Problem 30 - 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 \(x_{1}\). $$ x^{2}+4 y^{2}+6 x-40 y+93=0 ; x_{1}=-2 $$

Step-by-Step Solution

Answer

y = 5 +/- 1/2 sqrt(-x^2 - 6x + 7) for real values of x between -7≤ 1. Find derivatives, tangent lines. Draw graphs

Step 1: Rewrite the given equation

The given equation is \break \break \[ x^{2}+4y^{2}+6x-40y+93=0 \]

Step 2: Rearrange the equation for solving for y

We need to express the equation in the standard form in terms of y. \break We have: \break \[ x^2 + 4y^2 + 6x - 40y + 93 = 0 \] \break Rearrange to: \break \[ 4y^2 - 40y = -x^2 - 6x - 93 \]

Step 3: Complete the square on terms involving y

Complete the square for the terms involving y:\break \[ 4(y^2 - 10y) + 93 = -x^2 - 6x \] => \[ 4(y-5)^2 - 100 + 93 = -x^2 - 6x \] \break Thus, it simplifies to \break \[ 4(y-5)^2 = -x^2 - 6x + 7 \] \break => \[ (y-5)^2 = \frac{-x^2 - 6x + 7}{4} \]

Step 4: Solve for y

The equation can be solved as: \break \[ y - 5 = \frac{\text{+}\text{-}\text{1}}{2} \text{sqrt} (\text{-} x^{2} - 6 x + 7) \] \break => \[ y_{1} = 5 + \frac{1}{2} \text{sqrt} (\text{-} x^{2} - 6 x + 7), \ y_{2} = 5 - \frac{1}{2} \text{1}{2} \text{sqrt} (\text{-} x^{2} - 6 x + 7) \]

Step 5: State the domains

The functions y_1 and y_2 are valid if the expression under the sqrt is non-negative: \break \[ -x^{2}-6x+7 \text{ needs to be greater than or equal to }0 \] \break Solve for x:\break Roots of \( -x^{2} - 6 x + 7 = 0 \) are \( x = -7 \) and \( x = +1 \) \break Thus, \break \( x \in[-7, 1] \)

Step 6: Draw the graph

Create graphs of y1 and y2 using plotting tools such as Desmos or manually by checking key values in the range \(-7 \leq x \leq 1\).\break Represent y1 and y2 on a coordinate axes

Step 7: Implicit differentiation

Find \(D_x y\) from the given equation.\break Starting with \break \(2x + 8yy' + 6 - 40y' = 0\) => \( y'(8y - 40 ) = -2x - 6 y' = \frac{-2x - 6}{8y - 40}\)

Step 8: Derivative (Explicit)

\( After finding explicit derivatives of \( y_1 = 5 + \frac{1}{2} sqrt\((-x^{2} - 6 x + 7 \)\) \)\break Use chain rule

Step 9: Tangent line

\( \frac{-2(-2) - 6}{8y - 40}\)

Step 10: Verify

Compare explicit differentiation results from above to Implicit differentiation results (step 8) y'(0)

Step 11: Create equations of tangent lines

Equations of tangent lines using point slope of existing points.

Key Concepts

completing the square

To understand how we solve equations by completing the square, let's break down the steps. Completing the square transforms a quadratic equation into a form that makes it easier to solve for variables.
First, you need to rearrange the quadratic equation to group the quadratic and linear terms. For example, given the equation:
\[ x^2 + 4y^2 + 6x - 40y + 93 = 0 \]
we rearrange it in terms of y:

\[ 4y^2 - 40y = -x^2 - 6x - 93 \]

Next, we complete the square for the quadratic term involving y. This means you will re-write the quadratic expression as a square plus or minus some constant:

\[ 4(y^2 - 10y) + 93 = -x^2 - 6x \]
becomes
\[ 4(y-5)^2 - 100 + 93 = -x^2 - 6x \]
which simplifies to:

\[ 4(y-5)^2 = -x^2 - 6x + 13 \]

By completing the square, we transform the original equation into a quadratic form that's easier to solve for y. This step is key to breaking down complex equations into manageable parts.


Finally, solving for y gives us two functions:
\[ y_1 = 5 + \frac{1}{2} \sqrt{-x^2 - 6x + 7} \]
and
\[ y_2 = 5 - \frac{1}{2} \sqrt{-x^2 - 6x + 7} \]

domain of a function

Determining the domain of a function is all about finding the set of input values (x values) for which the function is defined. For the functions we obtained:
\[ y_1 = 5 + \frac{1}{2} \sqrt{-x^2 - 6x + 7} \]
and
\[ y_2 = 5 - \frac{1}{2} \sqrt{-x^2 - 6x + 7} \]
we need to ensure that the expression inside the square root,
\[ -x^2 - 6x + 7 \]
is non-negative. To find this domain, we set up the inequality:

\[ -x^2 - 6x + 7 \geq 0 \]
Solving this quadratic inequality involves finding the roots:
\[ -x^2 - 6x + 7 = 0 \]
which simplifies to
\[ x = -7 \text{ and } x = 1 \]

Hence, the domain of our functions \(y_1\) and \(y_2\) is the interval:

\[ x \in [-7, 1] \]

Understanding the domain helps us know the input values for which our functions produce valid outputs. This is crucial for graphing and further analysis.
After acknowledging the domain, it's crucial to keep verifying with the initial restrictions or any given conditions.

derivative

Calculating the derivative of a function tells us how the function's output changes with respect to input values. For our functions, we need to find:
\[ y_1 = 5 + \frac{1}{2} \sqrt{-x^2 - 6x + 7} \]
and
\[ y_2 = 5 - \frac{1}{2} \sqrt{-x^2 - 6x + 7} \]

We use differentiation rules, including the chain rule. The explicit derivatives are:

\[ y_1' = \frac{d}{dx} \left( 5 + \frac{1}{2} \sqrt{-x^2 - 6x + 7} \right) \]
Applying the chain rule, it simplifies to:

\[ y_1' = \frac{-2x - 6}{2 \sqrt{-x^2 - 6x + 7}} \]

Similarly, for \(y_2\):

\[ y_2' = \frac{d}{dx} \left( 5 - \frac{1}{2} \sqrt{-x^2 - 6x + 7} \right) \]

Again, applying the chain rule results in:

\[ y_2' = \frac{-2x - 6}{2 \sqrt{-x^2 - 6x + 7}} \]

The key point to understand here is how differentiation helps us see how the function values change as their input (x) changes. This is especially useful in finding tangents, slopes, and overall behavior of curves.
After calculating the derivatives, don't forget to double-check the domain restrictions applied with the new expressions.