help with rate of change qn please?

A circular cylinder is expanding in such a way that at a time t seconds, the length of the cylinder is 20x cm and the area of the cross section is x cm^2. Given that when x=5, the area of the cross section is increasing at a rate of 0.025 cm^2 s^-1, find the length of the cylinder.

1 Answer

  • Mary
    Lv 7
    8 years ago
    Favorite Answer

    As we know that x is expanding with respect to time, x can be given by the equation,

    x = at + b

    The derivative of x with respect to t is

    x' = a = 0.025 cm^2/s

    Since the value of t is not given when x=5, b can take on any value. Let's say that when t=0, x=0 then b=0. We have

    x = 0.025t

    When x=5

    5 = 0.025t

    t = 5/0.025 = 200

    However, this only tells us that area of the cross section of the cylinder is 5 at 200 seconds. This fact has nothing to do with what we want, namely the length of the cylinder, nor other values like the volume. This is a trick question!

    Since x=5 at whatever the time is, the length is 20x, so,

    length of the cylinder

    = 20*5 = 100 cm

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