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An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a rate of 24 m per year at a time when the region is 300 m wide. How fast is the area changing at that point in time

Respuesta :

Space

Answer:

[tex]\frac{dA}{dt} = 28800 \ m^2/year[/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

Equality Properties

Geometry

  • Area of a Rectangle: A = lw

Algebra I

  • Exponential Property: [tex]w^n \cdot w^m = w^{n + m}[/tex]

Calculus

Derivatives

Differentiating with respect to time

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Explanation:

Step 1: Define

Area is A = lw

2w = l

w = 300 m

[tex]\frac{dw}{dt} = 24 \ m/year[/tex]

Step 2: Rewrite Equation

  1. Substitute in l:                    A = (2w)w
  2. Multiply:                              A = 2w²

Step 3: Differentiate

Differentiate the new area formula with respect to time.

  1. Differentiate [Basic Power Rule]:                                                                   [tex]\frac{dA}{dt} = 2 \cdot 2w^{2-1}\frac{dw}{dt}[/tex]
  2. Simplify:                                                                                                           [tex]\frac{dA}{dt} = 4w\frac{dw}{dt}[/tex]

Step 4: Find Rate

Use defined variables

  1. Substitute:                    [tex]\frac{dA}{dt} = 4(300 \ m)(24 \ m/year)[/tex]
  2. Multiply:                        [tex]\frac{dA}{dt} = (1200 \ m)(24 \ m/year)[/tex]
  3. Multiply:                        [tex]\frac{dA}{dt} = 28800 \ m^2/year[/tex]
Supernova

Answer:

28,800 m²/yr

Explanation:

This rectangle has dimensions such that:

  • width = w
  • length = 2w  

We are given [tex]\displaystyle \frac{dw}{dt} = \frac{24 \ m}{yr}[/tex] and want to find [tex]\displaystyle \frac{dA}{dt} \Biggr | _{w \ = \ 300 \ m} = \ ?[/tex] when w = 300 m.

The area of a rectangle is denoted by Area = length * width.

Let's multiply the width and length (with respect to w) together to have an area equation in terms of w:

  • [tex]A=2w^2[/tex]

Differentiate this equation with respect to time t.  

  • [tex]\displaystyle \frac{dA}{dt} =4w \cdot \frac{dw}{dt}[/tex]

Let's plug known values into the equation:

  • [tex]\displaystyle \frac{dA}{dt} =4(300) \cdot (24)[/tex]

Simplify this equation.

  • [tex]\displaystyle \frac{dA}{dt} =1200 \cdot 24[/tex]
  • [tex]\displaystyle \frac{dA}{dt} =28800[/tex]

The area is changing at a rate of 28,800 m²/yr at this point in time.

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