Yes that is exactly what you do
A 1000 L tank loses water so that, after t days, the remaining volume is
v(t) = 1000[1-(t/10)]^2 for 0<t<10.
How rapidly is the water being lost when the tank is half full?
I got the derivative fine. -200(1-(t/10))
but then I am having troubling solving.
if the volume is half than v(t) will equal 500, correct? because half of 1000 is 500.
so I solve v(t) = 500 for t?
then input this value of t into the v'(t)?
your help is appreciated!