Problem 26 - Calculus Single Variable

Question

Use implicit differentiation to find the normal line to the given curve at the given point \(P_{0}\). \(x^{4}-8 y+y^{6}=9 \quad P_{0}=(2,1)\)

Step-by-Step Solution

Answer

The normal line is \(y = -\frac{1}{16}x + \frac{9}{8}\).

Step 1: Differentiate implicitly

First, take the derivative of both sides of the equation \(x^4 - 8y + y^6 = 9\) with respect to \(x\). When differentiating \(y\)-terms, apply the chain rule since \(y\) is a function of \(x\). The derivative is: \(4x^3 - 8 \frac{dy}{dx} + 6y^5 \frac{dy}{dx} = 0\).

Step 2: Solve for \(\frac{dy}{dx}\)

Rearrange the differentiated equation to solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx}(6y^5 - 8) = -4x^3\). Then, divide both sides by \(6y^5 - 8\) to isolate the derivative: \(\frac{dy}{dx} = \frac{-4x^3}{6y^5 - 8}\).

Step 3: Find \(\frac{dy}{dx}\) at point \(P_0=(2,1)\)

Substitute \(x = 2\) and \(y = 1\) into the derivative: \(\frac{dy}{dx} = \frac{-4(2)^3}{6(1)^5 - 8}\). Calculate: \(= \frac{-32}{6 - 8} = \frac{-32}{-2} = 16\). This is the slope of the tangent line at point \(P_0\).

Step 4: Determine the slope of the normal line

The slope of the normal line is the negative reciprocal of the tangent slope. Since the tangent slope is \(16\), the normal slope is \(-\frac{1}{16}\).

Step 5: Equation of the normal line

Use the point-slope form of a line to find the equation of the normal line using point \((2, 1)\) and the normal slope \(-\frac{1}{16}\). The equation is: \(y - 1 = -\frac{1}{16}(x - 2)\). Simplify this to get the normal line equation: \(y = -\frac{1}{16}x + \frac{1}{8} + 1\). Final equation: \(y = -\frac{1}{16}x + \frac{9}{8}\).

Key Concepts

Normal Line

The concept of a normal line is fundamental in understanding the relationships between curves and the lines that intersect them at specific points. A normal line to a curve at a particular point is perpendicular to the tangent line at that point. To determine the equation of the normal line, we first need to find the slope of the tangent line using implicit differentiation, which helps in dealing with curves defined implicitly, like in this exercise. Once we find the slope of the tangent line, the slope of the normal line is simply the negative reciprocal. This is because the product of the slopes of two perpendicular lines is -1. In our exercise, the tangent slope at the point (2,1) was found to be 16. Therefore, the slope of the normal line is -\(\frac{1}{16}\). Afterwards, by using the point-slope formula of a line, we can easily derive the equation of the normal line at the specified point.

Calculus

Calculus forms the backbone of mathematical techniques used in this exercise. The process of finding the tangent or normal lines involves differentiating equations, often requiring implicit differentiation when dealing with equations not explicitly solved for one variable in terms of the other. In this example, the equation \(x^4 - 8y + y^6 = 9\) is not easily rearranged to express \(y\) explicitly as a function of \(x\), hence implicit differentiation is necessary. This technique involves differentiating both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\). By applying implicit differentiation, we can find the derivative \(\frac{dy}{dx}\), which represents the slope of the tangent line. Understanding basic calculus principles like derivatives and their applications in geometry allows us to solve complex problems involving curves.

Chain Rule

In calculus, the chain rule is a powerful tool for differentiating composite functions, like those where \(y\) depends on another variable such as \(x\). In the context of implicit differentiation, we frequently encounter expressions that involve both \(x\) and \(y\), necessitating the chain rule.For the given equation \(x^4 - 8y + y^6 = 9\), when differentiating with respect to \(x\), each term with \(y\) requires the application of the chain rule because \(y\) is treated as a function of \(x\). This means that every direct differentiation of \(y\) with respect to \(x\) is also multiplied by \(\frac{dy}{dx}\), reflecting the dependence of \(y\) on \(x\). This application highlights the importance of the chain rule, allowing us to effectively handle complex forms and extract necessary information like derivatives for implicit functions.