Respuesta :
Answer:
[tex]\displaystyle y - 5 = 2(x - 8)[/tex]
General Formulas and Concepts:
Symbols
- e (Euler's number) ≈ 2.7182
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
Distributive Property
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
Algebra I
Functions
Point-Slope Form: y - y₁ = m(x - x₁)
- x₁ - x coordinate
- y₁ - y coordinate
- m - slope
Slope-Intercept Form: y = mx + b
- m - slope
- b - y-intercept
Algebra II
- Logarithms - ln and e
Calculus
Derivatives
Derivative Notation
Derivative of a constant is 0
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Slope Fields
Solving Differentials - Integrals
Integration Constant C
U-Substitution
ln Integration: [tex]\displaystyle \int {\frac{1}{x}} \, dx = ln|x| + C[/tex]
Step-by-step explanation:
*Note:
When solving differential equations in slope fields, disregard the integration constant C for variable y.
Step 1: Define
[tex]\displaystyle \frac{dy}{dx} = \frac{y - 5}{x - 8}[/tex]
x = 9, y = 7
Step 2: Rewrite Differential
Rewrite Leibniz Notation using Separation of Variables.
- [Separation of Variables] Isolate x's together: [tex]\displaystyle dy = \frac{y - 5}{x - 8}dx[/tex]
- [Separation of Variables] Isolate y's together: [tex]\displaystyle \frac{1}{y - 5}dy = \frac{1}{x - 8}dx[/tex]
Step 3: Integrate Pt. 1
Solving general form of differential using integration.
- [Differential] Integrate both sides: [tex]\displaystyle \int {\frac{1}{y - 5}} \, dy = \int {\frac{1}{x - 8}} \, dx[/tex]
Step 4: Identify Variables
Set up u-substitution for right integral.
Integral w/ respect to y
u = y - 5
du = dy
Integral w/ respect to x
z = x - 8
dz = dx
Step 5: Integrate Pt. 2
- [Integrals] U-Substitution: [tex]\displaystyle \int {\frac{1}{u}} \, du = \int {\frac{1}{z}} \, dz[/tex]
- [Integrals] ln Integration: [tex]\displaystyle ln|u| = ln|z| + C[/tex]
- Back-Substitute: [tex]\displaystyle ln|y - 5| = ln|x - 8| + C[/tex]
- [Equality Property] Raise e on both sides: [tex]\displaystyle e^{ln|y - 5|} = e^{ln|x - 8| + C}[/tex]
- Simplify: [tex]\displaystyle |y - 5| = C|x - 8|[/tex]
General Form: [tex]\displaystyle |y - 5| = C|x - 8|[/tex]
Step 6: Solve Particular Solution
Since both sides have absolute value, assume that the particular solution will be positive.
- Substitute in variables [General Form]: [tex]\displaystyle |7 - 5| = C|9 - 8|[/tex]
- [Particular] |Absolute Value| Subtract: [tex]\displaystyle |2| = C|1|[/tex]
- [Particular] Evaluate absolute values: [tex]\displaystyle 2 = C(1)[/tex]
- [Particular] Multiply: [tex]\displaystyle 2 = C[/tex]
- [Particular] Rewrite: [tex]\displaystyle C = 2[/tex]
Substituting integration constant C into the general form:
Particular Solution (in Point-Slope Form): [tex]\displaystyle y - 5 = 2(x - 8)[/tex]
Particular Solution (in Slope-Intercept Form): [tex]\displaystyle y = 2x - 11[/tex]
Topic: AP Calculus AB/BC (Calculus I/II)
Unit: Differentials and Slope Fields
Book: College Calculus 10e
