Question

The number of gallons of water in a swimming pool t minutes after the pool has started
to drain is Q(t) = 50(20 - x)2. How fast is the water running out at the end of 12 minutes?

Answer 1

I got -800 gal/min
but the key states +800 gal/min
What am I doing wrong?

Answer 2

calculus problem

Answer 3

you found the rate as -800 g/min
in English you could say this means the water leaves at a rate of 800 g/min

Answer 4

ok so you have to find Q'(12)...... the derivative function where t=12
use chain rule....inside = 20-t , derivative of inside is -1
outside = 50u^2 , derivative = 100u
\[Q'(t) = (-1)100(20-t)\]
plug in t=12
\[Q'(12) = -100(20-12) = -800\]

Answer 5

oh crap i didnt need to do that :{

Answer 6

ok so I am correct the negative represent the water leaving just like in related rates when we add negative and positive signs when needed

Answer 7

thanks for helping me

Answer 8

yes, your math is correct (as confirmed by dumb cow). So it comes down to semantics
water leaving versus water coming in at a negative rate.