Respuesta :
Answer:
[tex]\displaystyle y = \frac{t^2}{16} + 18[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
Algebra I
- Functions
- Function Notation
- Coordinates (x, y)
Calculus
Derivatives
Derivative Notation
Antiderivatives - Integrals
Integration Constant C
Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Step-by-step explanation:
Step 1: Define
Identify
Point (0, 18)
[tex]\displaystyle \frac{dy}{dt} = \frac{1}{8} t[/tex]
Step 2: Find General Solution
Use integration
- [Derivative] Rewrite: [tex]\displaystyle dy = \frac{1}{8} t\ dt[/tex]
- [Equality Property] Integrate both sides: [tex]\displaystyle \int dy = \int {\frac{1}{8} t} \, dt[/tex]
- [Left Integral] Integrate [Integration Rule - Reverse Power Rule]: [tex]\displaystyle y = \int {\frac{1}{8} t} \, dt[/tex]
- [Right Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle y = \frac{1}{8}\int {t} \, dt[/tex]
- [Right Integral] Integrate [Integration Rule - Reverse Power Rule]: [tex]\displaystyle y = \frac{1}{8}(\frac{t^2}{2}) + C[/tex]
- Multiply: [tex]\displaystyle y = \frac{t^2}{16} + C[/tex]
Step 3: Find Particular Solution
- Substitute in point [Function]: [tex]\displaystyle 18 = \frac{0^2}{16} + C[/tex]
- Simplify: [tex]\displaystyle 18 = 0 + C[/tex]
- Add: [tex]\displaystyle 18 = C[/tex]
- Rewrite: [tex]\displaystyle C = 18[/tex]
- Substitute in C [Function]: [tex]\displaystyle y = \frac{t^2}{16} + 18[/tex]
Topic: AP Calculus AB/BC (Calculus I/II)
Unit: Integration
Book: College Calculus 10e
