Oil is pumped continuously from a well at a rate...

Name: Problem 15, Chapter 9: Applied Calculus
Uploaded: 2020-07-16T02:00:22-08:00
Duration: 4 min 24 s
Channel: Cyrielle Lorio
Description: Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 1 million barrels of oil in the well; six years later 500,000 barrels remain. (a) At what rate was the amount of oil in the well decreasing when there were 600,000 barrels remaining? (b) When will there be 50,000 barrels remaining?

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 1 million barrels of oil in the well; six years later 500,000 barrels remain. (a) At what rate was the amount of oil in the well decreasing when there were 600,000 barrels remaining? (b) When will there be 50,000 barrels remaining?

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 1 million barrels of oil in the well; six years later 500,000 barrels remain.
(a) At what rate was the amount of oil in the well decreasing when there were 600,000 barrels remaining?
(b) When will there be 50,000 barrels remaining?

Key Concepts

Separation of Variables

Separation of variables is a method used for solving differential equations by isolating the variables on different sides of the equation. In exponential decay models, this technique allows the equation dQ/dt = -kQ to be rewritten in a form where integration can be applied to both sides, leading to an explicit solution. This method is fundamental in deriving the exponential decay formula.

Parameter Estimation

Parameter estimation involves determining the constants in a mathematical model that best fit the observed data. For an exponential decay process, this typically means finding the decay constant 'k' based on known values at specific times. Accurate estimation of model parameters is crucial for making reliable predictions and understanding the underlying dynamics of the system.

Exponential Decay

Exponential decay describes processes in which the rate of decrease of a quantity is proportional to the current amount. In such processes, as the amount decreases, the rate of decrease slows down, leading to a predictable, exponentially declining behavior over time. This concept is widely applicable in physics, chemistry, and engineering for modeling decay phenomena such as radioactive decay or depreciation.

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In the context of decay, the differential equation dQ/dt = -kQ models how a quantity Q decreases at a rate proportional to its current value. This formulation is essential for modeling continuous change in systems and provides a framework for solving many real-world dynamic problems.

Transcript

00:01 Okay, so what we have here is an oil that is being pumped from a well with a rate proportional to the amount of oil left in the well.

00:11 So we are going to define a as the amount of oil left in the well.

00:18 So we are also given that at t is equal to zero, the amount of a is equal to 10 raised to 6 or 1 million barrels.

00:26 And at t is equal to 6, a is equal to 500 ,000 barrels.

00:30 So we are required to get d -a of d -t and the amount of time wherein a would be 50 ,000 barrels.

00:40 So for our solution, we can represent d -y, d -a, i mean d -a over d -t as negative k -a.

00:58 So negative sign indicates that the amount of oil in the well is decreasing as you pump it out.

01:04 So k is the proportionality constant.

01:06 So we can rearrange our equation as d -a over a is equal to negative k -d -t.

01:16 So getting the integral of both sides will have ln of a is equal to negative k -t plus c.

01:27 So we can raise both sides to the power of e and we'll have a is equal to e -raise to the negative k -t plus c.

01:38 But remember that c here is constant, so it can take any value.

01:43 So we can represent our equation as a is equal to c, e raise to the negative kt.

01:49 So this is our general solution.

01:54 So we make use of the initial conditions given a while ago to solve for c and k.

01:59 So we start by at t is equal to 0, a is equal to 10 raised to 6.

02:07 So substituting this, so our general solution, we have.

02:10 10 raised to 6 is equal to c, e raised to the negative k of 0...

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