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The diagonal of a square is increasing at a rate of 3 inches per minute. When the area of the square is 18 square inches, how fast(in inches per minute) is the perimeter increasing?

  • calculus -

    The sides are .707 of the diagonal, and area is s squared, so
    Area= .5 diagonal^2

    or diagonal= sqrt (2*area)

    Perimeter= 4s=4(.707)diagonal

    dP/dt= 4*.707 * diagonal/dt
    = 4*.707 3in/min

    I don't see it vary with the size of the square. check my thinking.

  • calculus -

    bobpursely is right, the area of 18 in^2 has nothing to do with it.

    Here is how I did it

    D^2 = 2s^2
    D = √2 s
    dD/dt = √2 ds/dt
    3 = √2 ds/dt ------> ds/dt = 3/√2

    P = 4s
    dP/dt = 4ds/dt=4(3/√2) = 12/√2 in/min

    which is the same as bobpursley's answer

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