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 May 1st, 2012, 08:37 PM #1 Newbie   Joined: Apr 2012 Posts: 18 Thanks: 0 Rate of changes Think this goes in this forum Speed of sound in air at 0 degrees celsius (or 273 kelvins) is 331.5m/s but speed v increases as the temperature T rises. we know that the rate of change of v in respect to T is (dv/dT) = (331.5) / (2 sqrt(273T) where v is metres and T kelvins. find a formula that expresses v as a function of the temperature (T) how would i do this? thanks
 May 1st, 2012, 10:01 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Rate of changes We are given the initial value problem: $\frac{dv}{dT}=\frac{331.5}{2\sqrt{273T}}$ where $v(273)=331.5$ and $0. Write the ordinary differential equation as: $dv=\frac{331.5}{2\sqrt{273}}T^{\small{-\frac{1}{2}}}\,dT$ Now integrate: $\int\,dv=\frac{331.5}{2\sqrt{273}}\int T^{\small{-\frac{1}{2}}}\,dT$ $v=\frac{331.5}{2\sqrt{273}}\cdot\frac{T^{\small{\f rac{1}{2}}}}{\frac{1}{2}}+C$ And we find the general solution: $v(T)=\frac{331.5}{\sqrt{273}}T^{\small{\frac{1}{2} }}+C$ Now, we need to use the initial value to determine the constant of integration, or parameter: $v(273)=331.5=\frac{331.5}{\sqrt{273}}(273)^{\small {\frac{1}{2}}}+C$ $331.5=331.5+C$ $C=0$ And so, the solution satisfying the given condition is: $v(T)=\frac{331.5}{\sqrt{273}}T^{\small{\frac{1}{2} }}=331.5\sqrt{\frac{T}{273}}$

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