Asked by ivy
a toy balloon is in the shape of a sphere. it is being inflated at the rate of 20 in^3 / min. at the moment that the sphere has volume 64 cubic inches. what is the rate of change of the radius ?
Answers
Answered by
Damon
64 = (4/3)pi R^3
from that find R
dV = surface area * dR (easy way :)
so
dV/dt = 4 pi R^2 dR/dt
20 in^3/min = 4 pi R^2 dR/dt
solve for dR/dt
from that find R
dV = surface area * dR (easy way :)
so
dV/dt = 4 pi R^2 dR/dt
20 in^3/min = 4 pi R^2 dR/dt
solve for dR/dt
Answered by
Reiny
V= (4/3)π r^3
dV/dt = 4π r^2 dr/dt
when V = 64
64 = (4/3)π r^3
48/π = r^3
r = (48/π)^(1/3) = appr 2.4814 ---> I stored the 10 decimal answer
in dV/dt = 4π r^2 dr/dt
20 = 4π(2.4814..)^2 dr/dt
dr/dt = appr .2585 inches/min
dV/dt = 4π r^2 dr/dt
when V = 64
64 = (4/3)π r^3
48/π = r^3
r = (48/π)^(1/3) = appr 2.4814 ---> I stored the 10 decimal answer
in dV/dt = 4π r^2 dr/dt
20 = 4π(2.4814..)^2 dr/dt
dr/dt = appr .2585 inches/min
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