Peak oil extraction The owners of an oil reserve begin...

Peak oil extraction The owners of an oil reserve begin extracting oil at time $t=0 .$ Based on estimates of the reserves, suppose the projected extraction rate is given by $Q^{\prime}(t)=3 t^{2}(40-t)^{2}$ where $0 \leq t \leq 40, Q$ is measured in millions of barrels, and $t$ is measured in years. a. When does the peak extraction rate occur? b. How much oil is extracted in the first $10,20,$ and 30 years? c. What is the total amount of oil extracted in 40 years? d. Is one-fourth of the total oil extracted in the first one-fourth of the extraction period? Explain.

Peak oil extraction The owners of an oil reserve begin extracting oil at time $t=0 .$ Based on estimates of the reserves, suppose the projected extraction rate is given by $Q^{\prime}(t)=3 t^{2}(40-t)^{2}$ where $0 \leq t \leq 40, Q$ is measured in millions of barrels, and $t$ is measured in years.
a. When does the peak extraction rate occur?
b. How much oil is extracted in the first $10,20,$ and 30 years?
c. What is the total amount of oil extracted in 40 years?
d. Is one-fourth of the total oil extracted in the first one-fourth of the extraction period? Explain.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, showing that the process of integrating a rate function over time yields the total accumulated quantity. It also implies that differentiating the accumulation function recovers the original rate function. This theorem is pivotal in solving real-world problems involving cumulative processes, such as calculating total oil extraction based on a given rate function.

Optimization of Functions

Optimization techniques are used to find the maximum or minimum values of a function. In the context of rate functions, determining when the extraction rate reaches its peak involves finding the time at which the derivative of the extraction rate (or the rate function itself, when it is already derived) is maximized. This is a common application of differentiating a function and analyzing its critical points.

Rate Function and Accumulation

A rate function describes how a quantity changes over time. Integrating this rate function over an interval gives the accumulation function, which represents the total amount accrued during that period. This concept is central in moving from a description of instantaneous change to the overall quantity gathered or depleted, such as total oil extraction over time.

Definite Integration

Definite integrals allow calculation of the total accumulation of a quantity when its rate of change is known. By computing the integral of a rate function over a specified interval, one can determine exactly how much of the quantity, like oil, has been extracted between two time points.

Transcript

00:03 So here we're looking at oil extraction.

00:06 So q prime of t is equal to 3t squared times 40 minus t squared.

00:12 And this is going to be our extraction rate.

00:16 So this is derivative of q.

00:19 And it's from time 0 to 40.

00:22 So the question at first asks is when does the peak extraction rate occur? and we know that the peaks are going to be when the extraction rate, when our derivative of q is equal to 0.

00:38 Our derivative q is going to be equal to 0.

00:42 So we can make that equal to 0.

00:50 And you can see that t equals 0 at t equals 0 and t equals 40.

00:57 So that's where our extraction rates are.

01:04 Next, we have, we're going to be figuring out how much oil is extracted in 10 years.

01:14 So in 10 years, so we're going to have our integral.

01:17 We're going to do 1 to 10.

01:19 And we have 3t squared, 40 minus t squared.

01:25 So the first thing we're going to want to do is we're going to want to integrate this.

01:28 So to integrate this, i would expand it.

01:32 So you would do 40 minus t squared, and you would get 3t squared times t squared minus 80t plus 1 ,600, which then becomes 3t to the power of 4 minus 240 t to the power of 3 plus 1 ,600 t squared.

02:14 And if we integrate that, we get 3t to the power of 5 over 5 minus 60 t to the power of 4 plus 1 ,600 t to the power of three.

02:35 Oh, yeah, sorry.

02:39 One quick stipulation.

02:41 This 1600 should be a 4 ,800.

02:54 Okay, so we have our integral.

02:58 So all we got to do is plug it in.

03:00 We got this.

03:01 We're going to do, we're going to replace every t with 10.

03:17 What i would suggest you do here is i would suggest that you use a calculator to calculate every single one of these numbers...

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