Understanding Thermodynamics: Demystifying the Chain Rule for Integration

  • Thread starter Thread starter th3plan
  • Start date Start date
  • Tags Tags
    Integrating
th3plan
Messages
93
Reaction score
0
I am reading a thermodynamics book. I am confused on how they say use the chain rule Here. it makes no sense to me how they go from dV/dt to (dV/ds)(ds/dt) . I know how the chain rule works ,just don't know where they got these values
Picture 2.png

 
Physics news on Phys.org
Well they use the chain rule in the normal way here. If V is a function of s and s is a function of t, V(s(t)), then the chain rule tells you that (V \circ s)'(t)=V'(s(t))s'(t). Is that more familiar? This is Identical to \frac{dV}{ds}\frac{ds}{dt}. To display this a bit more clearly let's start with the function V(s(t)). We then use the chain rule by substituting u=s(t). So (V \circ s)'(t)=V'(u)u'=\frac{d V}{d u} \frac{d u}{d t}=\frac{d V}{d s} \frac{d s}{d t}.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top