Problem 29 - The Calculus with Analytic Geometry

Question

An equation is given. Do the following in each of these problems: (a) Find two functions defined by the equation, and state their domains. (b) Draw a sketch of the graph of each of the functions obtained in part (a). (c) Draw a sketch of the graph of the equation. (d) Find the derivative of each of the functions obtained in part (a) and state the domains of the derivatives. (e) Find \(D_{x} y\) by implicit differentiation from the given equation, and verify that the result so obtained agrees with the results in part (d). (f) Find an equation of each tangent line at the given value of \(x_{1}\). $$ x^{2}+y^{2}-2 x-4 y-4=0 ; x_{1}=1 $$

Step-by-Step Solution

Answer

Functions: y1 = 2 + √(9 - (x - 1)^2), y2 = 2 - √(9 - (x - 1)^2), Derivatives: y1' = -(x - 1)/√(9 - (x - 1)^2), y2' = -(x - 1)/√(9 - (x - 1)^2). Tangent lines: y -(2+√8) = 0, y -(2-√8) = 0.

Step 1: Rewrite the given equation

Start by rewriting the given equation to make it more manageable. The given equation is: \[ x^{2} + y^{2} - 2x - 4y - 4 = 0 \] Group the x and y terms together: \[ (x^{2} - 2x) + (y^{2} - 4y) = 4 \]

Step 2: Complete the square

Complete the square for both the x and y terms: For x:\[ x^{2} - 2x = (x - 1)^{2} - 1 \] For y:\[ y^{2} - 4y = (y - 2)^{2} - 4 \] Substitute back into the equation:\[ (x - 1)^{2} - 1 + (y - 2)^{2} - 4 = 4 \] Simplify to get:\[ (x - 1)^{2} + (y - 2)^{2} = 9 \]

Step 3: Identify the functions and their domains

From the equation of the circle, the functions can be written as: \[ y_1 = 2 + \sqrt{9 - (x - 1)^2} \] \[ y_2 = 2 - \sqrt{9 - (x - 1)^2} \] These functions are defined for \[ -2 \leq x \leq 4 \] because the radicand must be non-negative.

Step 4: Sketch the graphs of each function

Draw the graph of each function obtained in Step 3: - For \( y_1\), plot points and draw the top half of the circle centered at (1,2) with radius 3. - For \( y_2\), plot points and draw the bottom half of the circle centered at (1,2) with radius 3.

Step 5: Sketch the graph of the equation

Combine the sketches from Step 4 to draw the full circle defined by the equation. The circle is centered at (1,2) with a radius of 3.

Step 6: Find the derivatives of the functions and state their domains

Use the chain rule to differentiate each function obtained in Step 3. For \( y_1\): \[ y_1' = \frac{d}{dx} [2 + \sqrt{9 - (x - 1)^2}] = \frac{-(x - 1)}{\sqrt{9 - (x - 1)^2}} \] For \( y_2\): \[ y_2' = \frac{d}{dx} [2 - \sqrt{9 - (x - 1)^2}] = \frac{-(x - 1)}{\sqrt{9 - (x - 1)^2}} \] The derivatives are defined for \[ -2 \leq x \leq 4, x ≠ 1 \]

Step 7: Implicit differentiation

Implicitly differentiate both sides of the original equation \[ x^{2} + y^{2} - 2x - 4y - 4 = 0 \] Differentiate:\[ 2x + 2yy' - 2 - 4y' = 0 \] Solving for y': \[ y' = \frac{-2x + 2}{2y - 4} = \frac{x - 1}{2 - y} \] Check consistency with previous results: Substitute solutions for y to verify.

Step 8: Find tangent lines at given x

At \( x_1 = 1 \), use y values and derivatives obtained: For \( y_1 \), point (1, 2 + \sqrt{8}):\[ y_1'(1) = \frac{-(1-1)}{\sqrt{9 - (1-1)^2}} = 0 \] Tangent line: \[ y - (2+\sqrt{8}) = 0 \] For \( y_2 \), point (1, 2 - \sqrt{8}):\[ y_2'(1) = \frac{-(1-1)}{\sqrt{9-(1-1)^2}} = 0 \] Tangent line: \[ y - (2-\sqrt{8}) = 0 \]

Key Concepts

Completing the Square

Completing the square is a key algebraic technique used in this problem to rewrite the given equation into a simpler, more manageable form. We start with the given equation and group the x terms together and the y terms together:
\[ x^2 + y^2 - 2x - 4y - 4 = 0 \]. We rewrite it as follow-to-get-back-formula:

Next, we complete the square for the grouped terms. For the x terms - we rewrite:
\[ x^2 - 2x = (x - 1)^2 - 1 \]
For the y terms - we rewrite:
\[ y^2 - 4y = (y - 2)^2 - 4 \]
By substituting back into the equation, we get a circle equation:
\[ (x - 1)^2 + (y - 2)^2 = 9 \]. Completing the square helps us identify the center and radius of the circle.

Domain of Functions

The domain of a function is the set of all possible input values (x-values) that make the function defined. In this problem, after completing the square, we get two functions defined by the equation of the circle:
\[ y_1 = 2 + \sqrt{9 - (x - 1)^2} \] \[ y_2 = 2 - \sqrt{9 - (x - 1)^2} \]
Both functions contain a square root, which means the expression inside the square root must be non-negative:
\[ 9 - (x - 1)^2 \geq 0 \]
This simplifies to \[ -2 \leq x \leq 4 \].
Hence, the domain for both functions is:
\-2 \leq x \leq 4

Equation of Tangent Line

To find the equation of a tangent line, we first need the derivative of the functions, as the slope of the tangent line at a given point is equal to the derivative evaluated at that point.
Using the chain rule, we find the derivatives of the functions:
For \[ y_1: \]
\[ y_1' = \frac{ - (x - 1) }{ \sqrt{9 - (x - 1)^2} } \]
\[ y_2: \]
\[ y_2' = \frac{ - (x - 1) }{ \sqrt{9 - (x - 1)^2} } \]
At \[ x1 = 1 \], they both simplify to 0. Now, using the point-slope form of the lines, for \[ y_1 \] with point \[ (1, 2+\sqrt{8}) \], the tangent line is:
\[ y - (2 + \sqrt{8}) = 0 \]. For \[ y_2 \] with point \[ (1, 2-\sqrt{8}) \], the tangent line is:
\[ y - (2 - \sqrt{8}) = 0 \]

Derivative Calculation

Differentiation is a fundamental process in calculus used to find the rate of change of a function. For this problem, we need to find the derivatives of the two functions obtained from the original equation. These functions represent the upper and lower halves of a circle:
\[ y_1 = 2 + \sqrt{9 - (x - 1)^2} \]
\[ y_2 = 2 - \sqrt{9 - (x - 1)^2} \]
We use the chain rule to differentiate these functions:
\[ y_1' = \frac{-(x - 1)}{\sqrt{9 - (x - 1)^2}} \]
\[ y_2' = \frac{-(x - 1)}{\sqrt{9 - (x - 1)^2}} \]
The domain of the derivatives excludes \[ x \eq 1 \], as the denominator becomes zero. This gives us the final domains:
\-2 \leq x \leq 4, x \eq 1.