I don't know if you are the same person, but this question is a cut-and-past of
http://www.jiskha.com/display.cgi?id=1422453360
areas are measured in cm^2 , not cm^3
I cannot understand 2 cm2 ´ s -1
The area, A cm3 , of a circle increases at a constant rate of 2 cm2 ´ s -1 . If the initial area of A is 1 cm2 , show that the radius of the circle at time t is given by r=sqrt( 2t+1/ pi) sqrt on both denominator and numerator
2 answers
we know that
dA/dt = 2
and A can be written as pi*r^2
inital value of A, meaning t=0, is 1
from that, we get the equation
A = 2t+1
example:
t = 2
A = 2(2) + 1 = 5
1 + 2 + 2 = 5
we can then write
pi*r^2 = 2t + 1
r^2 = 2t + 1/pi
r = sqrt. 2t+1/pi
dA/dt = 2
and A can be written as pi*r^2
inital value of A, meaning t=0, is 1
from that, we get the equation
A = 2t+1
example:
t = 2
A = 2(2) + 1 = 5
1 + 2 + 2 = 5
we can then write
pi*r^2 = 2t + 1
r^2 = 2t + 1/pi
r = sqrt. 2t+1/pi