# 1. In a chemical reaction, a certain compound changes into another compound at a rate...

## Question:

1. In a chemical reaction, a certain compound changes into another compound at a rate proportional to the unchanged amount. There are 528 grams of the original compound initially and 330 grams after 1 hour. When will 75 percent of the compound be changed? (Round your answer to two decimal places.) hr 2. Suppose that you meet 40 new people each year, but each year you forget 20% of all of the people that you know. If y(t) is the total number of people who you remember after t years, then y satisfies the differential equation y' = 40 - 0.2y. (Do you see why?) Solve this differential equation subject to the condition y(0) = 0 (You knew no one at birth). y =

## Exponential growth:

The chemical reaction and the compounds involved is an example of an exponential decay. The exponential decay equation is given by {eq}A=A_0e^{kt} {/eq} where {eq}k {/eq} determines the rate of the decay.

1

We start initially with {eq}528 {/eq} grams of original compound. Then after an hour we have {eq}330 {/eq}. Writing this into the exponential decay formula

\begin{align} \left(330\right)&=\left(528\right)e^{k\left(1\right)} \\ \frac{330}{528}&=e^k \\ ln\left(e^k\right)&=ln\left(\frac{330}{528}\right) \\ k&=ln\left(\frac{330}{528}\right) \\ k&=ln\left(\frac{5}{8}\right) \\ \end{align}

Now to solve for the time when the compound is {eq}75 \% {/eq}, the final compound should be {eq}25 \% {/eq} of the initial amount

\begin{align} \left(0.25\right)528&=\left(528\right)e^{\left(ln\left(\frac{5}{8}\right)\right)t} \\ \left(0.25\right)&=e^{\left(ln\left(\frac{5}{8}\right)\right)t} \\ \left(0.25\right)&=\left(\frac{5}{8}\right)^t \\ ln\left(0.25\right)&=ln\left(\left(\frac{5}{8}\right)^t\right) \\ ln\left(0.25\right)&=tln\left(\frac{5}{8}\right) \\ t&=\frac{ln\left(0.25\right)}{ln\left(\frac{5}{8}\right)} \\ t \approx 2.94953 \end{align}

$$\boxed{t \approx 2.95 \hspace{1mm} hr }$$

2

The goal of this problem to solve for the differential equation {eq}y'=40-0.2y {/eq}. Now, to see why this is the differential equation and not something else. Let's recall what exactly is {eq}y' {/eq}. Well {eq}y' {/eq} refers to the increase or decrease of the number of people you know, the {eq}40 {/eq}, since it is positive, means every year you are increasing the number of people you know by {eq}40 {/eq} and the {eq}-0.2y {/eq} means that each year we are decreasing or forgetting, so to say, the number of people we know by {eq}0.2 {/eq} or {eq}20 \% {/eq} of all the people we initially known. Now, to solve for the differential equation. Since this is a separable ODE, we can let {eq}y' {/eq} be {eq}\frac{dy}{dt} {/eq}.

\begin{align} y'&=40-0.2y \\ \frac{dy}{dt}&=40-0.2y \\ \frac{dy}{\left(40-0.2y\right)}&=dt \\ \frac{dy}{40-0.2y}&=dt \\ \int \frac{dy}{40-0.2y}&=\int dt \\ \end{align}

Using integration by substitution, let {eq}u = 40-0.2y {/eq}, then {eq}du = -0.2dy {/eq}

\begin{align} \int \frac{dy}{40-0.2y}\times \frac{-0.2}{-0.2}&=\int dt \\ \frac{1}{-0.2}\int \frac{-0.2dy}{40-0.2y}\times &=\int dt \\ \frac{1}{-0.2}\int \frac{du}{u}&=\int dt \\ -\frac{1}{0.2}\int ln\left|u\right|&=\int dt \\ -\frac{1}{0.2}ln\left|u\right|&=t + C \\ -\frac{1}{0.2}ln\left|40-0.2y\right|&=t + C \\ e^{\left(-\frac{1}{0.2}ln\left|40-0.2y\right|\right)}&=e^{\left(t+C\right)} \\ \frac{1}{e^{\left(\frac{1}{0.2}ln\left|40-0.2y\right|\right)}}&=Ce^{\left(t\right)} \\ \frac{1}{e^{\left(ln\left|40-0.2y\right|^{\left(\frac{1}{0.2}\right)}\right)}}&=Ce^{\left(t\right)} \\ \frac{1}{\left(40-0.2y\right)^5}&=Ce^t \\ 1&=Ce^t\left(40-0.2y\right)^5 \\ \frac{1}{Ce^t}&=\left(40-0.2y\right)^5 \\ \frac{1}{\left(Ce^t\right)^{\left(\frac{1}{5}\right)}}&=\left(40-0.2y\right)^{5\times \frac{1}{5}} \\ \frac{C}{e^{\frac{1}{5}t}}&=40-0.2y \\ \frac{C}{e^{\frac{1}{5}t}}-40&=-0.2y \\ y&=\frac{\frac{C}{e^{\frac{1}{5}t}}-40}{-0.2} \\ y&=-\frac{5C}{e^{\frac{1}{5}t}}+200 \end{align}

Using the {eq}y(0) = 0 {/eq} and then solving for {eq}C {/eq}

\begin{align} y&=-\frac{5C}{e^{\frac{1}{5}t}}+200 \\ 0&=-\frac{5C}{e^{\frac{1}{5}\left(0\right)}}+200 \\ 0&=-\frac{5C}{e^1}+200 \\ \frac{5C}{e^1}&=200 \\ 5C&=200e \\ C&=\frac{200e}{5} \\ C&=40e \end{align}

Finally,

$$y=-\frac{5\left(40e\right)}{e^{\frac{1}{5}t}}+200 \\$$

$$\boxed {y=-\frac{200e}{e^{\frac{1}{5}t}}+200 }$$