Problem 20 - Calculus of a Single Variable

Question

(a) find two explicit functions by solving the equation for in terms of (b) sketch the graph of the equation and label the parts given by the corresponding explicit functions, (c) differentiate the explicit functions, and (d) find \(d y / d x\) implicitly and show that the result is equivalent to that of part (c). \(x^{2}+y^{2}-4 x+6 y+9=0\)

Step-by-Step Solution

Answer

The explicit functions are \(y=-3+\sqrt{4-(x-2)^{2}}\) and \(y=-3-\sqrt{4-(x-2)^{2}}\). After differentiation, for the upper half of the circle we have \(\frac{dy}{dx}=-\frac{x-2}{\sqrt{4-(x-2)^{2}}}\) and for the lower part, \(\frac{dy}{dx}=\frac{x-2}{\sqrt{4-(x-2)^{2}}}\). Implicit differentiation of the initial equation also provides the same results hence verifying our solutions.

Step 1: Rearrange the equation

First, rewrite the given equation, \(x^{2}-4 x+y^{2}+6 y+9=0\), in standard forms of a circle equation \(x^{2}-4 x+4+y^{2}+6 y+9=4\). This is achieved by completing the square. It can be further simplified to \((x-2)^2+(y+3)^2=4\).

Step 2: Determine explicit functions

Solving for \(y\) in terms of \(x\) gives two explicit functions in accordance with the form of the circle equation. This gives: \(y=-3+\sqrt{4-(x-2)^{2}}\) which represents the 'upper' half of the circle and \(y=-3-\sqrt{4-(x-2)^{2}}\) which represents the 'lower' half of the circle.

Step 3: Differentiate the explicit functions

Taking derivative of both functions using chain rule we get \(\frac{dy}{dx}=-\frac{x-2}{\sqrt{4-(x-2)^{2}}}\) for the 'upper' half and \(\frac{dy}{dx}=\frac{x-2}{\sqrt{4-(x-2)^{2}}}\) for the 'lower' half.

Step 4: Implicit differentiation and verification

Differentiate the original equation implicitly, treat \(y\) as a function of \(x\). The derivative of \(x^{2}\) with respect to \(x\) is \(2x\), and derivative of \(y^{2}\) with respect to \(x\) is \(2y\frac{dy}{dx}\) due to the chain rule. Similarly, find the derivates of other terms. Combine the derivatives, then solve for \(\frac{dy}{dx}\) to get : \(\frac{dy}{dx}=-\frac{x-2}{\sqrt{4-(y+3)^{2}}}\), which matches the earlier differentiated explicit functions, hence confirming they are equivalent.

Key Concepts

Explicit Functions

When we discuss explicit functions, we refer to mathematical expressions where the dependent variable, often denoted as 'y', is written explicitly in terms of the independent variable usually represented by 'x'. This means that you can clearly see the formula to calculate y for any given x without needing to solve for y each time.

For example, in the given exercise, when rearranging the circle's equation into the form \( (x-2)^2 + (y+3)^2 = 4 \), the aim was to express y explicitly in terms of x. The resulting functions \( y = -3 + \sqrt{4 - (x-2)^2} \) and \( y = -3 - \sqrt{4 - (x-2)^2} \) are explicit functions. They represent the upper and lower halves of the circle, respectively, allowing us to easily determine y for any x value lying on the circle's circumference.

To further clarify, the value of y is obtained directly by substituting any x-value into the function. This is particularly helpful for plotting graphs and performing subsequent calculus operations such as differentiation.

Graphing Equations

Graphing equations is a fundamental skill in mathematics that translates algebraic expressions into a visual form. This not only makes interpreting the relationship between variables easier but also helps discover properties of the function that may not be obvious from the equation alone.

In the context of the problem, the equation represents a circle transposed from the origin. By finding the explicit functions and plotting them, you recreate the circle on a coordinate plane. Here,

Labeling the Parts of the Graph

is an impactful step for visual learners, as it can make the connection between the algebraic and geometric representations clearer.

For each explicit function, the square root term dictates the curvature of the circle's halves. After sketching these on a graph, you'll notice the coordinates \( (x, y) \) fitting into either the upper or the lower half, making the learning experience more comprehensive and the concepts easier to understand.

Chain Rule

The chain rule is a powerful tool in calculus, which is used to differentiate composite functions. Essentially, if you have a function inside another function, the chain rule allows you to take the derivative of both, one after the other. Mathematically, if \( h(x) = f(g(x)) \), then \( h'(x) = f'(g(x)) \cdot g'(x) \).

In our exercise, when differentiating the explicit functions with respect to x, we encountered square root terms which are functions of \( (x-2)^2 \), creating a composite function situation. Hence, we applied the chain rule, differentiating the outer function (the square root) and multiplying by the derivative of the inner function \( (x-2) \).

The same principle was used in implicit differentiation of the original circle's equation. There, we treated y as an implicit function of x and used the chain rule to differentiate \( y^2 \) with respect to x. The chain rule is a cornerstone for problems like this, connecting deep concepts of calculus, and is essential for solving more complex differentiation tasks.