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

`\displaystyle (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 (d)/(dx) [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)`

Explanation:

Step 1: Define

`\displaystyle l = 10 \ meters`

`\displaystyle (dl)/(dt) = 4 \ m/day`

`\displaystyle w = 23 \ meters`

`\displaystyle (dw)/(dt) = 1 \ m/day`

Step 2: Differentiate

1. [Area of Rectangle] Product Rule:
   `\displaystyle (dA)/(dt) = l(dw)/(dt) + w(dl)/(dt)`

Step 3: Solve

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

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

Unit: Implicit Differentiation

Book: College Calculus 10e