1. ## Rate problem

1. A large tank of canola oil ruptures at a time
t = 0 minutes, and the oil leaks out at a rate of r(t) = 100e^-0.01(t) liters per minute. How much oil leaks out within the first hour?

I am assuming that you would set u = 100, and dv = e^-0.01(t), so then du = 0 and v = e^t/(-0.01)...right? But I have no idea how to go from there..someone please help?

2. Originally Posted by Jgirl689
1. A large tank of canola oil ruptures at a time
t = 0 minutes, and the oil leaks out at a rate of r(t) = 100e^-0.01(t) liters per minute. How much oil leaks out within the first hour?

I am assuming that you would set u = 100, and dv = e^-0.01(t), so then du = 0 and v = e^t/(-0.01)...right? But I have no idea how to go from there..someone please help?

liters of leaked oil in the first hour = $\int_0^{60} r(t) \, dt$

3. You don't have to use integration by parts to integrate this function. I would recommend reviewing the rules when integrating e^nx. If it still doesn't make sense, come back please

4. $\int_a^b e^{kt} \, dt = \frac{1}{k}\left[e^{kt} \right]_a^b$

5. Originally Posted by Jgirl689
1. A large tank of canola oil ruptures at a time
t = 0 minutes, and the oil leaks out at a rate of r(t) = 100e^-0.01(t) liters per minute. How much oil leaks out within the first hour?

I am assuming that you would set u = 100, and dv = e^-0.01(t), so then du = 0 and v = e^t/(-0.01)...right? But I have no idea how to go from there..someone please help?

With $dv= e^{-0.01t}$, $v= -\frac{e^{-0.01t}}{0.01}= -100e^{-0.01t}$.

And, if you are using integration by parts with du=0,

$\int udv= uv- \int vdu= uv$.

In other words, with u a constant, the integral of udv is just uv! The integral of "a constant time f(x)" is that constant times the integral of f(x). I'll bet you learned that before you learned integration by parts!

6. Ok, I came with an approach, I did the integral part:

u = 100, du = 0, dv = e^(-.01t), v = e^(.01t)/ .01 making the integral of vdu = 0 (because du = 0), so integral of udv = uv = 100 * (e^(.01t)/.01)

then I just substituted t with 0, and 60 and substracted, since that is the integral....

100(e^(-0.01*0)/-.01 = 100(e^0)/-.01 = 100/-.01 = -10000
100(e^-0.01*60)/-.01 = 100(e^-.6)/-.01 = 54.88/-0.01 = -5488

-5488 - (-10000) = -5488 + 10000 = 4511 liter per minute.. when it is done with a calculator..or

-10000 e^(-. 6) + 10000 liters per minute

Is that correct?

7. integrating liters per minute w/r to minutes yields liters ...

$\int_0^{60} 100 e^{-.01t} \, dt \approx 4512$ liters