Question

The length of a rectangle is increasing at a rate of 4 meters per day and the width is increasing at a rate of 1 meter per day. When the length is 10 meters and the width is 23 meters, then how fast is the AREA changing

Answer 1

Answer:

$\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

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

Geometry

Area of a Rectangle: A = lw

• l is length
• w is width

Calculus

Derivatives

Derivative Notation

Implicit Differentiation

Differentiation with respect to time

Derivative Rule [Product Rule]:                                                                              $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle l = 10 \ meters$

$\displaystyle \frac{dl}{dt} = 4 \ m/day$

$\displaystyle w = 23 \ meters$

$\displaystyle \frac{dw}{dt} = 1 \ m/day$

Step 2: Differentiate

1. [Area of Rectangle] Product Rule:                                                                 $\displaystyle \frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

Step 3: Solve

1. [Rate] Substitute in variables [Derivative]:                                                    $\displaystyle \frac{dA}{dt} = (10 \ m)(1 \ m/day) + (23 \ m)(4 \ m/day)$
2. [Rate] Multiply:                                                                                                $\displaystyle \frac{dA}{dt} = 10 \ m^2/day + 92 \ m^2/day$
3. [Rate] Add:                                                                                                      $\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

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

Unit: Implicit Differentiation

Book: College Calculus 10e