A chain smoker smokes five cigarettes every hour. From each...

A chain smoker smokes five cigarettes every hour. From each cigarette, 0.4 mg of nicotine is absorbed into the person's bloodstream. Nicotine leaves the body at a rate proportional to the amount present, with constant of proportionality -0.346 if $t$ is in hours. (a) Write a differential equation for the level of nicotine in the body, $N,$ in $\mathrm{mg},$ as a function of time, $t,$ in hours. (b) Solve the differential equation from part (a). Initially there is no nicotine in the blood. (c) The person wakes up at 7 am and begins smoking. How much nicotine is in the blood when the person goes to sleep at 11 pm ( 16 hours later)?

A chain smoker smokes five cigarettes every hour. From each cigarette, 0.4 mg of nicotine is absorbed into the person's bloodstream. Nicotine leaves the body at a rate proportional to the amount present, with constant of proportionality -0.346 if $t$ is in hours.
(a) Write a differential equation for the level of nicotine in the body, $N,$ in $\mathrm{mg},$ as a function of time, $t,$ in hours.
(b) Solve the differential equation from part (a). Initially there is no nicotine in the blood.
(c) The person wakes up at 7 am and begins smoking. How much nicotine is in the blood when the person goes to sleep at 11 pm ( 16 hours later)?

Key Concepts

Exponential Decay

Exponential decay describes processes in which a quantity decreases at a rate proportional to its current value. This concept is key in modeling situations where substances or populations decrease over time, and it appears as part of the natural loss term in many differential equations.

Differential Equation Modeling

This concept involves formulating a mathematical model that describes how a quantity changes over time. In many real-world scenarios, such as drug or pollutant levels in the bloodstream, the rate of change is determined by both inputs (sources or production) and outputs (losses or elimination), leading to an equation that balances these processes.

First-Order Linear Differential Equations

First-order linear differential equations are equations involving the first derivative of a function and the function itself, where the equation can be written in the form dN/dt + p(t)N = g(t). Such equations are fundamental in modeling processes that include proportional decay or growth combined with constant or variable sources.

Initial Value Problems

An initial value problem specifies the value of the solution at a particular point, allowing one to find a unique solution to a differential equation. This is critical in real-world applications where the state of a system is known at the beginning of observation, guiding the evolution of the process over time.

Integrating Factor Method

The integrating factor method is a technique used to solve first-order linear differential equations. It involves multiplying the equation by a strategically chosen function (the integrating factor) to transform the left side into the derivative of a product, making it easier to integrate and solve for the unknown function.

Transcript

00:01 Okay, so what we have here is a problem involving nicotine in a cigarette that is absorbed in the bloodstream of a smoker.

00:08 So suppose each cigarette contains 0 .4 milligram of nicotine and a person smokes five cigarettes per hour, we multiply this by 5 and will have 2 milligrams.

00:20 So we are also given a proportionality constant which is equal to 0 .3 .46, where in the rate at which the nicotine, in exits the body is proportional to the amount present in it.

00:34 So we can express it as d n over dp is equal to 2 minus 0 .346 n.

00:45 So we want to isolate the proportionality constant and we do this by d n over d t is equal to negative 0 .346 times n minus 5 .7.

01:00 So this is now a separable differential equation.

01:05 So what we're going to do now is to rearrange the whole equation.

01:10 So we have dn over n minus 5 .78 is equal to negative 0 .346 bt.

01:21 So integrating both sides, we now have ln of n minus 5 .78 is equal to negative 0 .346 bt.

01:30 460 plus c.

01:34 So to further simplify this, we want to get the power of p of both sides, and we do this by doing e raise to ln of n minus 5 .78 is equal to e raise to negative 0 .346t plus c.

01:50 So we know that we can just write the left side of the equation as n minus 5 .78, and the c here is just a constant, so it can take any value.

02:02 So now we have ce raised to negative 0 .346t...

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