Integral Applications: Rate of Change

In summary, after the valve is opened, approximately 0.4997 cubic feet of air is released per second.
  • #1
olicoh
24
0

Homework Statement


The volume of an air mattress is 15 cubic feet. Air escapes at a rate of r(t)=0.25e^(−0.05t), where r is in cubic feet per second. Assuming the mattress is still completely inflated when the valve is opened, how much air is released in the first 30 seconds?

Homework Equations


Here is what I have so far:
V(15) = V(0) + integral[0,15] (0.25e^(-0.05t))
= [-(1/2) (e^(-1/2t))]150 <-- Sorry, I don't know any other way to write the upper/lower bound.

The Attempt at a Solution


= -2.765E-4 - -0.5

Ugh... My problem with this question is I don't even know what to do for the first step. I just used a formula that was in my textbook and inserted numbers. What is the V(15) and V(0) for anyways? What formula should I be using for this problem?EDIT: I redid the problem and got 0.4997 cubic feet of air is released per second, roughly. Is that right?
 
Last edited:
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  • #2
welcome to pf!

hi olicoh! welcome to pf! :smile:

(have an integral: ∫ and try using the X2 icon just above the Reply box :wink:)
olicoh said:
… What is the V(15) and V(0) for anyways?

i've no idea :confused:

V(t) should be the volume at time t …

so V(0) = 15, and dV/dt = r

start again :smile:
 
  • #3


tiny-tim said:
hi olicoh! welcome to pf! :smile:

(have an integral: ∫ and try using the X2 icon just above the Reply box :wink:)


i've no idea :confused:

V(t) should be the volume at time t …

so V(0) = 15, and dV/dt = r

start again :smile:

Thank you! Oh ok... I don't know. My textbook seriously does not explain any of this stuff to me. they give me equations but they don't provide me an explanation. I'm basically always confused.
 
  • #4
olicoh said:
… they give me equations but they don't provide me an explanation.

that's to encourage you to work out the equations for yourself …

have a go :smile:
 

1. What is the concept of rate of change in integral applications?

Rate of change in integral applications refers to the rate at which a certain quantity is changing over a given interval. It is typically measured by finding the slope of a curve on a graph or by calculating the derivative of a function.

2. How is rate of change related to integrals?

The relationship between rate of change and integrals is that integrals are used to calculate the average rate of change of a function over a given interval. This is done by finding the area under the curve of the function.

3. What is the difference between instantaneous and average rate of change?

Instantaneous rate of change refers to the rate of change of a function at a specific point, while average rate of change refers to the rate of change over a given interval. Average rate of change can be calculated using integrals, while instantaneous rate of change is found by taking the derivative of the function at a specific point.

4. How can integrals be used to solve real-world problems involving rate of change?

Integrals can be used to solve real-world problems involving rate of change by representing the problem as a function and finding the integral of that function. The resulting value will represent the total change over the given interval, and can be used to answer questions about the rate of change at specific points within that interval.

5. Can rate of change be negative?

Yes, rate of change can be negative. A negative rate of change indicates that the quantity is decreasing over time or over a given interval, while a positive rate of change indicates that the quantity is increasing. This can be seen on a graph as a downward or upward slope, respectively.

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