{dx}[0.8 x^2] = 2 \cdot 0.8 \cdot x = 1.6 x\]
This derivative indicates how fast carbon concentration increases with depth, which is essential for understanding hardness and diffusion behavior.
7.1.2 Porosity Evolution
During the heating stage of a thermal treatment process (such as sintering or annealing), the porosity of a material changes due to densification, grain growth, or diffusion-driven mechanisms. The objectives of this study are to:
- Model the evolution of porosity \(P(t)\) as a function of time \(t\) and/or temperature \(T(t)\).
- Determine the rate of change (derivative) of porosity with respect to time.
- Evaluate the suitability of simple models (linear and exponential) and examine the direct relationship between porosity and temperature.
Provided Data
| Time \(t\) (min) | Porosity \(P(t)\) | Temperature \(T(t)\) (°C) |
|---|---|---|
| 0 | 0.30 | 800 |
| 10 | 0.28 | 830 |
| 20 | 0.26 | 860 |
Note: The dataset consists of three measurement points within a 0–20 minute interval, during which the temperature increases from 800 °C to 860 °C.
Density model:
\[\rho(t) = P(t)\,T(t)\]
Derivative of Pruduct Rule: \[\rho'(t) = P'(t)T(t) + P(t)T'(t)\]
Your Task: Shows how pellet density changes as porosity decreases and temperature rises.
7.1.3 Strength vs Density
A materials laboratory is performing a heat-treatment experiment on a steel sample. During heating, the mechanical properties of the material change over time, particularly tensile strength and density.
The researchers recorded the following measurements:
| Time \(t\) (min) | Strength \(\sigma\) (MPa) | Density \(\rho\) (g/cm³) |
|---|---|---|
| 0 | 380 | 7.90 |
| 30 | 410 | 7.85 |
| 60 | 440 | 7.80 |
They want to compute the specific strength, defined as:
\[R(t) = \frac{\sigma(t)}{\rho(t)}\]
Use Derivative Quotient Rule:
\[R'(t) = \frac{\sigma'(t)\rho(t) - \sigma(t)\rho'(t)}{[\rho(t)]^2}\]
Your Task: Shows how efficiently strength improves compared to weight reduction.
7.1.4 Thermal Grain Growth
A materials research group studies grain growth in a metal during annealing. They report effective annealing time and grain size at three furnace temperatures:
| Temperature \(T\) (°C) | Effective Time \(t\) (min) | Grain Size \(d\) (µm) |
|---|---|---|
| 600 | 8 | 12 |
| 650 | 12 | 15 |
| 700 | 20 | 19 |
Grain growth during annealing is often modeled by a time-power law (at a fixed temperature):
\[ d(t) = k \, t^{\,n} \]
where \(k\) and \(n\) are material- and temperature-dependent constants. Temperature itself typically influences the kinetics through an Arrhenius factor in \(k\):
\[ k(T) = k_0 \exp\!\left(-\frac{Q}{RT}\right), \]
with \(Q\) the activation energy, \(R\) the gas constant, and \(T\) the absolute temperature (K).
Empirical models:
- Time depends on temperature:
\[t = 2T^3\]
- Grain size depends on time:
\[d = k\,t^{1/2}\]
Derivative (Chain Rule):
\[\frac{dd}{dT} = \frac{dd}{dt} \times \frac{dt}{dT}\]
Your Task: Shows how grain size changes when annealing temperature changes.
7.1.5 Molten Metal Vibration
Data:
| Time t (s) | Position x (mm) |
|---|---|
| 0.00 | 0 |
| 0.02 | 4.3 |
| 0.04 | 3.1 |
| 0.06 | -2.5 |
Fitted oscillation model:
\[x(t) = 5\sin(60t)\]
Derivative Trigonometric:
Velocity:
\[x'(t) = 300\cos(60t)\]
Your Task: Tells how fast molten metal is moving inside the furnace.
7.1.6 Cooling Curve of Hot Steel
Data:
| Time t (s) | Temperature T (°C) |
|---|---|
| 0 | 900 |
| 20 | 800 |
| 40 | 710 |
| 60 | 630 |
Cooling model:
\[T(t) = 900e^{-0.03t}\]
Derivative Exponential:
\[T'(t) = -27e^{-0.03t}\]
Your Task: Shows the cooling rate, which affects final microstructure.
7.1.7 Gold Leaching Concentration
Data:
| Time t (h) | Gold in Solution C (mg/L) |
|---|---|
| 0 | 0 |
| 1 | 27 |
| 2 | 36 |
| 3 | 41 |
Fitted model:
\[C(t) = \ln(1 + 4t)\]
Derivative of Logarithmic:
\[C'(t) = \frac{4}{1 + 4t}\]
Your Task: Shows how fast gold dissolves over time.
7.1.8 Ball Motion in Ball Mill
Data (Position):
| Time t (s) | x (cm) |
|---|---|
| 0.0 | 0.0 |
| 0.1 | 1.5 |
| 0.2 | 0.0 |
| 0.3 | -1.5 |
Fitted motion model:
\[x(t) = 2\sin(10t)\]
Velocity:
\[x'(t) = 20\cos(10t)\]
Acceleration: \[x''(t) = -200\sin(10t)\]
Your Task: Acceleration determines grinding impact energy.
7.2 Derivatives in Petroleum
7.2.1 Reservoir Pressure and Porosity
A petroleum company has collected the following reservoir data:
| Depth \(x\) (m) | Pressure \(P\) (MPa) | Notes |
|---|---|---|
| 500 | 15 | Shallow zone |
| 1000 | 25 | Mid zone |
| 1500 | 40 | Deep zone |
The porosity of the reservoir decreases with depth due to compaction.
7.2.2 Pressure Function
From the data, pressure increases faster than linear with depth.
We use a power-law model:
\[ P(x) = 0.01\,x^{1.3} \quad \text{(MPa)} \]
Derivative:
\[ P'(x) = \frac{d}{dx}\left(0.01 x^{1.3}\right) = 0.013\,x^{0.3} \]
This indicates the rate of pressure increase with depth.
7.2.3 Porosity Function
Assume porosity decreases linearly with depth:
\[ \phi(x) = 0.25 - 0.00005\,x \]
Derivative:
\[ \phi'(x) = -0.00005 \]
Indicating a constant decrease in porosity per meter.
7.2.4 Economic Value Function
Economic value per barrel depends on pressure:
\[ V(P) = 10 P^2 \quad \text{(USD per barrel)} \]
Using the chain rule:
\[ \frac{dV}{dx} = \frac{dV}{dP} \cdot \frac{dP}{dx} = 20 P(x) \, P'(x) \]
Substitute values:
\[ \frac{dV}{dx} = 20(0.01 x^{1.3})(0.013 x^{0.3}) \approx 0.0026\,x^{1.6} \]
7.2.5 Total Reservoir Value per Meter
Assume fluid volume per meter is proportional to porosity:
\[ T(x) = 1000\,\phi(x) \]
Total value per meter:
\[ W(x) = V(P(x)) \, T(x) = 1000\, P(x)^2\, \phi(x) \]
Derivative:
\[ \begin{aligned} W'(x) &= 1000\left[2 P(x)P'(x)\phi(x) + P(x)^2 \phi'(x)\right] \\[6pt] &= 1000\left[ 2(0.01 x^{1.3})(0.013 x^{0.3})(0.25 - 0.00005x) + (0.01 x^{1.3})^2(-0.00005) \right] \\[6pt] &= 0.065\,x^{1.6} - 0.00051\,x^{2.6} \end{aligned} \]
7.2.6 Reservoir Flow Analysis
Observed Data
| Depth \(x\) (m) | Pressure (MPa) | Temp (°C) | Mobility (Pa·s) |
|---|---|---|---|
| 500 | 15 | 60 | 0.12 |
| 700 | 20 | 70 | 0.10 |
| 1000 | 28 | 85 | 0.08 |
| 1300 | 35 | 100 | 0.06 |
| 1500 | 40 | 115 | 0.05 |
Pressure Function
Assume sinusoidal heterogeneity:
\[ P(x) = 25 + 15 \sin\left(\frac{\pi x}{2000}\right) \]
Temperature Function
Assume logarithmic increase:
\[ T(x) = 10 + 17 \ln(x) \]
Oil Mobility Function
Assume exponential decay:
\[ \mu(x) = 0.15 e^{-0.0007 x} \]
Total Effective Flow
\[ F(x) = \frac{P(x)}{\mu(x)}\, \cos\left(\frac{T(x)\pi}{180}\right) \]
7.3 Derivatives in Mining
7.3.1 Ore Grade and Density
| Depth \(x\) (m) | Ore Grade (%) | Notes |
|---|---|---|
| 50 | 0.8 | Upper layer |
| 100 | 1.6 | Mid layer |
| 150 | 3.6 | Deep layer |
7.3.2 Ore Grade Function
\[ G(x) = 0.00637\,x^{1.2} \]
Derivative:
\[ G'(x) = 0.00764\,x^{0.2} \]
7.3.3 Ore Density Function
\[ \rho(x) = 2.7 - 0.002 x \]
Derivative:
\[ \rho'(x) = -0.002 \]
7.3.4 Economic Value Function
\[ V(G) = 50 G^2 \]
Chain rule:
\[ \frac{dV}{dx} = 100\,G(x)\,G'(x) \approx 0.00486\,x^{1.4} \]
7.3.5 Total Ore Value per Meter
\[ T(x) = 10 \rho(x) \]
\[ W(x) = 500 G(x)^2 \rho(x) \]
Derivative:
\[ W'(x) \approx 0.1314\,x^{1.4} - 0.0405\,x^{2.4} \]
7.3.6 Copper Grade and Profit Analysis
Copper Grade Model
Model \(C(x)\) using power-law (choose parameters based on regression):
\[ C(x) = a x^b \]
Derivative:
\[ C'(x) = ab x^{b-1} \]
Extraction Cost
\[ K(x) = 20 + 0.5 x \]
Derivative:
\[ K'(x) = 0.5 \]
Profit Per Ton
\[ P(x) = 100\,C(x) - K(x) \]
Derivative:
\[ P'(x) = 100\,C'(x) - 0.5 \]
Total Profit Per Meter
\[ T(x) = 50 - 0.2 x \]
\[ TP(x) = P(x) T(x) \]
Derivative:
\[ TP'(x) = P'(x)T(x) + P(x)T'(x) \]
7.3.7 Groundwater Flow
Water Table Height
\[ H(x) = 10 + 2 \sin\left(\frac{\pi x}{100}\right) \]
Derivative:
\[ H'(x) = 2 \frac{\pi}{100}\cos\left(\frac{\pi x}{100}\right) \]
Slope Function
\[ \theta(x) = 5 + 0.5 \ln(x) \]
Derivative:
\[ \theta'(x) = \frac{0.5}{x} \]
Infiltration Efficiency
\[ E(x) = 0.9 e^{-0.02 x} \]
Derivative:
\[ E'(x) = -0.018 e^{-0.02x} \]
Total Infiltration
\[ F(x) = H(x)\,E(x)\, \cos\left(\frac{\theta(x)\pi}{180}\right) \]
7.4 References
[1]
Stewart, J., Calculus: Early transcendentals, Cengage Learning, 2016
[2]
Apostol, T. M., Calculus, volume i: One-variable calculus with an introduction to linear algebra, Wiley, 1967