Question

Water is pouring into a container at a rate given by ri = t^3(gallons/hour), and water is pouring out of the same container at a rate given by ro(t) = e^t(gallons/hour). If there are 100 gallons of water at time t = 0, how many gallons of water will be in the container at t=4 hours?

Answer 1

@ganeshie8

Answer 2

rate of change of water in container = (in rate) - (out rate)

Answer 3

say "y" is the amount of water present in container at any given time

Answer 4

then you can setup a differential equation :

\(\large \dfrac{dy}{dt} = t^3 - e^t\)

Answer 5

solve it

Answer 6

@ganeshie8 64 - e^4

Answer 7

doesnt look right

Answer 8

whats ur solution to above equation ?

Answer 9

Ahh I see where the mistake was !

Answer 10

9.6, but I think that's wrong

Answer 11

you need to solve the differential equation first, before plugging in t = 4

Answer 12

\(\large \dfrac{dy}{dt} = t^3 - e^t\)

Solving this means, solving the function "\(y\)"

Answer 13

Integrate both sides

Answer 14

\(\large \dfrac{dy}{dt} = t^3 - e^t\)

\(\large y = \int t^3 - e^t dt\)

Answer 15

what do u get ?

Answer 16

t^4/4 -e^t, then we plug in 4?

Answer 17

I still got 9.4 lol

Answer 18

Yes, but not yet...
where is the arbitrary constant ?

Answer 19

\(\large \dfrac{dy}{dt} = t^3 - e^t\)

\(\large y = \int t^3 - e^t dt\)


\(\large y = \dfrac{t^4}{4} - e^t + c\)

Answer 20

right ?

Answer 21

we're given : ` If there are 100 gallons of water at time t = 0,`

Answer 22

Oh, yes. Lol, I always forgot the constant and my teacher marks points off for it -_-

Answer 23

hahah happens :) plugin t = 0, y = 100 and solve \(c\)

Answer 24

100 = (0^4)/4 + (e^0) + C
= 101

Answer 25

Then we do it for t - 4 and subtract?

Answer 26

I meant t = 4

Answer 27

c = 101 is correct,
so the equation for amount water in tank is :

\(\large y = \dfrac{t^4}{4} - e^t + 101\)

Answer 28

yes... simply plugin t = 4
that gives amount of water in tank after 4 hours

Answer 29

110.402

Answer 30

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