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Math Help - reciprocal cm?

  1. #1
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    reciprocal cm?

    hi all,

    im confused with the way an answer is given in reciprocal cm format the questions was:

    The temperature \theta C measured at a distance x cm is given by:

    \theta=16+\frac{450}{x^\frac{3}{2}}

    for distances x\geq 1

    Find the temeparture gradient (i.e. the rate at which temperature decreases with distance) 10cm from the flame.

    my answer was

    t=f(d)=16+450x^{-\frac{3}{2}}, where t(emp), d(istance)

    f'(d)=-675x^{-\frac{5}{2}}

    f'(10)=-675.\frac{1}{100\sqrt{x}}

    =-\frac{6.75}{\sqrt{10}} \equiv -0.675\sqrt{10}

    =-2.13C/cm to 3 sf

    but the book (Edexcel A level maths} gives the answer as

    =-2.13C/cm^{-1}

    can someone please explain what the reciprocal cm is all about.

    thanks.
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  2. #2
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    Quote Originally Posted by sammy28 View Post
    hi all,

    im confused with the way an answer is given in reciprocal cm format the questions was:

    The temperature \theta C measured at a distance x cm is given by:

    \theta=16+\frac{450}{x^\frac{3}{2}}

    for distances x\geq 1

    Find the temeparture gradient (i.e. the rate at which temperature decreases with distance) 10cm from the flame.

    my answer was

    t=f(d)=16+450x^{-\frac{3}{2}}, where t(emp), d(istance)

    f'(d)=-675x^{-\frac{5}{2}}

    f'(10)=-675.\frac{1}{100\sqrt{x}}

    =-\frac{6.75}{\sqrt{10}} \equiv -0.675\sqrt{10}

    =-2.13C/cm to 3 sf

    but the book (Edexcel A level maths} gives the answer as

    =-2.13C/cm^{-1}

    can someone please explain what the reciprocal cm is all about.

    thanks.
    Your book has got it wrong. The unit is either ^0 \text{C/cm} or ^0 \text{C cm}^{-1} but not a mixture of both.
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  3. #3
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    thanks for yor answer MF. it was me who got it wrong sorry. the book does say ^0\text{C cm}^{-1} now it makes sense. what an idiot!
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