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What is the rate of change of the area of a rectangle?

  1. May 27, 2017 #1

    a78

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    1. The problem statement, all variables and given/known data
    Find the rate of change of the area of a rectangle whose area is 75cm^2. The length is 3 times the width. The rate of change of the width is 2cm/second.

    2. Relevant equations


    3. The attempt at a solution
    A=75 A'=?
    L=3x= 15 L'=6
    W=x=5 W'=2

    A'=L'W+LW'
    A'= (6)(5)+ (15)(2)
    A'=60cm/sec
     
    Last edited: May 27, 2017
  2. jcsd
  3. May 27, 2017 #2

    BvU

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    And it looks like homework, so use the template and show your attempt!

    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution
     
  4. May 27, 2017 #3

    BvU

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    Much better ! The problem statement helps you to formulate what you actually need to solve.
    easiest if you express the rate of change as a ##d\over dt##, as you did.
    Do not forget to keep track of the units: an area is cm2, so a growth rate for an area can not be cm/s !!!
    Other than that, you are doing quite well !
     
  5. May 27, 2017 #4

    LCKurtz

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    While you seem to understand how to work this problem, your writeup could be much better. I will intersperse comments below.

    A rectangle whose area is ##75## has constant area, so that isn't what you mean. What you mean is a rectangle has length 3 times its width, so its area is ##A = lw =3w^2##. The rate of change of ##w## is ##2##, and you want the rate of change of ##A## when its area is ##75##.

    Again, the derivative of constants are zero.

    Since you already have the correct answer, I am going suggest a better way to write it up. Since you have ##A = 3w^2## you know that ##A' = 6ww'##, where ##' = \frac d {dt}##. That is called the related rate equation -- everything is a function of ##t## so the derivatives aren't ##0##. At the instant when ##A = 75##, you have figured out that ##w = 5## and ##w'## is given as ##2##. Just put those numbers in the related rate equation and you have your answer.
     
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