Water bubbles up at a rate of 1 cm^3/s, forimging a circular pond of depth 0.5 cm in his yard. How quickly is the surface area of the pond covering his lawn?
how would i find r to plug into the volume equation?
The volume of the pond is:
$\displaystyle
V=A d
$
where $\displaystyle d$ is the depth of the water, and $\displaystyle A$ is the surface area of the pond.
So the rate of change of volume is:
$\displaystyle
\frac{dV}{dt}=\frac{dA}{dt}\ d
$
but we are told that this is $\displaystyle 1\ \rm{ cm^2/s}$, and as $\displaystyle d=0.5 \mbox{ cm}$ we have:
$\displaystyle \frac{dA}{dt}=1/d=2\ \rm{cm^2/s}$
RonL