SOLUTION: The level of water in a rain gutter is given by the following V(t) = {{{ 10t sqrt( 4-t^2 ) }}}. Note : The square root contains 4-t^2 a. Determine dV/dt b. Is the water lev

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Question 1163278: The level of water in a rain gutter is given by the following
V(t) = . Note : The square root contains 4-t^2
a. Determine dV/dt
b. Is the water level rising or falling at t = 1.5 seconds?
Thank you.
10t(√4-t^2)

Found 2 solutions by solver91311, Boreal:
Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Use the product rule and the chain rule:

Evaluate

If then rising

If then falling

And, of course, if it is equal to zero it is doing neither, it is either at a minimum or a maximum.

John

My calculator said it, I believe it, that settles it

Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
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

-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.