a = pi r^2
da/dr = 2pi r
The area A = πr2 of a circular puddle changes with the radius. At what rate does the area change with respect to the radius when r = 5ft?
2 answers
A = pi r^2
A/dr = 2 pi r
(which is the circumference of course, draw a circle and a ring of thickness dr on the outside and look at the added area)
if r = 5
dA/dr = 10 pi
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this is most interesting if the radius is changing with time
r = k t
dr/dt = k
dA/dt = dA/dr * dr/dt
dA/dt = k (2 pi r)
A/dr = 2 pi r
(which is the circumference of course, draw a circle and a ring of thickness dr on the outside and look at the added area)
if r = 5
dA/dr = 10 pi
===================
this is most interesting if the radius is changing with time
r = k t
dr/dt = k
dA/dt = dA/dr * dr/dt
dA/dt = k (2 pi r)
1 answer