MathsDifferential Calculus and Approximations – Explanation and Solved Example

Differential Calculus and Approximations – Explanation and Solved Example

Approximation of Differential Equation

Differential equations are mathematical equations that describe the change in a certain quantity, or function, over time. In order to approximate a differential equation, a numerical method is used to approximate the solution to the equation. This involves breaking the differential equation down into a series of smaller, simpler equations, which can then be solved using a computer. The approximation of a differential equation is used to calculate the approximate solution to the equation, which can be used to predict the behavior of the quantity or function over time.

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    Approximation Example

    In the approximation example, we will approximate the value of pi. Pi is the ratio of a circle’s circumference to its diameter and is approximately 3.14.

    We will use the following equation to approximate pi:

    pi ≈ (C/d)^2

    In this equation, C is the circumference of the circle and d is the diameter of the circle.

    We can use the value of C and d from a circle with a diameter of 10 cm and a circumference of 31.4 cm.

    pi ≈ (31.4 cm/10 cm)^2

    pi ≈ 3.14

    Riemann Sum Example

    A Riemann sum is a way to estimate the value of a function over an interval by adding up the function values at evenly spaced points within the interval.

    The example below illustrates how to find a Riemann sum for a function over the interval [0, 3].

    The function being estimated is f(x) = x^3.

    The points at which the function is evaluated are:

    0, 1, 2, and 3.

    The width of the interval is 1.

    The sum of the function values at the points evaluated is:

    0 + 1 + 4 + 9 = 14.

    Thus, the Riemann sum for the function over the interval [0, 3] is 14.

    Problems for Practice

    Problem 1

    A plumber charges $25 for a service call plus $50 per hour of service. Write an equation in slope-intercept form for the cost, C, after h hours of service.

    C = 25 + 50h

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