Problem 76 - Thomas Calculus

Question

a. Given that \((x-2)^{2}+y^{2}=4\) find \(d y / d x\) two ways: \((1)\) by solving for \(y\) and differentiating the resulting functions with respect to \(x\) and \((2)\) by implicit differentiation. Do you get the same result each way? b. Solve the equation \((x-2)^{2}+y^{2}=4\) for \(y\) and graph the resulting functions together to produce a complete graph of the equation \((x-2)^{2}+y^{2}=4 .\) Then add the graphs of the functions' first derivatives to your picture. Could you have predicted the general behavior of the derivative graphs from looking at the graph of \((x-2)^{2}+y^{2}=4 ?\) Could you have predicted the general behavior of the graph of \((x-2)^{2}+y^{2}=4\) by looking at the derivative graphs? Give reasons for your answers.

Step-by-Step Solution

Answer

Yes, both methods give the same derivative. Predominantly circular shapes show symmetry, aligned with derivations. Derivatives reflect the shape's tangent behaviors.

Step 1: Solve for y

Start by isolating \(y^2\) in the equation \((x-2)^2 + y^2 = 4\). Subtract \((x-2)^2\) from both sides to get \(y^2 = 4 - (x-2)^2\). Take the square root of both sides: \(y = \pm \sqrt{4 - (x-2)^2}\).

Step 2: Differentiate y with respect to x

Now differentiate \(y = \pm \sqrt{4 - (x-2)^2}\) with respect to \(x\). Using the chain rule, \(\frac{dy}{dx} = \pm \frac{-1}{\sqrt{4 - (x-2)^2}} \cdot (2(x-2))\). Simplify to get \(\frac{dy}{dx} = \pm \frac{-(x-2)}{\sqrt{4 - (x-2)^2}}\).

Step 3: Use implicit differentiation

Start with the original equation \((x-2)^2 + y^2 = 4\). Differentiate both sides with respect to \(x\): \(2(x-2) + 2y \frac{dy}{dx} = 0\). Solve for \(\frac{dy}{dx}\): \(2y \frac{dy}{dx} = -2(x-2)\), so \(\frac{dy}{dx} = \frac{-(x-2)}{y}\). Substitute \(y = \pm \sqrt{4 - (x-2)^2}\) to verify \(\frac{dy}{dx} = \pm \frac{-(x-2)}{\sqrt{4 - (x-2)^2}}\).

Step 4: Confirm the results

Both solving for \(y\) and differentiating directly, and using implicit differentiation, yield the same expression for \(\frac{dy}{dx}\): \(\pm \frac{-(x-2)}{\sqrt{4 - (x-2)^2}}\).

Step 5: Graph the complete function

Graph the equations \(y = \sqrt{4 -(x-2)^2}\) and \(y = -\sqrt{4 -(x-2)^2}\) on the x-y plane. These are semi-circles centered at \((2, 0)\) with a radius of 2.

Step 6: Graph derivative functions

Graph the derivative equations \(\frac{dy}{dx} = \pm \frac{-(x-2)}{\sqrt{4 - (x-2)^2}}\). This shows the slopes of the tangents to the circle at various points. Close to the circle's outer edges, the slope is steeper.

Step 7: Analyze the behavior

The derivative graphs give the rate of change of \(y\) with respect to \(x\). Circles have symmetry which is reflected in the behavior of their derivatives, as these become more extreme (- or +) towards the sides, similar to the flattening observed in the circle's visual shape.

Key Concepts

Derivatives

Understanding derivatives is crucial in calculus as they represent the rate of change of a function with respect to a variable. In simpler terms, a derivative tells you how fast something is changing. In the case of our exercise, we want to find \( \frac{dy}{dx} \), which is the derivative of \( y \) concerning \( x \). This means we aim to determine how \( y \) changes as \( x \) changes along the curve defined by the circle equation.There are two main ways to deduce this:

• First, by solving the expression for \( y \) explicitly in terms of \( x \), differentiating the function directly.
• Second, by using implicit differentiation, a technique used when it is difficult or impossible to solve for \( y \) directly.

In our example, regardless of the method, the derived result for \( \frac{dy}{dx} \) remains the same: \( \pm \frac{-(x-2)}{\sqrt{4 - (x-2)^2}} \). This confirms that both routes accurately describe the rate of change of \( y \) with respect to \( x \) for the circle equation.

Circle Equation

A circle's equation in coordinate geometry often comes in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the circle's center, and \(r\) is its radius. In this scenario, our circle equation is \((x-2)^{2}+y^{2}=4\). This matches our standard form where the center is at \((2, 0)\), and the radius \(r\) is 2 (since \(r^2 = 4\)).By solving for \( y \) in terms of \( x \), we yield two equations:

• \(y = \sqrt{4 - (x-2)^2}\)
• \(y = -\sqrt{4 - (x-2)^2}\)

These represent the upper and lower semi-circles, respectively.
Graphing these functions in the x-y plane reveals a full circle split into two halves. Each half is symmetric about the x-axis, representing the top and bottom parts of the circle, centered at (2, 0).

Coordinate Geometry

In coordinate geometry, we study figures like lines, curves, and circles in the plane using coordinates, which are numbers that define their position. The exercise makes use of coordinate geometry by plotting the circle and analyzing the derivative graphs.The circle \((x-2)^{2}+y^{2}=4\) represents a symmetric geometric shape that can be viewed in the coordinate system as a continuous loop. Each point on this circle corresponds to an \((x, y)\) pair that satisfies the equation.
As we assess the derivative graph, plotted from the expression \(\frac{dy}{dx} = \pm \frac{-(x-2)}{\sqrt{4 - (x-2)^2}}\), it illustrates the slope of the tangent line at any given point on the circle's circumference.Understanding these graphs provides an insight into how the rate of change varies along the curve. Thus, through the lens of coordinate geometry, not only can we visualize the round shape, but we can also decipher the nature of changes and slopes at every point on the circle.