The amount of water in a tank t minutes after it has started to drain is given by
W=100(t-15)^2 gal. a) At what rate is the water running out at the end of 5 minutes? B) What is the average rate at which the water flows out during the first 5 minutes?
2007-11-25 13:46:35 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Rate of change is the derivative.
a)
w' = 100*2 (t-15)
w' = 200 (t-15)
w'(5) = 200 (5-15) = -2000 gal/min
b)
(w'(0) + w'(5) ) / 2
(-3000 + -2000 ) / 2
average rate = -2500 gal/min
2007-11-25 13:56:45 · answer #1 · answered by ? 3 · 0⤊ 0⤋
Assuming t is in minutes:
W(t) = 100(t-15)^2 = 100 t^2 - 3000 t + 22500
W'(t) = 200t - 3000
W'(5) = 1000 - 3000 = -2000 gpm
W'(0) = -3000 gpm
Since it's flowing out at 3000 gpm at the start and 2000 gpm after five minutes, the average rate is the average of thise values: -2500 gpm.
2007-11-25 13:53:27 · answer #2 · answered by poorcocoboiboi 6 · 0⤊ 0⤋
a. Take dW/dt and evaluate at t=5
b. Find W at 5 minutes time. W at time=0 is 22500 gal. Take the difference and divide by 5 min.
2007-11-25 13:59:02 · answer #3 · answered by cattbarf 7 · 0⤊ 0⤋
bear in concepts that the region function is the necessary of v(t) It is going like this: s(t) v(t) a(t) jerk besides: combine the equation v(t). i'm getting: s(t) = t^3/3 -t^2/2 -6t Now right here's the section i do not truly bear in mind... You upload a consistent t0. on condition that they grant you with that the initial position is 7 the basically top result's: s(t) = t^3/3 -t^2/2 -6t +7 << answer desire that enables.
2016-10-25 02:02:16 · answer #4 · answered by corujo 4 · 0⤊ 0⤋