
#1
Mar1609, 06:21 PM

P: 79

1. The problem statement, all variables and given/known data
Show that the rate of change of the area of a circle with respect to its radius is the same as the circumference of the circle. Can you suggest why? 2. Relevant equations A = [tex]\pi[/tex]r[tex]^{2}[/tex] = f(r) L = 2[tex]\pi[/tex]r = g(r) 3. The attempt at a solution I have showed that the derivative of f(r) is equal to g(r). But I have no idea why the area and the circumference of the circle are related in such a way. Any suggestions greatly appreciated. Thank you. 



#2
Mar1609, 06:26 PM

P: 52

Start by thinking about what any derivative of a function is describing in general, and then how it applies here specifically.




#3
Mar1609, 06:31 PM

P: 79

The derivative describes the slope of a tangent to the circle which is perpendicular to the radius... but I don't seem to go anywhere from here... hmmm



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