How Differential Calculus Helps You Estimate and Solve Problems
Differential calculus and approximation is a sub-branch of calculus which is a part of mathematics. Integration, differentiation, limits, and functions are dealt with in calculus. Calculus has various science and technology applications and in economics, too, where algebra alone is insufficient to apply. The basic understanding of differential calculus approximations is to make smaller segments of something to study the rate of changes. Differentiation is nothing but finding the derivative of the function. A differential equation is an equation that has the derivative of a dependent variable concerning an independent variable.
The differential equation is given by f(x)=dy/dx
Where ‘y’ is the dependent variable and ‘x’ is the independent variable
Usually, the following are used to find out the derivatives.
With respect to other variables, the rate of change of quantity can be determined.
For certain quantities, approximate values can be determined.
For the given function, intervals in which increase or decrease of function can be studied.
Derivatives are used to find out the equation of tangent.
Consider the example, r=10cm find the rate of change of circle per second for the given radius.
It is known that Area of circle A is given by A = πr²
So, the rate of change of area with respect to radius will be
dA/dr=(d/dr) πr²
dA/dr= 2πr
it is given r=10cm then
dA/dr=20 π
so, at 20 π cm2/s the area of the circle is changing.
Approximation of Differential Equation
Certain values are approximated using differentials. Consider a function f(x) is defined as f:D tends to R where D⊂R . Say y=f(x). An increment in x can be noted as ∆x. When the x is increased by ∆x, corresponding y has to increase by ∆y=f(x+∆x)-f(x). it is depicted in the below picture:
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The following conclusions are derived from the above-mentioned points.
A differential of the independent variable always equals the increment of the variable. On the other side, the differential of the dependent variable does not equal the increment of the variable.
In some cases, like when dx=∆x is too negligible a value to consider compared to x, ∆y is the best approximation of dy and dy≈∆y.
So now dx=∆x
dy=f ‘(x)dx=(dy/dx) ∆x
For better understanding let’s consider an example
Approximation Example
1. Using the differential approximate √36.5
Ans. Say y=√x where x=36 and ∆x=0.5
∆y = \[\sqrt{x}\] + ∆x - \[\sqrt{x}\]
∆y = \[\sqrt{36}\] + 0.5 - \[\sqrt{36}\]
∆y = \[\sqrt{36.5}\] - 6
∆y + 6 = \[\sqrt{36.5}\]
As dy≈∆y
Now, dy=(dy/dx) ∆x
dy=½ \[\sqrt{x}\](0,5)=0.05
\[\sqrt{36.5}\] = 6 + 0.05 = 6.05
Riemann Sum Example
Consider approximating the area under the graph of f(x) = \[\sqrt{x}\] between x=0.5 and y=3.5
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& when we consider doing it with the expression of Riemann right sum with 4 equal subdivisions then,
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Assume A(i) represent the area of ith rectangle in this approximation
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A(1)+A(2)+A(3)+A(4)=i=1∑4A(i)
Now we need to find the expression for A(i)
The width of the interval [0.5,3.5] is 3 units but we need 4 so ¾=0.75
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So now f(xi) = \[\sqrt{xi}\] = 0.5 + 0.75i
A(i)=width⋅height
= \[\sum_{i=1}^{4}\] A(i)
= \[\sum_{i=1}^{4}\] 0.75 ∗ \[\sqrt{0.5}\] + 0.75
Problems for Practice
Determine the approximate for the \[\sqrt{25.5}\] using the differential.
Using differentials find the approximate value for (26)1/3.
FAQs on Differential Calculus and Approximations Made Simple
1. What is the core idea of differential calculus?
Differential calculus is a major branch of mathematics that studies the rates at which quantities change. It is primarily built around the concept of the derivative, which measures the instantaneous rate of change of a function, geometrically interpreted as the slope of the tangent line to the function's graph at a specific point.
2. How are derivatives used for approximation in Class 12 Maths?
In Class 12 Maths, derivatives are used to find an approximate value of a function near a known point. The principle is to use the tangent line at the known point as a linear approximation of the function's curve. Since the derivative gives the slope of this tangent, it allows us to accurately estimate the small change in the function's value for a small change in its input.
3. What is the main formula for finding approximations using differentials?
The primary formula for approximation as per the CBSE Class 12 syllabus is:
f(x + Δx) ≈ f(x) + f'(x)Δx.
In this formula:
- f(x) is the value of the function at a known point 'x'.
- f'(x) is the derivative of the function at 'x'.
- Δx is the small change in 'x' for which we want to find the new approximate value.
4. What is the difference between the actual change in a function, Δy, and the approximate change, dy?
The key difference lies in what they measure and how they are calculated:
- Actual Change (Δy): This is the precise change in the function's value as `x` changes by `Δx`. It is calculated as Δy = f(x + Δx) - f(x) and represents the change along the actual curve.
- Approximate Change (dy): This is the estimated change calculated using the differential. It is calculated as dy = f'(x)Δx (or dy = (dy/dx)Δx) and represents the change along the tangent line to the curve at point `x`.
5. Can you show an example of using differentials to approximate a value like the square root of 25.3?
Certainly. To approximate √25.3, we can follow these steps:
- Step 1: Define the function as f(x) = √x.
- Step 2: Choose a nearby 'easy' point, x = 25, and the small change, Δx = 0.3.
- Step 3: Find the derivative: f'(x) = 1 / (2√x).
- Step 4: Calculate the values at x = 25: f(25) = 5 and f'(25) = 1 / (2√25) = 1/10 = 0.1.
- Step 5: Apply the approximation formula: f(25.3) ≈ f(25) + f'(25)Δx.
- Result: √25.3 ≈ 5 + (0.1)(0.3) = 5 + 0.03 = 5.03.
6. What is the geometric meaning of approximation using differentials?
Geometrically, approximation using differentials means we are using the tangent line at a known point (x, f(x)) to estimate the function's value at a nearby point (x+Δx, f(x+Δx)). Instead of calculating the value on the function's curve, which might be complex, we calculate the corresponding value on the straight tangent line, which is much simpler and provides a close estimate for small Δx.
7. How do differentials help in estimating errors in measurement?
Differentials are very useful for error estimation in science and engineering. If a calculated quantity `y` depends on a measured quantity `x` (i.e., y = f(x)), and there's a small error `Δx` in measuring `x`, we can approximate the resulting error `Δy` in `y` using the formula Δy ≈ (dy/dx)Δx. This helps us understand how small measurement inaccuracies can propagate and affect the final calculated result.
8. In which cases does the approximation method using differentials become less accurate?
The approximation method using differentials is less accurate in two main situations:
- When the change in x, i.e., Δx, is relatively large. The tangent line diverges from the curve as we move further from the point of tangency.
- When the function has a high curvature near the point of approximation. A high curvature means the function bends sharply, and the straight tangent line is a poor representation of the curve even for a small Δx.
9. Why is the derivative (dy/dx) so crucial for making these approximations?
The derivative, dy/dx, is crucial because it represents the instantaneous rate of change of the function at a single point. This value is precisely the slope of the tangent line to the function's graph at that point. Without the derivative, we would not know the slope of this line, and we couldn't construct the linear model that is used to make the approximation.
10. What is the difference between approximation using differentials and simple rounding?
Simple rounding is a basic arithmetic process of replacing a number with a shorter, simpler, approximate value (e.g., 2.89 becomes 3). It doesn't consider the behaviour of any function. In contrast, approximation using differentials is a calculus-based technique that uses the rate of change (derivative) of a function to systematically estimate its value near a known point. It is a more sophisticated and precise method used for functional values, not just standalone numbers.
