# Solving the derivative for a variable star's period and radius

I am trying to solve for $$R$$ as this is the radius of a RR Lyrae type of variable star from the formula: $$P^{-1}\frac{dP}{dT}=1.5R^{-1} \frac{dP}{dT}$$ $$P$$ is the star's brightness variability period and of course here we are to be taking the derivative of the period with respect to time.

Here are my questions: The period IS a rate of change, so isn't the value I get from my data for the period the derivative already? As such could I just replace the $$P^{-1}\frac{dP}{dT}$$ with that value? But then how do I take the derivative of the $$R^{-1}\frac{dP}{dT}$$? Surely I don't just replace this with the period value as well; that would result in turning this into a simple algebraic equation. Do I need to take an integral to solve for $$R$$, and if so can someone show me how that is done in this instance? I am using a spreadsheet to sort my data and eventually determine the period of several of these types of stars, so is there a different value I need to put in for $$P$$ and then let $$T$$ be the overall length of time for imaging (for instance, suppose I imaged a star for $$2$$ weeks; would $$T$$ be $$14$$ days)?

Thanks to all who are working on this!
-Melanie and Dave

• What's $T$, temperature or time? – Nosrati Aug 1 '18 at 17:03
• T is time, the unit I believe is in days – MPGDS APU Aug 3 '18 at 1:05
• I presume there's a typo in your equation: shouldn't it be $dR/dT$ on the right hand side? – Intelligenti pauca Jan 25 '19 at 20:29

Your equation (I corrected a typo) $$P^{-1}\frac{dP}{dT}=1.5R^{-1} \frac{dR}{dT}$$ can be rewritten as $$\frac{d}{dT}\ln P=\frac{d}{dT}(1.5\ln R),$$ that is: $$\ln P=\ln R^{1.5}+\text{constant},$$ and finally: $$P=k\cdot R^{1.5},$$ where $$k$$ is a constant, to be determined from initial conditions.