Re: 30 -- CCS AP Calculus

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Subject: Re: 30

Author:
Benny

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Date Posted: 20:15:53 09/25/03 Thu
In reply to: kenny 's message, "30" on 20:01:17 09/25/03 Thu

>it takes 12 hrs to drain a storage tank by opening the
>valve at the bottom. the depth y of fluid in the tank
>t hrs after the valve is opened is given by the
>formula: y=6[1-(t/12)]^2
>
>a.) find the rate dy/dt(m/h) at which the water level
>is changing at t time.
>
>b.) when is the fluid level in the tank falling
>fastest? slowest? What are the values of dy/dt at
>these times?
>
>c.) graph y and dy/dt together and discuss the
>behavior of y in relation to the signs and values of
>dy/dt?

a) just multiply out the expression and take the derivative.

b) you know this is all happening in 12 hours, when you look at the derivative, the rate is decreasing the fastest at t=0 and the slowest at t=12.

c) just graph y and y' on the same axis, the graph will show as y goes to the right, the negative value of y' goes toward 0, meaning, as time goes on, water is draining at at slower rate.

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