Why take the integral of this instead of just plugging in the number?

  • Thread starter Thread starter Xetman
  • Start date Start date
  • Tags Tags
    Integral
Xetman
Messages
8
Reaction score
0

Homework Statement



An oil storage tank ruptures at time t = 0 and oil leaks from the tank at a rate of r(t) = 100 e^(-0.01t) liters per minute. How much oil leaks out during the first hour?

Homework Equations



r(t) = 100 e^(-0.01t)

The Attempt at a Solution



Eh, I already got the answer, I just did a definite integral from 0 to 60 for the equation.
My question is: Why would you find the integral instead of just plugging in 60 for t into that equation? Is it because that equation is a rate, so it's a derivative and you have to take the integral to get the original equation?
 
Physics news on Phys.org
Rate multiplied by Time give the amount of oil leaked, but the rate is not constant. As you guessed, you integrate because the rate changes with time. NOT constant.

Another explanation: What if you just plug in the number?
If you just plug in, what... 60? Then this give you r(60)=100e^(-0.01*60)=54.88 liters per minute. But from time 0 to time 60, the rate was NOT 54.88, so you would be calculating the liters for 60 minutes at the mostly wrong rate of 54.88 liters per minute. This is a very large underestimation. The actual rate is different at every point in time from 0 to 60, and this rate VARIES with time. So,... integrate.
 
Last edited:
Xetman said:

Homework Statement



An oil storage tank ruptures at time t = 0 and oil leaks from the tank at a rate of r(t) = 100 e^(-0.01t) liters per minute. How much oil leaks out during the first hour?

Homework Equations



r(t) = 100 e^(-0.01t)

The Attempt at a Solution



Eh, I already got the answer, I just did a definite integral from 0 to 60 for the equation.
My question is: Why would you find the integral instead of just plugging in 60 for t into that equation? Is it because that equation is a rate, so it's a derivative and you have to take the integral to get the original equation?

Why did you integrate? That was the correct thing to do, but you claim you did it without knowing why. Did somebody tell you to do it, or what?
 
Nah lol the question was in the U-Substitution section of my book so I figured it wouldn't be that easy just to plug in a number.
 
Xetman said:
Is it because that equation is a rate, so it's a derivative and you have to take the integral to get the original equation?

You've kind of answered your own question here. Yes, the expression you're given is a rate (an instantaneous volume per unit time), so you need to integrate with respect to time to get the total volume that leaks out over a time interval. The definite integral just sets the bounds (i.e. answers the question: what's the total volume leaking out from the start [t=0] till t = 60?).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top