Mathematics Notes for Class 12 Application Of-Derivatives

Application of Derivatives

Welcome to the chapter on Application of Derivatives for Class 12. In this chapter, you will learn how derivatives are used to solve real-world problems, including finding rates of change, maxima and minima, tangents and normals, and more. By the end of this chapter, you will be able to apply derivatives to various mathematical and practical situations.

Key Concepts

Derivative: Measures the rate at which a function changes.
Rate of Change: How one quantity changes with respect to another.
Tangent: A line that touches a curve at one point without crossing it.
Normal: A line perpendicular to the tangent at the point of contact.
Maxima and Minima: The highest and lowest points on a curve, respectively.

1. Rate of Change

The derivative of a function gives the rate of change of one variable with respect to another. For example, if \( y = f(x) \), then \( \frac{dy}{dx} \) gives the rate of change of \( y \) with respect to \( x \).

If \( s(t) \) is the position of an object at time \( t \), then \( \frac{ds}{dt} \) is its velocity.
If \( V(r) \) is the volume of a sphere of radius \( r \), then \( \frac{dV}{dr} \) gives how the volume changes as the radius changes.

2. Tangents and Normals

The slope of the tangent to the curve \( y = f(x) \) at \( x = a \) is \( f'(a) \).

Equation of tangent: \( y - f(a) = f'(a)(x - a) \)
Equation of normal: \( y - f(a) = -\frac{1}{f'(a)}(x - a) \) (if \( f'(a) \neq 0 \))

3. Increasing and Decreasing Functions

- A function is increasing where its derivative is positive.
- A function is decreasing where its derivative is negative.

If \( f'(x) > 0 \) for all \( x \) in an interval, \( f(x) \) is increasing there.
If \( f'(x) < 0 \) for all \( x \) in an interval, \( f(x) \) is decreasing there.

4. Maxima and Minima

Derivatives help find the highest (maximum) and lowest (minimum) values of a function.

Set \( f'(x) = 0 \) to find critical points.
Use the second derivative test: If \( f''(x) > 0 \), it's a minimum; if \( f''(x) < 0 \), it's a maximum.

Example: Find the maximum or minimum value of \( f(x) = x^2 - 4x + 3 \).

\( f'(x) = 2x - 4 \). Set \( 2x - 4 = 0 \Rightarrow x = 2 \).
\( f''(x) = 2 > 0 \), so \( x = 2 \) is a minimum point.

5. Application in Real Life

Finding the speed of a moving car (rate of change of distance).
Maximizing profit or minimizing cost in business problems.
Finding the best dimensions for a box with maximum volume.

Practice Questions

Find the equation of the tangent to the curve \( y = x^2 \) at \( x = 1 \).
If \( y = 3x^3 - 5x^2 + 2x \), find the intervals where the function is increasing.
Find the maximum or minimum value of \( f(x) = -x^2 + 4x + 1 \).
A box with a square base and open top must have a volume of 32 cm³. Find the dimensions that minimize the surface area.
If the radius of a circle increases at a rate of 2 cm/s, how fast is the area increasing when the radius is 5 cm?

Challenge Yourself

Prove that the minimum value of \( x^2 + \frac{1}{x^2} \) for \( x > 0 \) is 2.
A wire of length 100 cm is cut into two pieces. One piece is bent into a square and the other into a circle. How should the wire be cut to minimize the total area?

Did You Know?

Derivatives are used in physics, engineering, economics, and biology to solve real-world problems.
The concept of maxima and minima is used in designing roller coasters and bridges!

Glossary

Derivative: The rate at which a function changes at any point.
Tangent: A straight line that touches a curve at one point.
Normal: A line perpendicular to the tangent at the point of contact.
Critical Point: A point where the derivative is zero or undefined.
Maxima/Minima: The highest/lowest value of a function in a given interval.

Answers to Practice Questions

Slope at \( x=1 \) is \( 2 \). Equation: \( y - 1 = 2(x - 1) \) or \( y = 2x - 1 \).
Find \( y' = 9x^2 - 10x + 2 \). The function is increasing where \( y' > 0 \).
Maximum at \( x = 2 \), \( f(2) = -4 + 8 + 1 = 5 \).
Let base = \( x \), height = \( h \). \( x^2 h = 32 \). Minimize surface area: \( x^2 + 4xh \). Use calculus to solve.
\( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi \times 5 \times 2 = 20\pi \) cm²/s.

Mastering the application of derivatives helps you solve real-life and advanced mathematical problems!

Mathematics Class 12 - Application Of-Derivatives Notes