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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

Sagot :

Space

Answer:

[tex]\displaystyle \frac{dA}{dt} = 102 \ m^2/day[/tex]

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]:                                                                              [tex]\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)[/tex]

Step-by-step explanation:

Step 1: Define

[tex]\displaystyle l = 10 \ meters[/tex]

[tex]\displaystyle \frac{dl}{dt} = 4 \ m/day[/tex]

[tex]\displaystyle w = 23 \ meters[/tex]

[tex]\displaystyle \frac{dw}{dt} = 1 \ m/day[/tex]

Step 2: Differentiate

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

Step 3: Solve

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

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

Unit: Implicit Differentiation

Book: College Calculus 10e

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