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Differential calculus and approximation is a branch of mathematics that is a subset of calculus. Calculus deals with integration, differentiation, limits, and functions. Calculus is used in a variety of science and technological fields, as well as in economics, where algebra alone is insufficient. Making smaller portions of anything to investigate the rate of change is the basic idea of differential calculus approximations. Finding the function's derivative is what differentiation is all about. A differential equation is one in which a dependent variable's derivative is related to an independent variable. The overall concept of approximation values, on the other hand, differs from its mathematical treatment.
| Table of Content |
Key Takeaways: Differential Calculus, Approximation, Error Calculation, Independent Variable, Dependent Variable, Derivatives, Differential Equation, Limits
What is Differential Calculus and Approximation?
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Differential calculus is a discipline of mathematics that deals with the process of determining a function's derivative. It is the study of how quickly quantities change in relation to other variables. Differentiation is the process of determining the derivative of a given function. Cutting anything into smaller pieces to find the rate of change is the core concept of differential calculus. In general, a differential equation is one that involves the derivative of the dependent variable with respect to the independent variable.
| dy/dx = f(x) |
"x" is an independent variable in this situation, while "y" is a dependent variable.
The derivatives are typically used to determine the following:
- It is used to calculate the rate at which quantities change.
- It's used to calculate the tangent and normal to a curve at a given point.
- It is used to locate the turning point on the graph function, which aids in determining where the function's highest or smallest value occurs.
- It's used to figure out whether the function is increasing or decreasing at different intervals.
- It's used to figure out what a specific quantity's approximate value is.
Making an approximation in mathematics is the act or process of obtaining a number that is acceptable close to an exact value; the number is then referred to as an approximation or approximate value. Because it is impossible to make absolutely exact measurements, approximating has always been an important procedure in the experimental sciences and engineering. The approximation is also necessary since some numbers cannot be fully stated in decimal notation. Approximations are employed in these circumstances. The overall concept of approximation values, on the other hand, differs from its mathematical treatment. While 9.99 is still close to 10, we'll be talking about approximating the values of quantities whose exact value is unknown or cannot be estimated using any known method.
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General Form of Approximation of Differential Calculus
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Let us write y(x = x′) = y(x = x0) + Δy in terms of notation, suggesting that y is the change in the value of the function y when x is given by x = x′ – x0.
Then we'll go on to the next step:
- Find a location x0 near the point x′ where the function's value is known.
- Differentiate the function as a function of x.
| \(\frac{dy}{dx}\) = \(\frac{d}{dx}\) (f (x)) dy = f′ (x) dx |
- Use approximations, such as dx = x = x′ – x0, then calculate the derivative at x = x0 to get dy, which is approximated as y:
| Δy = f′(x0)Δx Δy = f′ (x0) (x′ - x0) |
- As x goes from x0 to x′, this is the change in the value of the function y. As a result,
| f (x′) = f (x0) + Δy [ f (x′) = f (x0) + f′ (x0) (x′ - x0)] |
The following is a typical format for writing the above-mentioned result:
| f (x + Δx) = f (x) + f′ (x) Δx |
This allows one to obtain the function's value at a point near x.
Take a look at the diagram below in relation to this formula:
It's worth noting that we've used a straight line labelled AB to approximate the function's curve. Using the slope of a straight line formula, m = tanθ = tan(∠ABC) = dy/dx, we arrive at the relation Δy = dy, which is approximately accurate when considering the small values of Δx that we will use.
Calculation of Errors
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If the exact dependency on the independent parameters is known, the concept of approximating functions can also be extended to approximating the errors in calculating specific functions. Consider the following function:
y = f (a, b, c…….)
Then there's the mistake in calculating y at the positions a = a', b = b'....., caused by inaccuracies in the values of a, b, and c…
Δy ≈ dy = [∂y/∂a] a = a′, b=b′, c=c′….. Δa + [∂y/∂b] a=a′, b=b′, c=c′….. Δb +……
where [∂y/∂a] a = a′, b=b′, c=c′
is the derivative of y with respect to a (at a = a'), with the rest of the variables held constant.
This formula derives immediately from the above-mentioned approximation method and the logical premise that the total error in the computation of y is the sum of the various mistakes in y due to errors in the measurement of independent variables a, b, and c.
Things to Remember
- The branch of mathematics known as differential calculus and approximation is a subset of calculus. The primary principle behind differential calculus approximations is to break things down into smaller pieces to look at the rate of change.
- The differential of the independent variable is equal to the rise in the variable, but the differential of the dependent variable is not equal to the increase in the variable.
- The derivative of the dependent variable with respect to the independent variable is a differential equation.
- Approximation refers to the act or process of attaining a number that is acceptable and close to an exact value; the number is then referred to as an approximation or approximate value.
- The approximation method and the logical premise that the overall error in the computation of y is the sum of the multiple errors in y due to errors in the measurement of independent variables a, b, and c…. give rise to the error calculation formula.
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Sample Questions
Ques. Calculate √36.5 using differential approximate formula. (5 marks)
Ans: Consider y = √x when x = 36 and Δx = 0.5
Δy = √(x + Δx) – √x
Δy = √(36 + 0.5) – √36
Δy = √36.5 – 6
Δy + 6 = √36.5
As, dy ≈ Δy
Now, dy will be = (dy/dx) Δx
dy = ½ √x(0.5)= 0.05
So √36.5 = 6 + 0.05 = 6.05
Ques. Calculate √25.5 using differential approximate formula. (5 marks)
Ans. Consider y = √x when x = 25 and Δx = 0.5
Δy = √(x + Δx) – √x
Δy = √(25.5) – √25
Δy = √25.5 – 5
Δy + 5 = √25.5
As, dy ≈ Δy
Now, dy will be = (dy/dx) Δx
dy = ½ √x(0.5)= 0.05
So √25.5 = 5 + 0.05 = 5.05
Ques. f(x) is given as 3x2+5x+3. Calculate the approximate value of the function f(3.02). (5 marks)
Ans. Given f(x) = 3x2+5x+3
Consider x = 3, and ∆x = 0.02.
We can also write the given function as:
f (3.02) = f (x + ∆x) = 3(x + ∆x)2 + 5(x + ∆x) + 3
As we know that, ∆y = f (x + ∆x) – f (x).
The above expression can also be written as-
f (x + ∆x) = f (x) + ∆y
As, dx = ∆x, it can be hence approximately written as f (x) + f ′(x) ∆x
So, f (3.02) ≈ (3x2 + 5x + 3) + (6x + 5) ∆x
Substituting the values of x and ∆x, we get
= (3(3)2 + 5(3) + 3) + (6(3) + 5) (0.02)
Now, in order to get the approximate value, simplify it-
= (27 + 15 + 3) + (18 + 5) (0.02)
= 45 + 0.46
= 45.46
Therefore, the approximate value of f(3.02) is 45.46.
Ques. Up to three decimal places, estimate the value of √51. (5 marks)
Ans: We can calculate √51 to be about equivalent to 7; but, such approximations are insufficient for this situation. A function is defined as follows using the differentials method:
f(x) = √x
The value of this function at x = 51 is what we're looking for. At x = 49, we know what this function's value is. [49 has a square root of 7]. In the formula for y, we may use x = (51-49) = 2 and the derivative of f(x) at x = 49 to obtain the answer.
f (x) = √x
f’ (x) = ½ √x
Coming to the approximation, f = f’ (x = 49) ∆x
∆f = ½ √49 * (51-49)
∆f = (1/14) * 2 = 1/7 = 0.143
The function's value at x = 51 can then be calculated as:
f (x + ∆x) = f (x) + ∆f
√51 = 7 + 0.143 = 7.143
Ques. If the error in measuring the major axis is 0.05 cm and the minor axis is 0.01 cm, what is the error in computing the area of an ellipse with a major axis of 10 cm and a minor axis of 2 cm? (5 marks)
Ans: The area of an ellipse with a major axis a and a minor axis b is calculated as follows:
A = πab
A is simply a function of a and b in this case. So, using the above-mentioned method, let's calculate the error in the area:
Ques. Calculate the approximate inaccuracy in calculating the surface area of a sphere with a radius of 9 cm and an error of 0.03 cm. (5 marks)
Ans. The sphere's radius is 9 cm.
Radius calculation error, r = 0.03 cm.
Let's say there's a rounding error in determining surface area. If S is the sphere's surface area, then S = 4πr2.
⇒ dS/dr = 4π.2r = 8πr
Now by definition, approximately-
Ques. Calculate the approximate value of f(2.01), where f (x) = 4x2 + 5x + 2. (5 marks)
Ans. Let x = 2 and ∆x = 0.01
Given f(x) = 4x2 + 5x + 2
f’(x) = 8x + 5
Now, ∆y = f’(x) ∆x = (8x + 5) 0.01
Also, ∆y = f (x + ∆x) - f(x)
f (x + ∆x) = f (x) + ∆y
f(2.01) = 4x2 + 5x + 2 + (8x + 5) (0.01)
Putting the value of x = 2
f (2.01) = 4 (2)2 + 5 (2) + 2 + (0.01) [8*2+5]
= (16 + 10 + 2) + (21) (0.01)
= 28 + 0.21 = 28.21
Ques. Calculate the approximate value of (26)⅓ using differentials. (5 marks)
Ans: Let y = (x)⅓
Differentiating with respect to x,
dy/dx = (d(x1/3))/dx = (⅓ x⅓ - 1) = ⅓ x -⅔ = 1/3 x⅔
Using ∆y = (dy/dx)* ∆x
∆y = (1/3 x⅔) ∆x
Putting values,
∆y = (⅓ * 27⅔) * (-1)
∆y = ((1/3 * (33)⅔)) * (-1)
∆y = - 1/3 ^3 = -1/27
∆y = -0.037037
As we know that
∆y = f (x + ∆x) - f(x)
∆y = (x + ∆x)⅓ - (x)⅓
Putting values-
-0.037037 = (27 + (-1))⅓ - (27)⅓
-0.037037 = (26)⅓ - (3)3*1/3
-0.037037 = (26)⅓ - 3
-0.037037 + 3 = (26)1/3
2.9629 = (26)1/3
Hence, the approximate value of (26)1/3 is 2.9629.
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