V= 20- 2t- 0.1t^2
a) v=0 when t=10(√3-1) = 7.32
How did you get 10? v(10) = 20-20-10
b) dv/dt = -2 - .2t
at t=5, dv/dt = -2-1 = -3
Apparently, you have some misunderstanding of how related rates work.
Volume in liters of antifreeze remaining in a leaking radiator after t minutes is V= 20- 2t- 0.1t^2
a) how long does it take to drain the antifreeze mixture from the radiator?
i got t= 10
b) how fast is volume changing when t=5?
i got 9.75
is this right. thanks
2 answers
I read more about related rates, and have now understood these types of problems. Thank you!