The amount of cubic feet of water in a tank "t" minutes after it has started to drain is given by G=125(t-25)^2
At what rate did the water flows out during the first 10 minutes?
At what rate is the water running out at the end of 10 minutes?
2006-11-05 05:02:48 · 5 answers · asked by al33191 1 in Science & Mathematics ➔ Mathematics
If you take the derivative of G (with respect to t), you will get an expression for the rate at which the water is flowing out. If you substitute a value for t into that expression, you'll get a numerical value for the rate of flow. Actually, you'll get a negative number, which tells you that the volume is DEcreasing, so you should remove the negative sign to get the rate at which water is flowing out.
You can find the rate "at the end of 10 minutes" by setting t equal to 10.
The rate "during the first 10 minutes" is difficult to answer, because the rate changes constantly (as a function of t). What you probably want is the AVERAGE rate during the first 10 minutes. That is easy to calculate, and doesn't require calculus. Just figure out the value of G when t is 0 and when t is 10. The difference is the amount of water that flowed out in 10 minutes. You'll need to divide it by 10 to get an average rate.
2006-11-05 05:12:27 · answer #1 · answered by actuator 5 · 0⤊ 0⤋
What I think would be to get the Equation to G in terms of "t" then do a differentiation on that equation in terms of time.
From that equation just open up the brackets and expand them.
This will hence make it in terms of dG/dt as in the change of total amount of water / the change of time.
during the 1st ten minutes.. or I would think would be at the starting time which is 0 minutes. Just place 0 into the "t" and you will get the flow rate over time.
Next for the end of 10 minutes. Just place in 10 into the "t" and you will the flow rate over the end of 10 minutes.
Hope this helps. Do correct me if I am wrong.
Cheers.. :)
2006-11-05 05:16:11 · answer #2 · answered by Kaydon 1 · 0⤊ 0⤋
Amount of water in tank initially (t=0) = 78125 cubic feet
So let amount of water flowed out of tank after 't' minutes,
f = 78125 - 125(t-25)^2
rate water flows out = df/dt
df/dt = 6250-150t = 50(125 - 3t)
At the end of 10 minutes the water is flowing out at a rate of 4750 ft^3/min.
2006-11-05 05:21:14 · answer #3 · answered by ? 7 · 0⤊ 0⤋
A hint: the first derivate of a function is the slope of that function.
Instantaneous slopes are defined as rise/run or change in y over change in x.
2006-11-05 05:06:51 · answer #4 · answered by Your Best Fiend 6 · 0⤊ 0⤋
get in touch with a college and ask the question.
2006-11-05 05:09:18 · answer #5 · answered by wetrapme 1 · 0⤊ 0⤋