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Math Help - Linear approximation

  1. #1
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    Linear approximation

    Q: Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.090000 cm thick to a hemispherical dome with a diameter of 60.000 meters.

    My solution:

    Area of hemispherical dome = 2(pi)(d/2)^2; d = 60; looking for the volume of the coat that is 0.09cm thick on top of the area.

    Therefore, I have the equation f(x) = 2(pi)(60/2)^2 * x = 1800(pi)(x).

    f'(x) = x
    a = 0.1
    f(a) = 565.2
    f'(a) = 0.1

    L(a) = 0.1 ( x - 0.1) + 565.2
    L(0.09) = 565.199

    Answer is wrong, and I can tell something is wrong when I get f'(x).

    Any hints?
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  2. #2
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    Quote Originally Posted by tttcomrader
    Q: Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.090000 cm thick to a hemispherical dome with a diameter of 60.000 meters.

    My solution:

    Area of hemispherical dome = 2(pi)(d/2)^2; d = 60; looking for the volume of the coat that is 0.09cm thick on top of the area.

    Therefore, I have the equation f(x) = 2(pi)(60/2)^2 * x = 1800(pi)(x).

    f'(x) = x
    a = 0.1
    f(a) = 565.2
    f'(a) = 0.1

    L(a) = 0.1 ( x - 0.1) + 565.2
    L(0.09) = 565.199

    Answer is wrong, and I can tell something is wrong when I get f'(x).

    Any hints?
    The derivative of,
    f(x)=1800\pi x is,
    f'(x)=1800 \pi not f'(x)=x
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