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Thread: After many hours of trying, I am stuck.

  1. #1
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    After many hours of trying, I am stuck.

    Theres this one question I was doing for practice that I just cannot do. Even though I know the answer is -0.94, I cant figure out how.

    The question goes as followed.

    In a computer game, the motion of a car is enhanced by showing headlights on a building, when the car approaches the building. The headlights form a parabola on the road. As the car approaches the building, the beam narrows. The equation of the parabola is y = x^2 - 4x +c, where c < 0 and dc/dt = 14. The x-axis represents the wall, let length be measured in meteres and time in seconds.


    Determine the rate at which the width of the beam is narrowing when c = -14.


    The answer at the back is -0.94. But I really cant figure out how to get this, and how to approach this question.

    help would be appreciated.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Dave J View Post
    Theres this one question I was doing for practice that I just cannot do. Even though I know the answer is -0.94, I cant figure out how.

    The question goes as followed.

    In a computer game, the motion of a car is enhanced by showing headlights on a building, when the car approaches the building. The headlights form a parabola on the road. As the car approaches the building, the beam narrows. The equation of the parabola is y = x^2 - 4x +c, where c < 0 and dc/dt = 14. The x-axis represents the wall, let length be measured in meteres and time in seconds.


    Determine the rate at which the width of the beam is narrowing when c = -14.


    The answer at the back is -0.94. But I really cant figure out how to get this, and how to approach this question.

    help would be appreciated.
    This is poorly worded. I will assume "the width of the beam" is actually the
    distance between the roots of the quadratic:

    $\displaystyle
    x^2 - 4x +c
    $

    The roots of this are:

    $\displaystyle
    x=\frac{4 \pm \sqrt{16-4c}}{2}
    $,

    and the distance between the roots is:

    $\displaystyle
    \Delta=\sqrt{16-4c}
    $

    Then:

    $\displaystyle
    \frac{d \Delta}{dt}=\frac{1}{2}(16-4c)^{-1/2}\ (-4) \frac{dc}{dt}
    $


    $\displaystyle
    \frac{d \Delta}{dt}=-2 \frac{1}{(16-4c)^{1/2}} \frac{dc}{dt}
    $

    Putting $\displaystyle c=-14$ and $\displaystyle \frac{dc}{dt}=14$, gives:

    $\displaystyle
    \frac{d \Delta}{dt}=-2 \frac{1}{(16-4c)^{1/2}} \frac{dc}{dt}\approx -3.30
    $

    which does not agree with your book.

    The possibilities are: that I have interpreted the problem incorrectly, I have
    made a mistake in my algebra and derivatives, I have made a mistake in the
    arithmetic, the book is wrong. No doubt my fellow helpers will comment on
    the first three of these.

    RonL
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  3. #3
    Super Member

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    Hello, Dave!

    In a computer game, the motion of a car is enhanced by showing headlights on a building
    when the car approaches the building. .The headlights form a parabola on the road.
    As the car approaches the building, the beam narrows.
    The equation of the parabola is: $\displaystyle y \:= \:x^2 - 4x +c,$ where $\displaystyle c < 0$ and $\displaystyle \frac{dc}{dt} = 14$
    The x-axis represents the wall. Length is measured in meters and time in seconds.

    Determine the rate at which the width of the beam is narrowing when $\displaystyle c = -14$

    I agree completely with Captain Black.
    Code:
                  |
           *      |                        *
                  |
            *     |                       *
          ---o----+----------------------o---
               *  |                    *
                 c*                 *
                  |   *         *
                  |       ***
                  |

    The width of the beam is the distance between the two x-intecepts.

    I have the same steps as the Captain . . . and the same answer.
    I don't see how we could have misread the problem.

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