C4 Further Differentiation small changes Watch

bigmansouf
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Please check I am right

Question:
if the pressure and volume of a gas are  \rho and  \nu then Boyle's law states  \rho\nu = \mathrm{constant (k)} . if  \delta \rho and  \delta \nu denote corresponding small changes  \rho and  \nu express  \frac{\delta \rho}{\rho} in terms of  \frac{\delta \nu}{\nu}


My attempt:
theory being used to answer this question:


 \frac{\delta y}{\delta x} \approx \frac{\mathrm{d} y}{\mathrm{d} x}
thus
 \delta y \approx \frac{\mathrm{d} y}{\mathrm{d} x}\times \delta x

thus

 \frac{\delta \rho}{\delta \nu} \approx \frac{\mathrm{d} \rho}{\mathrm{d} \nu}
thus
 \delta \rho \approx \frac{\mathrm{d} \rho}{\mathrm{d} \nu}\times \delta \nu


 \rho\nu = \mathrm{constant (k)}
 \rho = \mathrm{constant (k) \nu^{-1}}
 \frac{\mathrm{d} \rho}{\mathrm{d} \nu} = - \mathrm{constant (k)\nu^{-2}}

 \delta \rho \approx - \mathrm{constant (k)\nu^{-2}} \times \delta \nu

\frac{\delta \rho}{\rho} \approx -\frac{\mathrm{constant (k)\nu^{-2}}}{constant (k)\nu^{-1}}\times \delta\nu
 \frac{\delta \rho}{\rho} \approx  - \frac{\delta\nu}{\nu}

I would like to know if the method used to find the answer

 \frac{\delta \rho}{\rho} =  - \frac{\delta\nu}{\nu} is right

Thank you
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mqb2766
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Looks ok to me
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bigmansouf
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(Original post by mqb2766)
Looks ok to me
thank you
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