Applications of Derivatives

The applications of derivatives is a core topic in calculus and mathematical analysis. It primarily involves using derivatives to understand and solve various mathematical problems. Key mathematical applications include:

1. Finding Maxima and Minima: Derivatives help identify the local maxima and minima of a function. This is done by finding the critical points where the first derivative f'(x) is zero or undefined and then using the second derivative f''(x) to determine whether the point is a maximum or minimum.
2. Rate of Change of Quantities: The derivative dy/dx measures the instantaneous rate of change of a dependent variable y with respect to an independent variable x. This concept is used in problems involving velocity, acceleration, and other changing quantities.
3. Increasing and Decreasing Functions: A function f(x) is increasing on an interval if f'(x) > 0 for all x in that interval and decreasing if f'(x) < 0. This helps in analyzing the monotonicity of functions.
4. Finding Tangents and Normal: The derivative f'(x) at a point gives the slope of the tangent to the curve at that point. The equation of the tangent line can be written as y – y₁ = f'(x₁)(x – x₁), and the normal line is perpendicular to the tangent.

1.0Rate of Change of Quantities

The rate of change of quantities is a fundamental concept in calculus and is a direct application of derivatives. It is used to measure how one quantity changes with respect to another. Understanding this concept allows students and professionals to solve a variety of real-world problems, such as determining how fast a car is accelerating, how the height of a balloon increases with time, or how a population changes over a certain period.

In mathematical terms, if y = f(x) is a function representing the relationship between y and x, then the derivative $\frac{dy}{dx}$ signifies the rate of change of y with respect to x. Here, x could represent time, distance, or any other independent variable, while y could represent a dependent variable like speed, volume, or temperature.

Mathematical Definition

The rate of change of a quantity y with respect to x is defined as: $\frac{dy}{dx}={\lim }_{\Delta x\to 0}\frac{\Delta y}{\Delta x}$

where:

• $\Delta y$ is the change in y.
• $\Delta x$ is the change in x.

The derivative $\frac{dy}{dx}$ tells us how much y changes when x changes by a small amount. If $\frac{dy}{dx}>0$ , then y is increasing with respect to x. If $\frac{dy}{dx}<0$, then y is decreasing.

2.0Increasing and Decreasing Functions

In mathematics, particularly in calculus, functions can be classified based on their behavior over certain intervals.

• Increasing Functions: A function f(x) is said to be increasing on an interval if, for any two points x₁ and x₂ within that interval where x₁ < x₂, the function values satisfy $f(x_1)\le f(x_2)$. This means that as x increases, f(x) either increases or remains constant.
• Decreasing Functions: Conversely, a function f(x) is decreasing on an interval if, for any two points x₁ and x₂ within that interval where x₁ < x₂, the function values satisfy $f(x_1)\ge f(x_2)$. This means that as ( x ) increases, ( f(x) ) either decreases or remains constant.

Understanding whether a function is increasing or decreasing is crucial for analyzing its behavior, finding local maxima and minima, and solving optimization problems.

3.0Maxima and Minima 

1. Definitions:

• Local Maximum: A function f(x) has a local maximum at x = a if f(a) is greater than or equal to f(x) for all x in some open interval around a. This means f(a) is the highest point in a small neighbourhood around a.
• Local Minimum: A function f(x) has a local minimum at x = b if f(b) is less than or equal to f(x) for all x in some open interval around b. This means f(b) is the lowest point in a small neighbourhood around b.
• Absolute Maximum: A function f(x) has an absolute maximum at x = c if f(c) is greater than or equal to f(x) for all x in the domain of f. This means f(c) is the highest point over the entire domain.
• Absolute Minimum: A function f(x) has an absolute minimum at x = d if f(d) is less than or equal to f(x) for all x in the domain of f. This means f(d) is the lowest point over the entire domain.

2. Finding Maxima and Minima:

To find the maxima and minima of a function, we typically follow these steps:

Step 1: Find the First Derivative: Compute the first derivative of the function.

Step 2: Find Critical Points: Set the first derivative equal to zero and solve for x. These values are called critical points. Also, include points where the derivative does not exist.

Step 3: Determine the Nature of Critical Points:

Second Derivative Test: Compute the second derivative, f’'(x). Evaluate the second derivative at each critical point.

If f’'(x) > 0 at a critical point, the function has a local minimum at that point.

If f’'(x) < 0 at a critical point, the function has a local maximum at that point.

If f’'(x) = 0, the test is inconclusive.

First Derivative Test: Alternatively, examine the sign of the first derivative before and after each critical point.

If f’(x) changes from positive to negative at a critical point, the function has a local maximum there.

If f’(x) changes from negative to positive at a critical point, the function has a local minimum there.

4.0Solved Examples on Rate of Change of Quantities

Example 1: A stone is dropped into a calm lake, creating circular waves that expand at a speed of 4 cm per second. At the moment when the radius of the circular wave reaches 10 cm, how quickly is the enclosed area increasing?

Solution:

The area A of a circle with radius r is given by $A=\pi r^2$ .To find the rate of change of the area A with respect to time t, we differentiate:

$\frac{dA}{dt}=\frac{d}{dt}\left(\pi r^2\right)$

$\frac{dA}{dt}=\pi \frac{d}{dt}\left(r^2\right)$

$\frac{dA}{dt}=\pi \left(2r\frac{dr}{dt}\right)$

$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$

Given that:

$\frac{dr}{dt}=4\mathrm{cm}/\mathrm{s}$

Therefore:

$\frac{dA}{dt}=2\pi (10)(4)=80\pi$

Thus, the enclosed area is increasing at a rate of $80\pi {\mathrm{cm}}^2/\mathrm{s}$ when r = 10 cm.

Example 2: A balloon is being inflated and its volume is increasing at a constant rate of 5 cubic centimeters per second. At what rate is the radius of the balloon expanding when the radius is 6 cm?

Solution:

The volume V of a sphere with radius r is given by $V=\frac{4}{3}\pi r^3$. To find the rate of change of the radius r with respect to time t, we differentiate the volume formula with respect to time:

$\frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$

$\frac{dV}{dt}=\frac{4}{3}\pi \cdot 3r^2\frac{dr}{dt}$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

Given that:

$\frac{dV}{dt}=5{\mathrm{cm}}^3/\mathrm{s}$

We need to find $\frac{dr}{dt}$ when r = 6 cm. Substituting the known values into the differentiated equation:

$5=4\pi (6{)}^2\frac{dr}{dt}$

$5=144\pi \frac{dr}{dt}$

$\frac{dr}{dt}=\frac{5}{144\pi }$

Therefore, the radius is increasing at a rate of $\frac{5}{144\pi }\mathrm{cm}/\mathrm{s}$ when the radius is 6 cm.

Example 3: Consider the function $f(x)=x^3-3x^2+4.$

Solution: 

Let $f(x)=x^3-3x^2+4$

Step 1: Find the first derivative: f’(x) = 3x² – 6x 

Step 2: Find critical points by setting the first derivative to zero: 

⇒ 3x² – 6x = 0 

⇒ 3x (x – 2) = 0 

⇒ x = 0 or x = 2 

Step 3: Use the second derivative test: f’'(x) = 6x – 6 

⇒ At x = 0: f’'(0) = 6(0) – 6 = –6 (negative, so local maximum)

⇒ At x = 2: f’'(2) = 6(2) – 6 = 6 (positive, so local minimum)

Therefore, x = 0 is a local maximum and x = 2 is a local minimum.

Example 4: Consider the function $g(x)=x^4-4x^3+6x^2.$

Solution: 

Let $g(x)=x^4-4x^3+6x^2$

Step 1: Find the first derivative: $g^{\prime }(x)=4x^3-12x^2+12x$

Step 2: Find critical points by setting the first derivative to zero: 

$\Rightarrow 4x\left(x^2-3x+3\right)=0$

$\Rightarrow x=0\text{ or solve }{x}^2-3x+3=0$

Step 3: Use the second derivative test: $g^{\prime \prime }(x)=12x^2-24x+12$

⇒ At x = 0: $g^{\prime \prime }(0)=12(0{)}^2-24(0)+12=12$ (positive, so local minimum)

Therefore, x = 0 is a local minimum. The quadratic has no real roots, so there are no other critical points.

Example 5: Show that the function $f(x)=3x^2-6x+4$ is increasing on the set of real numbers $\mathbb{R}$.

Solution: 

To determine if the function $f(x)=3x^2-6x+4$ is increasing on $\mathbb{R}$, we need to find its first derivative and analyze its sign.

1. Find the first derivative: $f^{\prime }(x)=\frac{d}{dx}\left(3x^2-6x+4\right)=6x-6$
2. Simplify the derivative: $f’(x)=6(x–1)$
3. Analyze the sign of the derivative: Since 6(x – 1) is always positive for all $x\in \mathbb{R}$ ,  the function f(x) is increasing on $\mathbb{R}$.

Therefore, the function $f(x)=3x^2-6x+4$ is increasing on the set of real numbers $\mathbb{R}$.

Example 6: Determine the intervals on which the function $f(x)=\sin x-\cos x$ is increasing or decreasing for $0\le x\le 2\pi$

Solution: 

To find the intervals where the function $f(x)=\sin x-\cos x$ is increasing or decreasing, we need to analyze its first derivative.

1. Find the first derivative: $f^{\prime }(x)=\cos x+\sin x$
2. Determine where f’(x) is positive or negative: $f^{\prime }(x)>0\Rightarrow \cos x+\sin x>0$

$f^{\prime }(x)<0\Rightarrow \cos x+\sin x<0$

3. Solve the inequality: The points $\frac{\pi }{4}$ and $\frac{5\pi }{4}$ divide the interval [0,2 \pi] into three subintervals:

$\left[0,\frac{\pi }{4}\right],\left[\frac{\pi }{4},\frac{5\pi }{4}\right]$and $\left[\frac{5\pi }{4},2\pi \right]$

• For $0<x<\frac{\pi }{4},\cos x+\sin x>0$, so f(x) is increasing.
• For $\frac{\pi }{4}<x<\frac{5\pi }{4},\cos x+\sin x<0$, so f(x) is decreasing.
• For $\frac{5\pi }{4}<x<2\pi ,\cos x+\sin x>0$, so f(x) is increasing.

Therefore, the function $f(x)=\sin x-\cos x$ is increasing on $\left[0,\frac{\pi }{4}\right]$ and $\left[\frac{5\pi }{4},2\pi \right]$ and decreasing on $\left[\frac{\pi }{4},\frac{5\pi }{4}\right]$.

5.0Practice Questions on Applications of Derivatives

1. A spherical balloon is being inflated, causing its volume to increase at a rate of 10 cubic centimeters per second. Determine how quickly the radius of the balloon is growing when the radius measures 5 cm.
2. The radius of a circular oil spill is increasing at a rate of 0.1 meters per minute. How fast is the area of the oil spill increasing when the radius is 20 meters?
3. A conical tank is being filled with water at a speed of 5 cubic meters per minute. The tank has a height of 10 meters and a base radius of 3 meters. How fast is the water level rising when the water is 6 meters deep?

6.0Sample Questions on Applications of Derivatives

1. What is meant by the rate of change of a function?

Ans: The rate of change of a function y = f(x) w.r.t x is the derivative $\frac{dy}{dx}$. It measures how y changes as x varies. For example, in physics, if s(t) represents displacement as a function of time t, then $\frac{ds}{dt}$ gives the velocity.

2. How do you find the equation of a tangent line using derivatives?

Ans: To find the tangent line to a curve y = f(x) at a point (x₁, y₁):

1. Find the derivative f'(x), which gives the slope of the tangent at any point x.
2. Evaluate f'(x) at x₁ to get the slope m.
3. Use the point-slope form to express the equation of a line:  $y-y_1=m\left(x-x_1\right)$