deriving expression help

• Oct 5th 2010, 08:22 AM
Tweety
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
• Oct 5th 2010, 08:31 AM
earboth
Quote:

Originally Posted by Tweety
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.
• Oct 5th 2010, 08:35 AM
Zamzen
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)
• Oct 5th 2010, 02:17 PM
Tweety
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
• Oct 6th 2010, 12:55 AM
earboth
Quote:

Originally Posted by Tweety
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.