Linear Approximation Formula and Solved Examples in Differential Calculus
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:
[Image will be Uploaded Soon]
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
[Image will be Uploaded Soon]
& when we consider doing it with the expression of Riemann right sum with 4 equal subdivisions then,
[Image will be Uploaded Soon]
Assume A(i) represent the area of ith rectangle in this approximation
[Image will be Uploaded Soon]
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
[Image will be Uploaded Soon]
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 Explained
1. What is differential calculus?
Differential calculus is the branch of mathematics that studies rates of change and slopes of curves using derivatives. It focuses on how a function changes as its input changes.
- The central concept is the derivative.
- It measures the slope of a curve at a specific point.
- It is used in optimisation, motion problems, and approximations.
2. What is a derivative in differential calculus?
A derivative is the instantaneous rate of change of a function with respect to its variable. It represents the slope of the tangent line at a point on the graph.
- Notation: f′(x) or dy/dx.
- Definition: f′(x) = lim(h→0) [f(x+h) − f(x)] / h.
- If f(x) = x², then f′(x) = 2x.
3. What is the formula for the derivative from first principles?
The derivative from first principles is defined as f′(x) = lim(h→0) [f(x+h) − f(x)] / h. This limit measures how the function changes as h approaches zero.
- Step 1: Substitute x + h into the function.
- Step 2: Subtract f(x).
- Step 3: Divide by h.
- Step 4: Take the limit as h → 0.
4. How do you use linear approximation in differential calculus?
Linear approximation uses the formula L(x) = f(a) + f′(a)(x − a) to estimate values of a function near x = a. It is based on the tangent line at a point.
- Choose a value a close to x.
- Find f(a) and f′(a).
- Substitute into L(x).
- f(4)=2, f′(x)=1/(2√x), so f′(4)=1/4.
- L(4.1)=2 + (1/4)(0.1)=2.025.
5. What is the difference between average rate of change and instantaneous rate of change?
The average rate of change measures change over an interval, while the instantaneous rate of change is the derivative at a point.
- Average rate of change: [f(b) − f(a)] / (b − a).
- Instantaneous rate of change: f′(a).
- Average rate → slope of a secant line.
- Instantaneous rate → slope of a tangent line.
6. What are the basic rules of differentiation?
The basic rules of differentiation simplify finding derivatives of functions. The main rules are:
- Power rule: d/dx (xⁿ) = nxⁿ⁻¹.
- Constant rule: d/dx (c) = 0.
- Sum rule: (f + g)′ = f′ + g′.
- Product rule: (fg)′ = f′g + fg′.
- Quotient rule: (f/g)′ = (f′g − fg′)/g².
- Chain rule: d/dx f(g(x)) = f′(g(x))g′(x).
7. How do you find the derivative of a function step by step?
To find a derivative, apply the appropriate differentiation rules to each term of the function. Example: Find the derivative of f(x) = 3x³ − 5x² + 2x.
- Apply the power rule to each term.
- d/dx (3x³) = 9x².
- d/dx (−5x²) = −10x.
- d/dx (2x) = 2.
8. What is differential approximation?
Differential approximation estimates small changes in a function using dy ≈ f′(x) dx. It uses derivatives to approximate how much y changes when x changes slightly.
- dx = small change in x.
- dy = approximate change in y.
- If dx = 0.1, then dy ≈ 6 × 0.1 = 0.6.
9. How are derivatives used for error estimation?
Derivatives estimate measurement errors using absolute error ≈ |f′(x)|·|dx|. This approach uses differential approximation.
- Compute the derivative f′(x).
- Multiply by the small input error dx.
- If r = 5 and dr = 0.1, then dA ≈ 2π(5)(0.1) = π.
10. What are the real-life applications of differential calculus and approximations?
Differential calculus and approximations are used to model rates of change, optimisation, and error analysis in real life. Common applications include:
- Physics: velocity and acceleration (derivatives of position).
- Economics: marginal cost and revenue.
- Engineering: maximising efficiency and minimising materials.
- Medicine: modelling growth rates of populations or bacteria.
