# 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 ${f}^{\mathrm{\prime }}\left(x\right)$ of a function $f\left(x\right)$ at a point represents the rate of change of $f\left(x\right)$ concerning $x$ at that point.
• Average Rate of Change: $\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}$ represents the average rate of change of $y$ concerning $x$ over an interval.
• Limit as Delta $\to$ 0: The derivative represents the instantaneous rate of change as the interval approaches zero.

### Example Problem:

#### Problem Statement:

Consider a function $f\left(x\right)=3{x}^{2}+2x-5$. Find the rate of change of $f\left(x\right)$ concerning $x$ at $x=2$.

#### Solution Steps:

1. Find the Derivative ${f}^{\mathrm{\prime }}\left(x\right)$:

• $f\left(x\right)=3{x}^{2}+2x-5$
• ${f}^{\mathrm{\prime }}\left(x\right)=\frac{d}{dx}\left[3{x}^{2}+2x-5\right]$
• ${f}^{\mathrm{\prime }}\left(x\right)=6x+2$
2. Evaluate the Derivative at $x=2$:

• ${f}^{\mathrm{\prime }}\left(x\right)=6x+2$
• ${f}^{\mathrm{\prime }}\left(2\right)=6\left(2\right)+2=12+2=14$
3. Interpretation:

• The rate of change of $f\left(x\right)$ 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 $s\left(t\right)=3{t}^{2}-2t+5$, 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 $s\left(t\right)$ to find the velocity function $v\left(t\right)$.
• $v\left(t\right)={s}^{\mathrm{\prime }}\left(t\right)=\frac{ds}{dt}=\frac{d}{dt}\left(3{t}^{2}-2t+5\right)$
• $v\left(t\right)=6t-2$ 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 $v\left(t\right)$ to find the acceleration function $a\left(t\right)$.
• $a\left(t\right)={v}^{\mathrm{\prime }}\left(t\right)=\frac{dv}{dt}=\frac{d}{dt}\left(6t-2\right)$
• $a\left(t\right)=6$ meters per second squared (m/s²)

### Interpretation:

• Velocity: The object's velocity is given by $v\left(t\right)=6t-2$. 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 $f\left(x\right)={x}^{2}+3x-5$. Find the average rate of change of $f\left(x\right)$ concerning $x$ between $x=2$ and $x=4$.

Solution Steps:

1. Calculate $f\left(x\right)$ at $x=2$ and $x=4$:

• $f\left(2\right)={2}^{2}+3\left(2\right)-5=4+6-5=5$
• $f\left(4\right)={4}^{2}+3\left(4\right)-5=16+12-5=23$
2. Calculate Average Rate of Change:

• Average Rate of Change $=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{f\left(4\right)-f\left(2\right)}{4-2}$
• Average Rate of Change $=\frac{23-5}{4-2}=\frac{18}{2}=9$

### Interpretation:

• The average rate of change of $f\left(x\right)={x}^{2}+3x-5$ 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.