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# 数学代写|微积分代写Calculus代考|MATH1023 A DIFFERENTIAL EQUATION

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## 数学代写|微积分代写Calculus代考|A DIFFERENTIAL EQUATION

If $B(t)$ represents the number of bacteria present in a given population at time $t$, then the preceding discussion suggests that
$$\frac{d B}{d t}=K \cdot B(t),$$
where $K$ is a constant of proportionality. This equation expresses quantitatively the assertion that the rate of change of $B(t)$ (that is to say, the quantity $d B / d t)$ is proportional to $B(t)$. To solve this equation, we rewrite it as
$$\frac{1}{B(t)} \cdot \frac{d B}{d t}=K$$
We integrate both sides with respect to the variable $t$ :
$$\int \frac{1}{B(t)} \cdot \frac{d B}{d t} d t=\int K d t$$
The left side is
$$\ln |B(t)|+C$$
and the right side is
$$K t+\tilde{C},$$

where $C$ and $\tilde{C}$ are constants of integration. We thus obtain
$$\ln |B(t)|=K t+D,$$
where we have amalgamated the two constants into a single constant $D$. Exponentiating both sides gives
$$|B(t)|=e^{K t+D}$$
or
$$B(t)=e^D \cdot e^{K t}=P \cdot e^{K t} .$$
Notice that we have omitted the absolute value signs since the number of bacteria is always positive. Also we have renamed the constant $e^D$ with the simpler symbol $P$. Equation $(\star)$ will be our key to solving exponential growth and decay problems. We motivated our calculation by discussing bacteria, but in fact the calculation applies to any function which grows at a rate proportional to the size of the function. Next we turn to some examples.

## 数学代写|微积分代写Calculus代考|BACTERIAL GROWTH

EXAMPLE $6.30$
A population of bacteria tends to double every four hours. If there are 5000 bacteria at 9:00 a.m., then how many will there be at noon?
SOLUTION
To answer this question, let $B(t)$ be the number of bacteria at time $t$. For convenience, let $t=0$ correspond to 9:00 a.m. and suppose that time is measured in hours. Thus noon corresponds to $t=3$.
Equation ( $\star$ ) guarantees that
$$B(t)=P \cdot e^{K t}$$
for some undetermined constants $P$ and $K$. We also know that
$$5000=B(0)=P \cdot e^{K \cdot 0}=P$$
We see that $P=5000$ and $B(t)=5000 \cdot e^{K t}$. We still need to solve for $K$.
Since the population tends to double in four hours, there will be 10,000 bacteria at time $t=4$; hence
$$10000=B(4)=5000 \cdot e^{K \cdot 4} .$$
We divide by 5000 to obtain
$$2=e^{K \cdot 4}$$

## 数学代写|微积分代写Calculus代考|A DIFFERENTIAL EQUATION

$$\frac{d B}{d t}=K \cdot B(t),$$

$$\frac{1}{B(t)} \cdot \frac{d B}{d t}=K$$

$$\int \frac{1}{B(t)} \cdot \frac{d B}{d t} d t=\int K d t$$

$$\ln |B(t)|+C$$

$$K t+\tilde{C},$$

$$\ln |B(t)|=K t+D,$$

$$|B(t)|=e^{K t+D}$$

$$B(t)=e^D \cdot e^{K t}=P \cdot e^{K t}$$

## 数学代写|微积分代写Calculus代考|BACTERIAL GROWTH

$$B(t)=P \cdot e^{K t}$$

$$5000=B(0)=P \cdot e^{K \cdot 0}=P$$

$$10000=B(4)=5000 \cdot e^{K \cdot 4} .$$

$$2=e^{K \cdot 4}$$

## MATLAB代写

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