Question 1163278
V=10t*(4-t^2)^(1/2)
dV/dt=10t *(1/2)(4-t^2)^(-1/2)*(-2t)+10*sqrt(4-t^2)
This is -10t^2/(sqrt(4-t^2))+10 sqrt(4-t^2).  (the 1/2 and 2 go away)
when t=1.5
dV/dt=-22.50/sqrt(1.75)+10*sqrt(1.75)=-17.01+13.23
=-3.78, which is negative

{{{graph(300,300,-2,3,-10,30,10x*sqrt(4-x^2))}}}

-10t^2/sqrt(4-t^2)=-10*sqrt(4-t^2) when derivative equals 0
t^2=4-t^2
2t^2=4
t^2=2
t=1.414 where the derivative is 0. It was rising until then but it is falling after that, which is where 1.5 seconds is.