Suppose r(t)=r0 e^-L t with k>0, is the rate at which a nation extracts oil, where r0=10^2 barre

Suppose r(t)=r0 e^-L t with k>0, is the rate at which a...

Key Concepts

Parameter Constraint Analysis in Modeling

In modeling scenarios, constraints on parameters (like the decay constant k) are analyzed to achieve certain outcomes, such as ensuring that cumulative extraction does not exceed total reserves. This involves setting up inequalities based on the derived expressions for cumulative extraction and solving for the parameter thresholds that guarantee sustainability or other desired conditions.

Improper Integrals and Convergence

Improper integrals occur when the bounds of integration extend to infinity. In cases where the integrand, such as an exponentially decaying function, decreases sufficiently fast, the integral converges to a finite limit. This convergence is crucial for determining the total accumulated value over an infinite time horizon and assessing long-term behaviors in resource models.

Limit of a Cumulative Function

Taking the limit of a cumulative extraction function as time approaches infinity reveals the ultimate total amount extracted over the lifetime of the resource. This limit helps to compare the modeled extraction with the available reserves, indicating whether the resource will eventually be exhausted or remain unextracted in the long term.

Definite Integration of Rate Functions

When a rate function r(t) describes how a quantity is changing over time, the total amount accumulated over an interval can be found via definite integration. In the context of extraction problems, integrating the rate function from the starting time to a specific time t yields the cumulative amount extracted up to that point, providing a bridge between the instantaneous rate and long-term totals.

Exponential Decay

Exponential decay refers to a process in which a quantity decreases at a rate proportional to its current value, typically modeled by functions of the form r(t) = r? e^(–kt). This concept is essential for understanding situations where the rate of change diminishes over time, such as radioactive decay, depreciation of assets, or, in this case, the decreasing rate of resource extraction.

Transcript

00:01 So giving r of t to be equal to r0 minus lt with k greater than 0, we have r0 to be equal to 10 squared barrels a in total oil reserve is 2 times 10 times 10 the power line was barrels okay so for a we want to find the total amounts of oil extracted by the nation after tea years so total amounts of oil extracted after tea yes so if we have a r of t to be equal to the q over the t this implies that the integral from zero to t of k of t is going to the t is equal to the integral to t of what r t be what the t b w t okay and this is going to be equal to integral 0 to t of what r not a minus k word t the word t so what do i have so this gives me we have what k of word t to be equal to this is equal to k of word t so k of word t because it's a constant from 0 to t e minus k t the word d t which is equal to what is r not i not is 10 so i have 10 over k but it is what minus e minus what k t from what 0 to 1 and this gives us what this is an interval 10 squared over key r is what then squared so then squared over key to write x1 minus words e exponent so this is the amounts of words or you extracted after saying with years amount of oil extracted after towards saying yes then we want to evaluate the limits of the amount extracted and explain the meaning of dates then so b if i evaluate the limits here of what c it's implies that the limits as c approaches words infinity of words 10 10 square over k 1 minus e k t is going to be the limit of words as t approaches what infinity of what k of what c which is equal to 10 square over what k okay so if i evaluate thing this is what the results i get then for if i want to find the minimum decay so let's i want to find the minimum weight decay constant key for which the total oil reserves will last what's forever so the minimum key the minimum decay constant okay so you are looking for decay the key the key constant okay constant okay for which the total oil will last forever so if i set 10 square over k to be equal to the total of your reserve which is two times 10 is a power 9 then this implies that one over k is going to be two times 10 is the power word 7 and what does k give us so from here you realize like k is going to be equal to 1 over 2 times 10 is the power 7 which is equally equals what 5 times what 10 to the power minus what 8 so hence the minimum decay we are looking at the minimum decay constants constant k for which the total oil raises will last forever is equal to 5 times 10 to the power was minus word 8 then consider if r is to the power 10 to suppose that r is equal to 2 times 10 to the power 7 and k we know it's what 5 times 10 to the power minus 8 then what is going to be q c is going to be minus 2 times 10 to the power minus 7 over 5 times 10 to the power minus 8 e exponent minus 5 times 10 to the power minus 8 words t plus 2 times 10 to the power 7 over 5 times 10 to the power was minus what 8 so that if i simplify then what i have what i have here is if i simplify you realize that you have two times 20 to the power 9 that is q of 2 then to be equal to minus 2 over 5 times 10 to the power 15e minus 5 times 10 to the power 15 e minus 5 times 10 to the power minus 8 still t plus 2 over 5 times 10 to the power was 15 okay, so we want to solve for the time.

09:05 So this is going to give us minus 4 times 10 to the power 14.

09:12 E 5 times 10 to power 8, t plus 4 times 10 to the power word 14.

09:25 Okay, so if it's simplified, so let's have ye here.

09:31 Here if i have e to the power minus 5 times 10 is the power 8 is going to give me what 4 times 10 to the power 14 minus 2 times 10 the power 9 or divided by minus 4 times 14 to the power 14 and this is going to be equal to 1 .0 0 .00 5...

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