Derivatives

• Ref: SC7E.2.8.3

Q Each side of a square is increasing at a rate of <math>{\color{green}\text{6 }cm/s}</math>. At what rate is the area of the square increasing when the area of the square is <math>{\color{green}\text{16 }cm^2}</math>?

A

Givens:

• The rate of increasing side length is <math>{\color{blue}\text{6 }cm/s}</math>, which can be expressed as <math>{\color{blue}\frac{ds}{dt}}</math> where <math>{\color{red}s}</math> represents the length, and <math>{\color{red}t}</math> represents the time.
• The area of square is <math>\text{16 }cm^2</math>. The formula for the area of square is <math>{\color{purple}A = s^2}</math>.

Unknowns and the rest

• The length of the side when the area is <math>\text{16 }cm^2</math>.
  • Using the formula for the area, we get
    <math>16 = s^2</math>
    <math>{\color{green}s} = \sqrt{16} = 4</math>
    The length of the side is <math>\color{green}\text{4 }cm</math>
• The problem is asking for <math>\frac{dA}{dt}</math>.
  • <math>A = s^2</math>
    <math>\frac{dA}{dt} = \frac{ds^2}{ds}{\color{blue}\frac{ds}{dt}}</math>
    <math>\frac{dA}{dt} = 2{\color{green}s}{\color{blue}\frac{ds}{dt}} = 2{\color{green}s}({\color{blue}6})

= 2({\color{green}4})({\color{blue}6}) = 48 \frac{cm^2}{s}</math>