Question

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?

Answer 1

sounds legit to me..
\[\Delta A = \int \limits_{r_0}^{r_1} \frac{dA}{dr} dr = \int \limits_{r_0}^{r_1} f(r) dr\]
\[r = g(t) \implies dr = g'(t) dt\] \[t=0 \implies r = r_0 , t=1 \implies r=r_1\]
\[\Delta A = \int \limits_{0}^1 f[g(t)]g'(t) dt\]

Answer 2

Thanks! I thought it was right to, but I figured it'd be better safe than sorry :p Thanks again! Much appreciated :)

Answer 3

By the way, that working out seemed much better for me to understand than what I did :p Keep up the great work!

Answer 4

Its a pleasure :)