Derivatives as Rate of Change

Derivatives are a mathematical tool used to analyze how quantities change. We can calculate derivatives for various, quotient, and chain rulesfunctions, including trigonometric, exponential, polynomial, and implicit functions. There are two main methods for calculating derivatives: using limits or applying formulas and rules such as the product rule, quotient rule, and chain rule. Derivatives have many real-world applications, making them valuable for understanding and solving physical phenomena. Let's explore some of these applications.

Rate of Change 

Derivatives are fundamental to differential calculus. They describe how a function behaves, such as increasing or decreasing. Suppose we have two quantities, x, and y, that vary together and are related by the function y=f(x). The derivative of this function, denoted as \frac{dy}{dx}, represents the rate of change of y with respect to x. This tells us how y changes as x changes.

For Example: Find the rate of change of volume of a cube whose sides are increasing at the rate of 2 m/s. 

Solution: 

Let's say the length of the side of cube is "a". The volume of cube is given by, V = a³. 

\frac{dV}{da} = \frac{d(a^3)}{da}

⇒ \frac{dV}{dt} = 3a^2\frac{da}{dt}

⇒\frac{dV}{dt} = 3a^2(2)

⇒\frac{dV}{dt} = 6a² m³/s. 

Increasing and Decreasing Functions 

Derivatives are also used in finding out whether the function is increasing or decreasing or none of them. The figure given below shows the function f(x) = x². 

Notice in the figure, the function is decreasing in the interval (-∞, 0) and increasing in the interval (0,∞). 

In an interval I contained in the domain of real valued function “f”. Then, f is said to be, 

• Increasing on I, if x₁ < x₂ in I ⇒ f(x₁) ≤ f(x₂) for all x₁, x₂ ∈ I.
• Strictly Increasing on I, if x₁ < x₂ in I ⇒ f(x₁) < f(x₂) for all x₁, x₂ ∈ I.

• Decreasing on I, if x₁ < x₂ in I ⇒ f(x₁) ≥ f(x₂) for all x₁, x₂ ∈ I.
• Strictly Decreasing on I, if x₁ < x₂ in I ⇒ f(x₁) > f(x₂) for all x₁, x₂ ∈ I.

Now we know the definitions for increasing and decreasing functions. Let's see how to recognize a function that is increasing or decreasing in an interval. 

Let's say f is continuous on [a, b] and differentiable on the open interval (a, b). Then, 

1. f is increasing in (a, b) if f'(x) > 0 in the interval [a, b].
2. f is decreasing in (a, b) if f'(x) < 0 in the given interval.
3. f is constant if f'(x) = 0.

Learn More , Increasing and Decreasing Functions

Sample Problems on Derivatives as Rate of Change

Question 1: Let's say we have a circle whose radius is increasing. Find the rate of change of area with radius when r = 4cm. 

Solution: 

Let's say "A" is the area of the circle and “r” be the radius of the circle.

A = πr²

Differentiating it with respect to radius. 

\frac{dA}{dr} = \frac{d(\pi r^2)}{dr}

⇒ \frac{dA}{dr} = 2\pi r

At r = 4. 

\frac{dA}{dr} = 8\pi

Question 2: Let's say we have a rectangle whose sides are changing every second. The length is increasing at the rate of 3 m/s while the breadth is increasing at 8 m/s. Calculate the rate at which the area of the rectangle is increasing when length = 8m and breadth = 5m. 

Solution: 

Let, x be the length of the rectangle and y be the breadth of rectangle. 

\frac{dx}{dt} = 3 And \frac{dy}{dt} = 8

The area of rectangle is given by, 

A = xy

Differentiating the equation w.r.t time. 

\frac{dA}{dt} = \frac{d(xy)}{dt}

⇒ \frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}

⇒ \frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}

⇒\frac{dA}{dt} = 8(\frac{dy}{dt}) + 5(\frac{dx}{dt})

⇒ \frac{dA}{dt} = 8(8) + 5(3)

⇒ \frac{dA}{dt} = 64 + 15

Question 3: For the given curve, find the points where the value of the rate of change of y is zero. 

y = x² + x

Solution: 

y = x² + x

\frac{dy}{dx}  = \frac{d(x^2 + x)}{dx}

⇒\frac{dy}{dx}  = 2x + 1

This rate of change must be zero, 

2x + 1 = 0 

⇒ x = \frac{-1}{2}

Thus, at x = \frac{-1}{2} the rate of change is zero. 

Question 4: Prove that the function discussed above, f(x) = x² is increasing in the interval (0, ∞). 

Solution: 

According to above definition, a function in increasing in any interval if its derivative is greater than zero in that interval. 

f(x) = x²

Differentiating with respect to x, 

f'(x) = 2x 

For the given interval (0,∞) f'(x) > 0. 

Thus, the function is increasing in the given interval. 

Question 5: Find the intervals where the function f(x) = x² + 5x + 6 is increasing or decreasing. 

Solution: 

Given f(x) = x² + 5x + 6

f'(x) = 2x + 5

We need to study the sign of the derivative to find the intervals where this function is increasing or decreasing. 

f'(x) < 0 

⇒ 2x + 5 < 0 

⇒ x < \frac{-5}{2}

f'(x) > 0 

⇒ 2x + 5 > 0 

⇒ x > \frac{-5}{2}

Thus, the function is increasing in (\frac{-5}{2}, ∞) and is decreasing in (-∞, \frac{-5}{2}). 

Practice Problems on Derivatives as Rate of Change

Question 1: If the radius of a circle is increasing at a uniform rate of 2 cm/sec, find the rate of increasing of area of circle,at the instant when the radius is 20 cm.

Question 2: The percentage error in calculating the volume of a cubical box if an error of 1% is made in measuring the length of edges of the cube is?

Question 3: if s= \frac{1}{2}t^{3}-6t , Calculate the acceleration at the time when the velocity becomes zero.

Question 4: Prove that the function f(x) = x/log x increases on the interval (e,∞).

Question 5: Find the intervals where the function y = \frac{3}{2}x^{64} -3x^2 + 1 is increasing or decreasing. 

Conclusion

Derivatives are crucial for understanding how quantities change in relation to each other. They reveal the behavior of functions and are used to determine whether functions are increasing or decreasing. By quantifying the rate of change, derivatives provide valuable insights for solving real-world problems across various fields. Derivatives not only help us understand the dynamic nature of various phenomena but also offer practical solutions in fields ranging from physics and engineering to economics and biology.