Clarence
  • Clarence
The area of a circular cell changes as a function of its radius, r, and its radius changes with time r = g(t). If dA/dr=f(r), then the total change in area, ΔA between t=0 and t=1 is ΔA=∫ f(g(t))g′(t)dt. This statement sounds true to me. Thoughts?
Mathematics
schrodinger
  • schrodinger
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BAdhi
  • BAdhi
sounds legit to me.. \[\Delta A = \int \limits_{r_0}^{r_1} \frac{dA}{dr} dr = \int \limits_{r_0}^{r_1} f(r) dr\] \[r = g(t) \implies dr = g'(t) dt\] \[t=0 \implies r = r_0 , t=1 \implies r=r_1\] \[\Delta A = \int \limits_{0}^1 f[g(t)]g'(t) dt\]
Clarence
  • Clarence
Thanks! I thought it was right to, but I figured it'd be better safe than sorry :p Thanks again! Much appreciated :)
Clarence
  • Clarence
By the way, that working out seemed much better for me to understand than what I did :p Keep up the great work!

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BAdhi
  • BAdhi
Its a pleasure :)

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