#Introduction
In the realm of mathematics, particularly in calculus and analytical geometry, curves defined by implicit equations present fascinating challenges. One such challenge involves finding the normal to a curve at a specific point and determining its y-intercept. This article delves into the problem of finding the value of c for the curve defined by the equation xe^(-y) + e^(y) = 1 + x, given that the normal to the curve at the point (c, ln c) has a y-intercept of c^2 + 1. This exploration will involve a step-by-step approach, utilizing concepts from differential calculus, including implicit differentiation and the properties of normal lines. Understanding the behavior of curves and their tangents is crucial in various fields, from physics and engineering to computer graphics and economics. This problem serves as an excellent example of how mathematical concepts intertwine to provide solutions to complex scenarios. We will meticulously analyze the given equation, differentiate it implicitly, and use the information about the y-intercept of the normal to establish a relationship that will allow us to solve for c. This exercise not only reinforces fundamental calculus skills but also highlights the power of analytical thinking in problem-solving. Let's embark on this mathematical journey to unravel the intricacies of the curve and discover the value of c that satisfies the given conditions.
Problem Statement
We are given the curve defined by the equation xe^(-y) + e^(y) = 1 + x. Our task is to find the value of c such that the normal to the curve at the point (c, ln c) has a y-intercept of c^2 + 1. This problem elegantly combines concepts from calculus and coordinate geometry. The equation of the curve is an implicit equation, meaning that y is not explicitly defined as a function of x. To find the slope of the tangent, and subsequently the normal, we will need to employ implicit differentiation. The point (c, ln c) lies on the curve, which means that substituting these coordinates into the equation of the curve should satisfy the equation. This provides us with an initial condition that can be used to verify our solution. The normal to the curve at a point is a line perpendicular to the tangent at that point. The slope of the normal is the negative reciprocal of the slope of the tangent. The y-intercept of a line is the point where the line intersects the y-axis, which occurs when x = 0. The given y-intercept of c^2 + 1 provides us with another condition that we can use to solve for c. By carefully combining these concepts and conditions, we can establish an equation that allows us to determine the value of c. This problem showcases the interplay between different mathematical concepts and the importance of a systematic approach to problem-solving.
Implicit Differentiation
To find the slope of the tangent to the curve, we need to differentiate the equation xe^(-y) + e^(y) = 1 + x implicitly with respect to x. Implicit differentiation is a technique used when y is not explicitly defined as a function of x. It involves differentiating both sides of the equation with respect to x, treating y as a function of x and using the chain rule when differentiating terms involving y. Let's begin by differentiating each term of the equation. The derivative of xe^(-y) with respect to x requires the product rule, which states that the derivative of uv is u'v + uv', where u and v are functions of x. In this case, u = x and v = e^(-y). The derivative of x with respect to x is 1, and the derivative of e^(-y) with respect to x is -e^(-y)(dy/dx) by the chain rule. Therefore, the derivative of xe^(-y) is e^(-y) - xe^(-y)(dy/dx). Next, we differentiate e^(y) with respect to x. Using the chain rule, the derivative is e^(y)(dy/dx). Now, we differentiate the right side of the equation. The derivative of 1 with respect to x is 0, and the derivative of x with respect to x is 1. Thus, the derivative of 1 + x is 1. Putting it all together, we have the equation e^(-y) - xe^(-y)(dy/dx) + e^(y)(dy/dx) = 1. This equation relates dy/dx to x and y. We can now solve for dy/dx, which represents the slope of the tangent to the curve at any point (x, y). This process of implicit differentiation is a cornerstone of calculus and is essential for analyzing curves defined by implicit equations. Understanding the chain rule and the product rule is crucial for performing implicit differentiation accurately.
Finding dy/dx
Now, let's isolate dy/dx in the equation e^(-y) - xe^(-y)(dy/dx) + e^(y)(dy/dx) = 1. Our goal is to express dy/dx as a function of x and y. We begin by rearranging the terms to group the terms containing dy/dx on one side of the equation and the other terms on the other side. This gives us e^(y)(dy/dx) - xe^(-y)(dy/dx) = 1 - e^(-y). Next, we factor out dy/dx from the left side of the equation: dy/dx(e^(y) - xe^(-y)) = 1 - e^(-y). To solve for dy/dx, we divide both sides of the equation by (e^(y) - xe^(-y)): dy/dx = (1 - e^(-y)) / (e^(y) - xe^(-y)). This expression gives us the slope of the tangent to the curve at any point (x, y). However, we can simplify this expression further to make it easier to work with. Multiplying the numerator and denominator by e^(y), we get: dy/dx = (e^(y) - 1) / (e^(2y) - x). This simplified expression for dy/dx is more compact and easier to evaluate. It represents the slope of the tangent line at any point (x, y) on the curve. This derivative is crucial for finding the equation of the tangent and the normal to the curve at a specific point. The ability to manipulate and simplify algebraic expressions is an important skill in calculus and is essential for obtaining accurate results. This simplified form of dy/dx will be used in the subsequent steps to find the slope of the normal and determine the value of c.
Slope of the Normal
Having found the slope of the tangent, dy/dx = (e^(y) - 1) / (e^(2y) - x), we can now determine the slope of the normal to the curve at the point (c, ln c). The normal to a curve at a point is perpendicular to the tangent at that point. The slopes of two perpendicular lines are negative reciprocals of each other. Therefore, if m_t is the slope of the tangent and m_n is the slope of the normal, then m_n = -1/m_t. To find the slope of the tangent at the point (c, ln c), we substitute x = c and y = ln c into the expression for dy/dx: m_t = (e^(ln c) - 1) / (e^(2 ln c) - c). Since e^(ln c) = c and e^(2 ln c) = e^(ln c^2) = c^2, we have: m_t = (c - 1) / (c^2 - c). We can simplify this expression by factoring out a c from the denominator: m_t = (c - 1) / (c(c - 1)). Assuming c ≠ 1, we can cancel the (c - 1) terms: m_t = 1/c. Now, we can find the slope of the normal, m_n, by taking the negative reciprocal of m_t: m_n = -1 / (1/c) = -c. Thus, the slope of the normal to the curve at the point (c, ln c) is -c. This result is crucial for finding the equation of the normal line, which we will use to determine the value of c. The relationship between the slopes of perpendicular lines is a fundamental concept in coordinate geometry and is essential for solving problems involving tangents and normals to curves.
Equation of the Normal
Now that we have the slope of the normal, m_n = -c, and the point (c, ln c) through which it passes, we can find the equation of the normal line. The point-slope form of a line is given by y - y_1 = m(x - x_1), where (x_1, y_1) is a point on the line and m is the slope of the line. In our case, (x_1, y_1) = (c, ln c) and m = -c. Substituting these values into the point-slope form, we get: y - ln c = -c(x - c). This is the equation of the normal line. We can rewrite this equation in slope-intercept form (y = mx + b) to find the y-intercept. Expanding the right side, we have: y - ln c = -cx + c^2. Adding ln c to both sides, we get: y = -cx + c^2 + ln c. This is the equation of the normal line in slope-intercept form, where the slope is -c and the y-intercept is c^2 + ln c. We are given that the y-intercept of the normal is c^2 + 1. Therefore, we can equate the two expressions for the y-intercept: c^2 + ln c = c^2 + 1. This equation allows us to solve for c. The ability to find the equation of a line given its slope and a point is a fundamental skill in coordinate geometry. The point-slope form is a versatile tool for finding the equation of a line, and the slope-intercept form provides a clear representation of the slope and y-intercept.
Solving for c
We have the equation c^2 + ln c = c^2 + 1, which we obtained by equating the two expressions for the y-intercept of the normal line. To solve for c, we can subtract c^2 from both sides of the equation: ln c = 1. To isolate c, we can take the exponential of both sides of the equation. The exponential function is the inverse of the natural logarithm function, so e^(ln c) = c. Thus, we have: c = e^1 = e. Therefore, the value of c that satisfies the given conditions is e, which is approximately 2.71828. This result indicates that the normal to the curve xe^(-y) + e^(y) = 1 + x at the point (e, ln e) = (e, 1) has a y-intercept of e^2 + 1. We can verify this result by substituting c = e back into the equation of the normal line: y = -ex + e^2 + ln e = -ex + e^2 + 1. The y-intercept is indeed e^2 + 1, which confirms our solution. This problem demonstrates how different mathematical concepts, such as implicit differentiation, coordinate geometry, and algebraic manipulation, come together to solve a complex problem. The ability to solve equations involving logarithms and exponentials is an essential skill in mathematics and its applications. The value of c = e is a significant result that provides a complete solution to the problem.
Conclusion
In conclusion, we have successfully determined the value of c for the curve xe^(-y) + e^(y) = 1 + x, given that the normal to the curve at the point (c, ln c) has a y-intercept of c^2 + 1. Through a step-by-step approach involving implicit differentiation, finding the slope of the normal, and using the point-slope form to determine the equation of the normal line, we arrived at the equation ln c = 1, which yielded the solution c = e. This problem showcased the power of calculus and analytical geometry in solving complex problems involving curves and their properties. The process involved a combination of techniques, including implicit differentiation, finding the slope of a perpendicular line, and solving equations involving logarithms and exponentials. The solution c = e is a specific value that satisfies the given conditions, providing a complete and accurate answer to the problem. This exercise reinforces the importance of a systematic approach to problem-solving and the ability to connect different mathematical concepts. The understanding of tangents and normals to curves is crucial in various applications, including optimization problems, curve sketching, and physical simulations. The skills developed in solving this problem are valuable in a wide range of mathematical and scientific contexts. This exploration serves as a testament to the beauty and power of mathematics in unraveling the intricacies of the world around us.