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Math Help - deriving expression help

  1. #1
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    deriving expression help

    Derive an expression that gives the density p of the sphere in terms of its mass m and diameter d .

    density =   p = \frac{m}{v} volume of sphere  v = \frac{4}{3}\pi r^{3}

    I cant just substitute in v into the density formula, as I need to have it in diameter instead of radius, been trying for sometime now, how do I get diameter into the formula?

    any help appreciated.

    thank you
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  2. #2
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    Quote Originally Posted by Tweety View Post
    Derive an expression that gives the density p of the sphere in terms of its mass m and diameter d .

    density =   p = \frac{m}{v} volume of sphere  v = \frac{4}{3}\pi r^{3}

    I cant just substitute in v into the density formula, as I need to have it in diameter instead of radius, been trying for sometime now, how do I get diameter into the formula?

    any help appreciated.

    thank you
    1. You are supposed to know that a diameter is twice as long as a radius:

    d = 2r~\implies~r=\frac12 \cdot d

    2. The volume becomes:

    V_{sphere} = \frac43 \cdot \pi \left( \frac12 \cdot d \right)^3

    3. ... and now it's your turn.
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  3. #3
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    The radius is equal to half of the diameter.
    Therefor r = d/2.
    r^3 = (d/2)^3
    r^3 = d^3 / 8
    so the volume of the cylinder is equal to (pi * d^3) / 6
    which leads to that p = m / ((pi * d^3) / 6 )
    rearrange it and u derive to that the density is equal to 6m / (pi * d^3)
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  4. #4
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    thanks guys, I thought d = r^2, so thats why I was stuck.

    I also need help on the second part, ;

    from this derive an expression that gives the resolution of the density measurement  \Delta p in terms of the resolution of the mass and diameter measurements  \Delta m and  \Delta d respectively.

    my booklet gives me this equation for resolution  \Delta p = |  (\frac{\delta p}{\delta m})_{v} \Delta m  |  +  |  (\frac{\delta p }{\delta v }) _{m} \Delta V  |


    I am not even sure what exactly is being asked here, but I did manage to work out  \frac{\delta p}{\delta m}

    density =  p = \frac{m}{v} and from the rearranged volume formula  v = \frac{1}{6} \pi d^{3}

    so if I jut plug in v into density formula, than differentiate I get  p = \frac{6m}{\pi d^{3}}

     \frac{\delts p}{\delta m} = \frac{-18m}{\pi d^{4}}

    If this is correct, than can someone please help me with the other part of the equation, deltap/deltaaV

    thank you
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  5. #5
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    Quote Originally Posted by Tweety View Post
    thanks guys, I thought d = r^2, so thats why I was stuck.

    I also need help on the second part, ;

    from this derive an expression that gives the resolution of the density measurement  \Delta p in terms of the resolution of the mass and diameter measurements  \Delta m and  \Delta d respectively.

    my booklet gives me this equation for resolution  \Delta p = |  (\frac{\delta p}{\delta m})_{v} \Delta m  |  +  |  (\frac{\delta p }{\delta v }) _{m} \Delta V  |


    I am not even sure what exactly is being asked here, but I did manage to work out  \frac{\delta p}{\delta m}

    density =  p = \frac{m}{v} and from the rearranged volume formula  v = \frac{1}{6} \pi d^{3}

    so if I jut plug in v into density formula, than differentiate I get  p = \frac{6m}{\pi d^{3}}

     \frac{\delts p}{\delta m} = \frac{-18m}{\pi d^{4}}

    If this is correct, than can someone please help me with the other part of the equation, deltap/deltaaV

    thank you
    I'm going to try to explain what the given formula describes:

     \Delta p = |  (\frac{\delta p}{\delta m})_{v} \Delta m  |  +  |  (\frac{\delta p }{\delta v }) _{m} \Delta V  |

    (\frac{\delta p}{\delta m})_{v} that's the rate of change in the density if the volume is constant. Obviously the density is linearly dependent on the mass if the volume is constant.

    (\frac{\delta p }{\delta v }) _{m} that's the rate of change in the density if the mass is constant. Obviously the density is reciprocally(?) dependent on the volume if the mass is constant.

    Write p = \dfrac mV as p(m)=\dfrac1V \cdot m~,~V \ const. then it is obvious that

    \dfrac{\partial p}{\partial m} = \dfrac1V

    Write p = \dfrac mV as p(m)=m \cdot {V^{-1}}~,~m \  const. then it is obvious that

    \dfrac{\partial p}{\partial V} = -m \cdot V^{-2}

    ... and now it's your turn.
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