Calculus Exploration Derivatives, Normals, And Differential Equations

#title: Derivatives of Arctangent, Normals to Curves, and Solving Differential Equations

This comprehensive exploration delves into the intricacies of calculus, tackling problems ranging from finding the derivative of the arctangent function to determining the intersection of a normal line with the y-axis and solving differential equations. Understanding these concepts is crucial for anyone pursuing studies or a career in mathematics, physics, engineering, or related fields. Let's embark on this mathematical journey, unraveling each problem step-by-step.

9. Determining dy/dx given y = arctan(x)

Arctangent Derivatives are a fundamental concept in calculus. When faced with the problem of finding the derivative of y = arctan(x), we venture into the realm of inverse trigonometric functions and their differentiation rules. The arctangent function, also known as the inverse tangent function, is the inverse of the tangent function. To find dy/dx, we will utilize the derivative rule for inverse trigonometric functions. Let's begin by recalling the fundamental relationship between a function and its inverse. If y = arctan(x), then tan(y) = x. This seemingly simple transformation is the key to unlocking the derivative we seek. Now, we differentiate both sides of the equation tan(y) = x with respect to x. This is where the chain rule comes into play, a cornerstone of differential calculus. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Applying the chain rule, we get: d/dx [tan(y)] = d/dx [x]. The derivative of tan(y) with respect to y is sec²(y). Thus, using the chain rule, we have sec²(y) * dy/dx = 1. Our goal is to isolate dy/dx, so we rearrange the equation to get dy/dx = 1 / sec²(y). While this expression is mathematically correct, it's not in the most convenient form. We need to express sec²(y) in terms of x. Recall the trigonometric identity that connects secant and tangent: sec²(y) = 1 + tan²(y). This identity is a powerful tool in simplifying trigonometric expressions. Since we know that tan(y) = x, we can substitute x into the identity: sec²(y) = 1 + x². Now we can substitute this back into our expression for dy/dx: dy/dx = 1 / (1 + x²). This elegant and concise formula represents the derivative of the arctangent function. It's a crucial result to remember in calculus and its applications. In summary, to find the derivative of y = arctan(x), we used the relationship between the arctangent and tangent functions, implicit differentiation, the chain rule, and a trigonometric identity. This process showcases the interconnectedness of various calculus concepts and highlights the importance of mastering these fundamental techniques. The final result, dy/dx = 1 / (1 + x²), is a cornerstone in calculus and finds applications in numerous mathematical and scientific fields. Mastering the derivation of this result deepens one's understanding of inverse trigonometric functions and the power of calculus.

10. Determining the Coordinates of Point P

This question deals with Normals to Curves, requiring us to find the equation of a normal line and its intersection with the y-axis. We are given the curve y = 16/x - 4√x and the point (4, -4). The normal to a curve at a given point is a line perpendicular to the tangent at that point. To find the equation of the normal, we first need to determine the slope of the tangent. This involves finding the derivative of the curve's equation. Let's differentiate y = 16/x - 4√x with respect to x. We can rewrite the equation as y = 16x⁻¹ - 4x¹/². Applying the power rule of differentiation, we get: dy/dx = -16x⁻² - 2x⁻¹/². This can be rewritten as dy/dx = -16/x² - 2/√x. Now, we need to find the slope of the tangent at the point (4, -4). We substitute x = 4 into the derivative: dy/dx |_(x=4) = -16/4² - 2/√4 = -1 - 1 = -2. So, the slope of the tangent at (4, -4) is -2. The slope of the normal is the negative reciprocal of the tangent's slope. Therefore, the slope of the normal is 1/2. Now we have the slope of the normal and a point it passes through, (4, -4). We can use the point-slope form of a line to find the equation of the normal: y - y₁ = m(x - x₁). Substituting the values, we get: y - (-4) = (1/2)(x - 4). Simplifying, we get: y + 4 = (1/2)x - 2. Further simplifying, we get the equation of the normal: y = (1/2)x - 6. To find the point P where the normal intersects the y-axis, we set x = 0 in the equation of the normal: y = (1/2)(0) - 6 = -6. Therefore, the coordinates of point P are (0, -6). This problem highlights the connection between derivatives, tangents, and normals. Finding the derivative is crucial for determining the slope of the tangent, which in turn allows us to find the slope of the normal. The point-slope form of a line is a valuable tool for finding the equation of a line when a point and slope are known. Understanding these concepts and techniques is essential for solving problems involving curves and their properties in calculus.

Section 2: Long Answer Questions (60 Marks)

1. a) Solving Differential Equations

Differential Equations are at the heart of many scientific and engineering applications, describing the relationships between functions and their derivatives. Solving these equations allows us to model and understand a wide range of phenomena, from the motion of objects to the flow of heat. This particular problem likely involves a first-order differential equation, as it is the most common type encountered in introductory calculus courses. To tackle this, we'll need to identify the type of differential equation and apply the appropriate solution technique. Common methods for solving first-order differential equations include separation of variables, integrating factors, and recognizing exact equations. The method of separation of variables is particularly useful when the equation can be written in the form dy/dx = f(x)g(y). In this case, we can separate the variables by dividing both sides by g(y) and multiplying both sides by dx, resulting in the equation (1/g(y)) dy = f(x) dx. We then integrate both sides with respect to their respective variables. If the differential equation is not separable, we might consider using an integrating factor. This technique involves multiplying both sides of the equation by a carefully chosen function, called the integrating factor, which transforms the equation into an exact differential equation. An exact differential equation is one that can be written in the form M(x, y) dx + N(x, y) dy = 0, where ∂M/∂y = ∂N/∂x. If we can identify an integrating factor, we can solve the equation by finding a function F(x, y) such that ∂F/∂x = M and ∂F/∂y = N. The solution to the differential equation is then given by F(x, y) = C, where C is a constant. Let's assume, for the sake of illustration, that the differential equation we are dealing with is dy/dx + y = e⁻ˣ. This is a linear first-order differential equation, which can be solved using an integrating factor. The integrating factor is given by e^(∫P(x) dx), where P(x) is the coefficient of y in the differential equation. In this case, P(x) = 1, so the integrating factor is e^(∫1 dx) = eˣ. Multiplying both sides of the differential equation by eˣ, we get: eˣ dy/dx + eˣ y = e⁻ˣ eˣ = 1. The left-hand side of this equation is the derivative of yeˣ with respect to x: d/dx (yeˣ) = eˣ dy/dx + eˣ y. Therefore, we can rewrite the equation as: d/dx (yeˣ) = 1. Integrating both sides with respect to x, we get: yeˣ = ∫1 dx = x + C, where C is the constant of integration. Finally, we solve for y by dividing both sides by eˣ: y = (x + C)e⁻ˣ. This is the general solution to the differential equation. To find a particular solution, we would need an initial condition, such as the value of y at a specific value of x. In summary, solving differential equations requires a careful understanding of the different types of equations and the appropriate solution techniques. Whether using separation of variables, integrating factors, or other methods, the key is to manipulate the equation into a form that can be integrated to find the solution. Differential equations are powerful tools for modeling and understanding the world around us, and mastering their solution is crucial for many scientific and engineering disciplines.