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RS Aggarwal Solutions

RS Aggarwal Class 12 Solutions Chapter-11 Applications of Derivatives

RS Aggarwal Class 12 Solutions Chapter-11 Applications of Derivatives

Class 12 RS Aggarwal Chapter-11 Applications of Derivatives Solutions - Free PDF Download

Students of CBSE preparing for the 12th Board examination can benefit from RS Aggarwal's application of derivatives solutions. It explains all the concepts outlined in the chapter in an easy and understandable manner. By taking references, students can adequately prepare for exams from this chapter. All the concepts, formulas, and exercises listed in the RS Aggarwal Class 12 Solutions Application of Derivatives make the topic more accessible to the students. Moreover, they can download the questions and solutions in an online PDF version. This will assist in their exam preparation and enable them to score outstanding marks in the examination. The subject matter experts are involved in designing the length-wise content of this chapter. With a proper review of these questions, students are going to benefit from it.

RS Aggarwal Class 12 Solutions free PDF for download is available both on the website of Vedantu and its mobile application.

RS Aggarwal Class 12 Solutions - Free PDF Download

Application of Derivatives

The application of derivatives is used for the determination of various things like Rolle’s Theorem, Lagrange’s Mean Value Theorem, derivatives, finding the rate of change, intervals decreasing or increasing, display increase or decrease in whole domain, finding the slope of the tangent, the approximate value of numbers and many other things. It will explain derivatives from composite function, inverse trigonometric function, implicit function, and exponential function. All these are explained in engineering and social sciences. Other concepts discussed here are:

Equation of tangent.
Maximum and minimum values of the function.
Interval increases and decreases.
Approximation and errors.

Advantages of using RS Aggarwal Class 12 Application of Derivatives Solutions to prepare for the Exams

RS Aggarwal Application of Derivatives Solutions provides a good resource and study material for exam preparation. All the questions and solutions designed in this textbook are in accordance with the CBSE exam pattern and syllabus. Students can take leverage from this RS Aggarwal book to understand the practical questions that come in the examination. One can also take help from this book in solving their daily homework, practice set, answering model questions, and others. Some of the advantages are explained to the students.

Easy to access and ready-made solutions for every chapter.
Simple and student-friendly explanations to help them in solving the problems quickly.
It incorporates all the chapters included in the Class 12 Maths books for CBSE.
Chapter-wise solutions are provided in detail for easy comprehension.
All the solutions are based on the last ten years of question patterns by the subject experts.

Exercise Questions in RS Aggarwal Class 12 Solutions Application of Derivatives

Question 1

Sol: The side of a square is c.

The rate of change of a side = dc/dt = 0.4 cm/s

Therefore, the perimeter of the square = 4c

The rate of change in the perimeter of the square= dc/dt = 4x 0.4

So, dc/dt = 1.6 cm/s

Question 2

Sol: The radius of a circle= m.

The circumference of the circle = 2πr

Given, the rate of change of a side = dm/dt = 0.76 cm/s

Therefore, the rate of change of circumference of the circle = 2πr dm/dt

The rate of change in the perimeter of the square= 2 x 1.3 x 0.76

So, dm/dt = 1.976 cm/s

Question 3

Sol: A globe is in the shape of a sphere

Let’s assume the radius of the globe is represented with s

The surface area of the globe sphere = 4πr²

ds/dt = 0.43 cm/s

Therefore, the rate of change of the surface area of the globe sphere = 8πr² ds/dt

= 8 x 1.7 x 7 x 0.43

ds/dt = 40.936 cm²/s

Question 4

The radius of a circle is expanding evenly at the rate of 0.2 cm/s. State at which rate is the area increasing when the radius is 20 cm? (Take π = 3.14.)

Sol: Consider the radius of the circle as c

dc/dt = 0.2 cm/s

We are already familiar, that area of the circle = πr2

Here the rate of change of area = 2πr ds/dt

By putting the values

= 2 × 3.14 × 20 × 0.2

dc/dt = 25.12 cm²/s

The RS Aggarwal Class 12 Solutions Application of Derivatives is also helpful for students making advance preparation in higher studies. The concept discussed here is really helpful for various examinations like JEE, NEET, BITSAT, etc.

FAQs on RS Aggarwal Class 12 Solutions Chapter-11 Applications of Derivatives

1. How do the RS Aggarwal Class 12 Solutions for Chapter 11 help in preparing for the CBSE board exams?

The RS Aggarwal Class 12 Solutions for Applications of Derivatives are highly beneficial for CBSE board exam preparation. They provide a wide variety of problems that are solved in a step-by-step format, which helps in understanding the correct methodology as per CBSE guidelines. By practising these solutions, students can master complex topics and improve their speed and accuracy in solving questions on tangents, normals, maxima, and minima.

2. What key topics from Applications of Derivatives are covered in the RS Aggarwal Class 12 Chapter 11 solutions?

The solutions for Chapter 11 cover all essential topics prescribed in the CBSE syllabus for Applications of Derivatives. Key areas explained with detailed solved examples include:

The rate of change of quantities
Determining intervals where functions are increasing or decreasing
Finding the equations of tangents and normals to a curve
Using derivatives for approximations
Solving problems on maxima and minima using the first and second derivative tests

3. Why is the step-by-step method in RS Aggarwal solutions crucial for solving maxima and minima problems?

Problems on maxima and minima involve multiple logical steps, and a small error can lead to an incorrect result. The step-by-step method is crucial because it breaks down the process into manageable parts:

Formulating the function to be maximised or minimised.
Finding the first derivative and identifying critical points.
Applying the second derivative test to classify these points as maxima, minima, or points of inflection.

This structured approach helps prevent common mistakes and builds a clear understanding of the underlying logic.

4. How do the RS Aggarwal solutions for Chapter 11 explain the process of finding the equations of tangents and normals?

The solutions provide a clear, methodical approach. First, they demonstrate how to find the slope of the tangent by calculating the derivative (dy/dx) of the curve's equation at the given point. Then, using the point-slope form, the equation of the tangent is derived. For the normal, the solutions explain how to find its slope by taking the negative reciprocal of the tangent's slope (-1/m) and then applying the same point-slope formula.

5. How do the solutions help differentiate between when to use the first derivative test versus the second derivative test?

The RS Aggarwal solutions clarify this common point of confusion through varied examples. They show that the Second Derivative Test is often more efficient when finding the second derivative is simple. However, if the second derivative is complex to compute or equals zero at a critical point, the solutions demonstrate the application of the First Derivative Test. This test involves checking the sign of the first derivative on either side of the critical point to determine if it's a local maximum or minimum.

6. What approach do the RS Aggarwal solutions take for problems involving the rate of change of quantities?

For rate-of-change problems, the solutions follow a systematic approach. They guide students to first identify the variables involved and establish a relationship between them (e.g., the formula for the area of a circle, A = πr²). Next, they demonstrate how to differentiate this equation with respect to time (t), applying the chain rule. Finally, they show how to substitute the given rates and values to find the required rate of change.

7. Beyond just finding answers, how do these solutions build a conceptual understanding of increasing and decreasing functions?

Instead of just providing a final interval, the solutions build conceptual clarity by illustrating the connection between the derivative and the function's behaviour. They show how finding where the derivative f'(x) is positive corresponds to the intervals where the function is increasing. Conversely, they show that where f'(x) is negative, the function is decreasing. This reinforces the core concept that the derivative represents the slope of the function's graph.

8. Are the questions in RS Aggarwal for Applications of Derivatives sufficient for competitive exams like JEE Main?

RS Aggarwal provides an excellent and comprehensive foundation for competitive exams like JEE Main. The extensive range of problems helps build strong problem-solving skills in all key areas of Applications of Derivatives. While it covers the necessary fundamentals thoroughly, students aiming for top ranks in exams like JEE Advanced might benefit from supplementing their practice with books that focus on more complex, multi-concept problems.