Question

Find the function y = f(t) passing through the point (0, 18) with the given first derivative.
dy/dt = 1/8 t

y = ?

Answer 1

Answer:

$\displaystyle y = \frac{t^2}{16} + 18$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. 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]:                                                                   $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Multiplied Constant]:                                                             $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

Point (0, 18)

$\displaystyle \frac{dy}{dt} = \frac{1}{8} t$

Step 2: Find General Solution

Use integration

1. [Derivative] Rewrite:                                                                                         $\displaystyle dy = \frac{1}{8} t\ dt$
2. [Equality Property] Integrate both sides:                                                        $\displaystyle \int dy = \int {\frac{1}{8} t} \, dt$
3. [Left Integral] Integrate [Integration Rule - Reverse Power Rule]:                 $\displaystyle y = \int {\frac{1}{8} t} \, dt$
4. [Right Integral] Rewrite [Integration Property - Multiplied Constant]:           $\displaystyle y = \frac{1}{8}\int {t} \, dt$
5. [Right Integral] Integrate [Integration Rule - Reverse Power Rule]:              $\displaystyle y = \frac{1}{8}(\frac{t^2}{2}) + C$
6. Multiply:                                                                                                             $\displaystyle y = \frac{t^2}{16} + C$

Step 3: Find Particular Solution

1. Substitute in point [Function]:                                                                         $\displaystyle 18 = \frac{0^2}{16} + C$
2. Simplify:                                                                                                             $\displaystyle 18 = 0 + C$
3. Add:                                                                                                                   $\displaystyle 18 = C$
4. Rewrite:                                                                                                             $\displaystyle C = 18$
5. Substitute in C [Function]:                                                                                $\displaystyle y = \frac{t^2}{16} + 18$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Integration

Book: College Calculus 10e