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Thread: More word problem help

  1. #1
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    More word problem help

    So my question is-

    A fire has started in a dry open field and is spreading in the form of a circle. If the radius of this circle increases at the rate of 6 feet/ minute express the total area A as a function of time t (in minutes).

    It seems to me that area would = time(6(pi)time^2)
    The answer in the book says the answer is= 36(pi)time^2

    It is not clear to me why the 6 must be squared.
    Help?

  2. #2
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    Re: More word problem help

    I'm a little puzzled why 't' was defined but never used. The variable "time" is quite cumbersone.

    Radius(t) = t*(6 ft/min)

    [tex]Area(t)\;=\;\pi\cdot\left[Radius(t)\right]^{2}\;=\;??[/tex]

    Are you seeing it, yet?

  3. #3
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    Re: More word problem help

    Hello, VP1!

    A fire has started in a dry open field and is spreading in the form of a circle.
    If the radius of this circle increases at the rate of 6 feet/minute, express the total area [tex]A[/tex] as a function of time [tex]t[/tex] (in minutes).

    It seems to me that area would be: [tex]A \,=\, 6\pi t^2[/tex]

    The answer in the book says the answer is: [tex]36\pi t^2[/tex]

    It is not clear to me why the 6 must be squared.

    We are told that the radius is a function of time: .[tex]r \,=\,6t[/tex]

    [tex]\text{Therefore: }\;A \;=\;\pi r^2 \;=\;\pi(6t)^2 \;=\;36\pi t^2[/tex]

    I'm the other of the two guys who "do" homework.

  4. #4
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    Re: More word problem help

    Thanks for the quick help. That makes so much sense when it's explained, I just can't see it sometimes when looking at the question.

    I definitely need to do more work with word problems to get more comfortable with them. Does anyone have any good websites that devote some reading to solving word problems?

    Thanks- Todd

  5. #5
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    Re: More word problem help

    The most important work and pratice you can do is learning to WRITE clear and concise definitions at the beginning.

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