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  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 = $\displaystyle p = \frac{m}{v} $ volume of sphere $\displaystyle 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 = $\displaystyle p = \frac{m}{v} $ volume of sphere $\displaystyle 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:

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

    2. The volume becomes:

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

    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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    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 $\displaystyle \Delta p $ in terms of the resolution of the mass and diameter measurements $\displaystyle \Delta m $ and $\displaystyle \Delta d $ respectively.

    my booklet gives me this equation for resolution $\displaystyle \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 $\displaystyle \frac{\delta p}{\delta m} $

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

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

    $\displaystyle \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 $\displaystyle \Delta p $ in terms of the resolution of the mass and diameter measurements $\displaystyle \Delta m $ and $\displaystyle \Delta d $ respectively.

    my booklet gives me this equation for resolution $\displaystyle \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 $\displaystyle \frac{\delta p}{\delta m} $

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

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

    $\displaystyle \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:

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

    $\displaystyle (\frac{\delta p}{\delta m})_{v}$ 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.

    $\displaystyle (\frac{\delta p }{\delta v }) _{m}$ 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 $\displaystyle p = \dfrac mV$ as $\displaystyle p(m)=\dfrac1V \cdot m~,~V \ const.$ then it is obvious that

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

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

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

    ... and now it's your turn.
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