Rate of change

Rate of Change Using Derivatives:

  • Derivatives in calculus measure the rate at which one quantity changes concerning another, representing instantaneous rates of change.

Derivative as Rate of Change:

  • Function Derivative: The derivative f(x) of a function f(x) at a point represents the rate of change of f(x) concerning x at that point.
  • Average Rate of Change: ΔyΔx represents the average rate of change of y concerning x over an interval.
  • Limit as Delta 0: The derivative represents the instantaneous rate of change as the interval approaches zero.

Example Problem:

Problem Statement:

Consider a function f(x)=3x2+2x5. Find the rate of change of f(x) concerning x at x=2.

Solution Steps:

  1. Find the Derivative f(x):

    • f(x)=3x2+2x5
    • f(x)=ddx[3x2+2x5]
    • f(x)=6x+2
  2. Evaluate the Derivative at x=2:

    • f(x)=6x+2
    • f(2)=6(2)+2=12+2=14
  3. Interpretation:

    • The rate of change of f(x) concerning x at x=2 is 14.

Interpretation:

  • Positive Rate of Change: Indicates an increasing function concerning x at that point.
  • Negative Rate of Change: Indicates a decreasing function concerning x at that point.

Example:

Problem Statement: Suppose the position of an object moving along a straight line is given by the function s(t)=3t22t+5, where s represents position in meters and t represents time in seconds. Find the object's velocity and acceleration functions.

Solution Steps:

  1. Velocity Function:

    • The velocity of the object is the rate of change of position concerning time.
    • Calculate the derivative of the position function s(t) to find the velocity function v(t).
    • v(t)=s(t)=dsdt=ddt(3t22t+5)
    • v(t)=6t2 meters per second (m/s)
  2. Acceleration Function:

    • The acceleration of the object is the rate of change of velocity concerning time.
    • Calculate the derivative of the velocity function v(t) to find the acceleration function a(t).
    • a(t)=v(t)=dvdt=ddt(6t2)
    • a(t)=6 meters per second squared (m/s²)

Interpretation:

  • Velocity: The object's velocity is given by v(t)=6t2. It changes at a rate of 6 m/s concerning time.
  • Acceleration: The object's acceleration is a constant 6 m/s². It represents the rate at which the velocity changes concerning time.

Example:

Problem Statement: Consider the function f(x)=x2+3x5. Find the average rate of change of f(x) concerning x between x=2 and x=4.

Solution Steps:

  1. Calculate f(x) at x=2 and x=4:

    • f(2)=22+3(2)5=4+65=5
    • f(4)=42+3(4)5=16+125=23
  2. Calculate Average Rate of Change:

    • Average Rate of Change =ΔyΔx=f(4)f(2)42
    • Average Rate of Change =23542=182=9

Interpretation:

  • The average rate of change of f(x)=x2+3x5 concerning x between x=2 and x=4 is 9 units.

Applications:

  • Physics: Analyzing velocity, acceleration, and force concerning time.
  • Economics: Determining marginal cost, revenue, and profit concerning production levels or sales.
  • Engineering: Optimizing designs, analyzing structural stability, and describing rates of flow.

Key Points:

  • Tangent Line: The slope of the tangent line to the graph of a function at a point represents the instantaneous rate of change at that point.
  • Derivative Calculation: Derivatives are calculated using differentiation rules to find the rate of change of a function concerning its variable.