Govt. College Of Engineering & Textile Technology, Serampore Mathematics III-A Past Paper PDF

GOVT. COLLEGE OF ENGINEEERING & TEXTILE TECHNOLOGY, SERAMPORE Topic: Implicit function, Explicit function & Chain Rule Name: Debosmita Maity Stream: CSE University Roll No: 11000123014 Year: 2nd...

GOVT. COLLEGE OF ENGINEEERING & TEXTILE TECHNOLOGY, SERAMPORE Topic: Implicit function, Explicit function & Chain Rule Name: Debosmita Maity Stream: CSE University Roll No: 11000123014 Year: 2nd Subject Name: Mathematics III-A Subject Code: BSC301  -ABSTRACT- The chain rule is a fundamental concept in calculus that relates the derivative of a composite function to the derivatives of its individual components. When dealing with implicit functions, the chain rule plays a critical role, as it allows for the differentiation of one variable with respect to another, even when the relationship between the variables is not explicitly defined. An implicit function is one where the dependent variable is not isolated, meaning it is defined by an equation involving both the dependent and independent variables (e.g., F(x, y) = 0 \)). In such cases, to differentiate the dependent variable with respect to the independent one, the chain rule must be applied to account for the indirect dependence. In contrast, an explicit function is directly expressed in the form y = f(x), where the dependent variable is explicitly written as a function of the independent variable. The chain rule for implicit differentiation involves treating the dependent variable as a function of the independent variable and applying the derivative to both sides of the equation, often leading to mixed partial derivatives. For explicit functions, the chain rule simplifies to the product of the derivative of the outer function and the inner function. This rule is essential for connecting the implicit relationships between variables and for solving complex differentiation problems involving composite functions.  -INTRODUCTION- -IMPLICIT FUNCTION- HISTORY: Implicit differentiation was developed by the famed physicist and mathematician Isaac Newton. He applied it to various physics problems he came across. In addition, the German mathematician Gottfried W. Leibniz also developed the technique independently of Newton around the same time period. INTRODUCTION: In calculus, functions are typically expressed in an explicit form, where the dependent variable is written directly in terms of the independent variable. However, not all functions can be neatly written in this way. Many important relationships between variables are given implicitly, meaning that the function isn't solved for one variable in terms of another but instead represents a general relationship between the two. For example, a circle can be expressed as x^2 + y^2 = r^2 , where y is not isolated as a function of x, but both are related through the equation. Implicit functions are crucial because many natural and physical phenomena are described using such relationships, where variables are interdependent in complex ways. For example, equations describing curves, surfaces, or constraints in physics and engineering often appear in implicit form. In these cases, differentiating the relationship to find rates of change between variables requires a special technique known as implicit differentiation.  Importance of Implicit Functions in Calculus: 1. Broader Application: Implicit functions allow us to work with equations that define curves or surfaces without the need to isolate one variable in terms of another. This expands the types of problems we can solve. 2. Implicit Differentiation: Implicit functions are differentiated using a technique called implicit differentiation, which allows us to find derivatives even when the function is not explicitly defined. This method is widely used in calculus, especially when dealing with relationships like circles, ellipses, and more complex curves that can't easily be solved for one variable. 3. Multivariable Calculus: Implicit functions become particularly important in multivariable calculus, where functions may depend on several variables at once. In such cases, solving for one variable explicitly is often impossible or impractical, making implicit methods essential. 4. Real-World Problems: Many real-world problems involve constraints or relationships that naturally occur in implicit form. For example, in physics, certain laws of motion and forces are defined implicitly, requiring the use of implicit differentiation to analyze their behaviour. Implicit functions form a foundational concept in calculus, helping to broaden our ability to model and solve complex problems that cannot be handled using explicit functions alone.  -EXPLICIT FUNCTION- HISTORY: The concept of explicit functions dates back to ancient civilizations, where basic relationships between quantities were studied. In the 17th century, with the development of calculus by Newton and Leibniz, explicit functions gained prominence as a way to describe one variable in terms of another. In the 18th century, Euler introduced the modern notation f(x), formalizing the concept. During the 19th century, mathematicians like Cauchy and Weierstrass rigorously developed function theory. Explicit functions remain fundamental in mathematics, particularly in calculus, physics, and engineering. INTRODUCTION: An explicit function is a function where the dependent variable is expressed directly in terms of the independent variable. For example, in the equation y = f(x), y is explicitly defined as a function of x. This form provides a clear and straightforward way to compute the value of one variable given another.  Importance of Explicit Functions in Calculus: 1. Ease of Differentiation and Integration: Explicit functions allow for direct application of differentiation and integration rules. For example, given y = 3x^2, finding the derivative (frac{dy}{dx}) is straightforward. 2. Simplification of Problem Solving: Explicit functions simplify the process of solving equations and analyzing functions. This makes it easier to find critical points, slopes, and areas under curves, which are fundamental operations in calculus. 3. Clear Representation of Relationships: They provide a clear and understandable way to express the relationship between variables, making it easier to model real-world phenomena. 4. Foundation of Calculus: Much of calculus relies on working with explicit functions, especially when applying rules like the power rule, chain rule, and product rule for differentiation, or basic integration techniques. In summary, explicit functions are vital in calculus because they simplify analysis, making differentiation, integration, and problem-solving more intuitive and efficient. -CHAIN RULE- HISTORY: The Chain Rule is thought to have first originated from the German mathematician Gottfried W. Leibniz. Although the memoir it was first found in contained various mistakes, it is apparent that he used chain rule in order to differentiate a polynomial inside of a square root. Guillaume de l'Hôpital, a French mathematician, also has traces of the chain rule in his Analyse des infiniment petits. INTRODUCTION: The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. It plays a crucial role in understanding how two or more functions relate to each other when composed. If you have two functions, say f(x) and g(x), and you form a composite function h(x) = f(g(x)), the chain rule helps you find the derivative of h(x) in terms of the derivatives of f(x) and g(x).  The Chain Rule Formula: If h(x) = f(g(x)), then the derivative of h(x), denoted as h'(x), is: h'(x) = f'(g(x)). g'(x) This can be interpreted as: - f'(g(x)): the derivative of the outer function evaluated at the inner function. - g'(x): the derivative of the inner function.  Why the Chain Rule is Important: 1. Composite Functions: Many real-world problems involve composite functions, where one variable depends on another, which in turn depends on a third. The chain rule provides a systematic way to differentiate such functions. 2. Applications in Physics and Engineering: In fields like physics, engineering, and economics, quantities often depend on multiple variables, where the chain rule helps compute rates of change (like velocity, acceleration, etc.) when these dependencies are linked. 3. Higher-order Derivatives: The chain rule is also useful in finding higher-order derivatives (like second or third derivatives) of composite functions. 4. Implicit Differentiation: In cases where functions are given implicitly, the chain rule is essential for differentiating them. In summary, the chain rule is a powerful tool for handling the differentiation of complex functions and is indispensable for solving a wide range of problems in calculus and its applications. -MAIN RESULTS- -IMPLICIT FUNCTION- An implicit function is a function that is defined not directly in terms of one variable as a function of another, but by an equation relating the variables. In contrast to an explicit function, where y is directly given as y = f(x), an implicit function has the form F(x, y) = 0. This means that the function y is not isolated on one side of the equation. For example, the equation of a circle ( x^2 + y^2 = r^2 ) is an implicit function. Here, y is not expressed explicitly in terms of x , but both variables are related by the equation.  Implicit Function Theorems Several of the problems in the text pertain to the Implicit Function Theorem.  The theorem give conditions under which it is possible to solve an equation of the form F(x, y) = 0 for y as a function of x. Implicit Function Theorem I Let F : D ⊂ ℝ2 → R and let (𝑥0 , 𝑦0 ) be an interior point of D with F(𝑥0 , 𝑦0 ) = 0. Suppose both first order partial derivatives of F exist in D and are continuous at (𝑥0 , 𝑦0 ) with 𝐹𝑌 (𝑥0 , 𝑦0 ) ≠ 0. Then there is an interval I ⊂ R with 𝑥0 an interior point of I and a function φ : I → R such that φ is differentiable on I, φ(𝑥0 ) = 𝑦0 and F(x, φ(x)) = 0 for each x ∈ I. The Chain Rule for Functions of Several Variables is then used to compute a formula for 𝜑 ′ (x). Because F(x, φ(x)) = 0 for all x ∈ I, the derivative of the function h(x) = F(x, φ(x)) is 0. But this derivative can also be computed using the Chain Rule for Functions of Several Variables. 0 = F1(x, φ(x)) + F2(x, φ(x)) 𝜑 ′ (x) Solving this equation for 𝜑 ′ (x) yields 𝜑 ′ (x) = −F1(x, φ(x))/F2(x, φ(x)) It is also possible to “solve” an equation of the form F(x, y, z) for z as a function of z and y. Implicit Function Theorem II Let F : D ⊂ ℝ3 → R and let (𝑥0 , 𝑦0 ,𝑧0 ) be an interior point of D with F(𝑥0 , 𝑦0 ,𝑧0 ) = 0. Suppose all first order partial derivatives of F exist in D and are continuous at (𝑥0 , 𝑦0 ,𝑧0 ) with 𝐹𝑧 (𝑥0 , 𝑦0 ,𝑧0 )≠0. Then there is a rectangle R ⊂ ℝ2 with (𝑥0 , 𝑦0 ) an interior point of R and a function φ : R → R such that φ has continuous first order partial derivatives on R, φ(𝑥0 , 𝑦0 ) =𝑧0 and F((x, y, φ(x, y)) = 0 for each (x, y) ∈ R. As before the Chain Rule for Functions of Several Variables is then used to derive a formulas for 𝜑1 (x, y) and for 𝜑2 (x, y). Because F((x, y), φ(x, y)) = 0 for all (x, y) ∈ R, the partial derivatives of the function h(x, y) = F((x, y, φ(x, y)) are 0. But these partial derivatives can also be computed using the Chain Rule for Functions of Several Variables. First compute the partial derivative with respect to x. 0 = F1((x, y, φ(x, y)) + F3((x, y, φ(x, y)) 𝜑1 (x, y) Solving this equation for 𝜑1 (x, y) yields 𝜑1 (x, y) = −F1((x, y, φ(x, y))/F3((x, y, φ(x, y)) As an exercise, you are encouraged to show that 𝜑2 (x, y) = −F2((x, y, φ(x, y))/F3((x, y, φ(x, y)) These types of applications of the Chain Rule for Functions of Several Variables are typical of its use. It isn’t used to compute specific partial derivatives. The Chain Rule from first semester calculus is sufficient for that purpose.  Example Find the slope of the tangent line to the circle 𝒙𝟐 + 𝒚𝟐 = 25 at the point (3, 4). What is the slope at the point (3, −4)? Answer. Differentiating both sides of the equation x 2 + y 2 = 25 with respect to x, you get  ⅆ𝑌 2x + 2y ⅆ𝑋= 0 ⅆ𝑌 = − x/y ⅆ𝑋 The slope at (3, 4) is the value of the derivative when x = 3 and y = 4: ⅆ𝑦 𝑥 𝑥−3 | = − 𝑦 𝑙𝑦=−4 − = − 3/4 ⅆ𝑥 (3,4) Similarly, the slope at (3, −4) is the value of dy dx when x = 3 and y = −4: ⅆ𝑦 | = − x/y = − 3/−4 = 3/4 ⅆ𝑥 (3,4) The graph of the circle is shown together with the tangent lines at (3, 4) and (3, −4). -EXPLICIT FUNCTION- An explicit function is a function in algebra in which the output variable (dependent variable) can be explicitly written only in terms of the input variable (independent variable). An explicit function usually involves two variables - dependent and independent variables. It is expressed more clearly and hence, it is easy to determine the values of the variables in an explicit function. On the other hand, a function that cannot be written as one variable in terms of the other variable is called an implicit function. In mathematics, an explicit function is defined as a function in which the dependent variable can be explicitly written in terms of the independent variable. In standard form, we can write an explicit function as y = f(x), where y is the output variable expressed completely in terms of the input variable x. In an explicit function, it is easy to find the value of the function corresponding to each value as the dependent variable is expressed clearly in terms of the input.  Derivative of Explicit Function The derivative of an explicit function is done regularly just like simple differentiation of algebraic functions. An explicit function is written as y = f(x), where x is an input and y is an output. The differentiation of y = f(x) with respect to the input variable is written as y' = f'(x). So, simple rules of differentiation are applied to determine the derivative of an explicit function. Let us solve a few examples to understand finding the derivatives. Example 1: Find the derivative of the explicit function y = 𝑥 2 + sin x - x + 4. Solution: To find the derivative of y = 𝑥 2 + sin x - x + 4, we will differentiate both sides w.r.t. x. dy/dx = 2x + cos x - 1 Hence, the differentiation of y = x2 + sin x - x + 4 is dy/dx = 2x + cos x - 1. Example 2: Find the derivative of the function xy - y = 0 Solution: First, we will express the given function explicitly. xy - y = 0 ⇒ y (x - 1) = 0 ⇒ y = 1/(x - 1) Now, we have expressed the given function as an explicit function. Next, we differentiate both sides of the function w.r.t. x. dy/dx = -1(𝑥 − 1)2 Hence, the derivative of the given function is dy/dx = -1(𝑥 − 1)2 IMPLICIT FUNCTION EXPLICIT FUNCTION An implicit function is a function with An explicit function is defined as a function in several variables, and one of the variables is which the dependent variable can be explicitly a function of the other set of variables. written in terms of the independent variable. General form of Implicit Function: f(x, y) = General Form of Explicit Function: y = f(x) 0 Example: xy + 2x - tan (xy) + 𝑦 2 = 0 Example: y = x + 2 -CHAIN RULE- The chain rule is a fundamental rule in calculus used to compute the derivative of a composite function. In essence, the chain rule states that if a function depends on an intermediate variable, and that intermediate variable depends on another variable, then the derivative of the function can be found by multiplying the derivative of the function with respect to the intermediate variable by the derivative of the intermediate variable with respect to the original variable. Chain Rule for a Function of Two Variables Suppose z = f(x, y), and both x and y are functions of another variable t, i.e., x = g(t) and y = h(t). The chain rule for a function of two variables gives the derivative of z with respect to t as: ⅆ𝑧 ⅆ𝑧 ⅆ𝑥 ⅆ𝑧 ⅆ𝑦 = ⅆ𝑥. + ⅆ𝑦. ⅆ𝑡 ⅆ𝑡 ⅆ𝑡  Example: ⅆ𝑧 Let z = 𝑥 2 + 𝑌 2 , where x = cos t and y = sin t. To find ⅆ𝑡 :  𝜕𝑧 𝜕𝑧 1. Partial derivatives of z: 𝜕𝑥 = 2x, 𝜕𝑦 = 2y ⅆ𝑥 ⅆ𝑦 2. Derivatives of x and y: = - sin t, = cos t ⅆ𝑡 ⅆ𝑡 ⅆ𝑧 3. Apply the chain rule: = 2x. (-\sin t) + 2y. cos t ⅆ𝑡 ⅆ𝑧 4. Substitute x = cos t and y = sin t: = 2cos t. (-\sin t) + 2sin t. cos t ⅆ𝑡 ⅆ𝑧 5. Simplify: =0 ⅆ𝑡 Chain Rule for a Function of Three Variables For a function z = f(x, y, u), where x = g(t), y = h(t), u = k(t), the chain rule gives the derivative of z with respect to t as: ⅆ𝑧 𝜕𝑧 ⅆ𝑥 𝜕𝑧 ⅆ𝑦 𝜕𝑧 ⅆ𝑢 = 𝜕𝑥. + 𝜕𝑦. + 𝜕𝑢. ⅆ𝑡 ⅆ𝑡 ⅆ𝑡 ⅆ𝑡  Example: ⅆ𝑧 Let z = 𝑥 2 + 𝑦 2 + 𝑢2 , where x = t, y = sin t , and u = cos t. To find ⅆ𝑡 : 𝜕𝑧 𝜕𝑧 𝜕𝑧 1. Partial derivatives of z: = 2x, 𝜕𝑦 = 2y, 𝜕𝑢 = 2u 𝜕𝑥 ⅆ𝑥 ⅆ𝑦 ⅆ𝑢 2. Derivatives of x, y, and u: = 1, = cos t, = -\sin t ⅆ𝑡 ⅆ𝑡 ⅆ𝑡 ⅆ𝑧 3. Apply the chain rule: = 2x. 1 + 2y. cos t + 2u. (-\sin t) ⅆ𝑡 ⅆ𝑧 4. Substitute x = t, y = sin t, and u = cos t: = 2t + 2sin t. cos t + 2cos t. (-\sin t) ⅆ𝑡 ⅆ𝑧 5. Simplify: ⅆ𝑡 = 2t -CONCLUSIONS- Explicit Functions: An explicit function is one where the dependent variable is expressed directly in terms of the independent variable(s). For example, y = f(x) is an explicit function because y is written explicitly as a function of x. In such cases, derivatives are straightforward and use regular differentiation techniques. Implicit Functions: In implicit functions, the dependent and independent variables are not separated. Instead, they are related by an equation like F(x, y) = 0, where solving explicitly for y in terms of x may not be easy or even possible. Implicit differentiation is used to find the derivative of one variable in terms of another, typically involving partial derivatives. Chain Rule: The chain rule is a powerful tool for finding derivatives of composite functions—whether they are explicitly defined or involve implicit relations between variables. The chain rule allows us to handle functions that depend on intermediate variables and helps compute derivatives when multiple variables are interconnected.  -REFERENCES- Here’s a brief explanation of each reference on implicit functions, explicit functions, and the chain rule: 1. Class Notes 2."Calculus: Early Transcendentals" by James Stewart - This textbook is well-known for its clear explanations and numerous examples. It covers both explicit and implicit functions, detailing how to differentiate them. The chain rule is discussed extensively, with applications in various contexts. 3. "Calculus" by Michael Spivak - A more rigorous and theoretical approach to calculus, this book includes detailed discussions on implicit and explicit functions. It provides proofs and problems that deepen understanding of the chain rule and its applications in calculus. 4. "Calculus: A Complete Course" by Robert A. Adams and Christopher Esse - This book provides a clear explanation of the differences between implicit and explicit functions, along with a strong focus on the chain rule. It includes numerous examples and applications to help students grasp these concepts. 5. Khan Academy - A free educational platform that offers videos and exercises covering a variety of calculus topics, including explicit and implicit differentiation and the chain rule. It's an excellent resource for visual and interactive learning. 6. Paul's Online Math Notes - This website provides comprehensive notes on calculus topics, including implicit differentiation and the chain rule. It includes explanations, examples, and practice problems, making it a great supplementary resource for students. 7. "Implicit Functions and the Chain Rule" - This article discusses the theoretical aspects of implicit functions and the chain rule, emphasizing their importance in calculus education. It may provide insights into teaching methods and practical applications. 8. "The Role of Implicit Differentiation in the Teaching of Calculus" - This research article examines the pedagogical approaches to teaching implicit differentiation, highlighting its significance in understanding the broader concepts of calculus. It discusses methods for improving student comprehension of these topics. These references collectively cover a range of explanations, examples, and applications of implicit functions, explicit functions, and the chain rule, providing a solid foundation for understanding these essential calculus concepts.