33) Suppose an experimental population of fruit flies increases according to the law of exponential growth. There were 100 flies after the second day of the experiment and 300 flies after the fourth day. Approximately how many flies were in the original population? 34) The rate of change of $$ y $$ is proportional to $$ y $$. When $$ t=0, y=2 $$. When $$ t=2, y=4 $$. What is the value of $$ y $$ when $$ t=3 $$.

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**33)**  
The population grows according to an exponential model  
$$
P(t) = P(0)e^{kt},
$$  
where $$P(0)$$ is the original population, and $$k$$ is the growth constant.

Given:  
$$
P(2)=100 \quad \text{and} \quad P(4)=300.
$$

Writing the equations:  
$$
P(2)=P(0)e^{2k}=100, \quad P(4)=P(0)e^{4k}=300.
$$

Divide the second equation by the first:  
$$
\frac{P(0)e^{4k}}{P(0)e^{2k}} = \frac{300}{100} \quad \Rightarrow \quad e^{2k}=3.
$$

Taking the natural logarithm of both sides:  
$$
2k=\ln 3 \quad \Rightarrow \quad k=\frac{\ln 3}{2}.
$$

Now substitute back into the equation for $$P(2)$$ to find $$P(0)$$:  
$$
P(0)e^{2k}=100 \quad \Rightarrow \quad P(0)e^{\ln 3}=100 \quad \Rightarrow \quad P(0)\cdot 3=100.
$$

Thus,  
$$
P(0)=\frac{100}{3}\approx33.33.
$$

**34)**  
The rate of change of $$y$$ is proportional to $$y$$, which gives the differential equation  
$$
\frac{dy}{dt}=ky,
$$  
with solution  
$$
y(t)=y(0)e^{kt}.
$$

Given:  
$$
y(0)=2 \quad \text{and} \quad y(2)=4.
$$

Substitute $$t=2$$:  
$$
2e^{2k}=4 \quad \Rightarrow \quad e^{2k}=2.
$$

Taking natural logs:  
$$
2k=\ln 2 \quad \Rightarrow \quad k=\frac{\ln 2}{2}.
$$

Now, find $$y(3)$$:  
$$
y(3)=2e^{3k}=2e^{\frac{3\ln2}{2}}=2\cdot2^{\frac{3}{2}}.
$$

Simplify $$2^{\frac{3}{2}}$$:  
$$
2^{\frac{3}{2}}=2\sqrt{2},
$$
so  
$$
y(3)=2\cdot 2\sqrt{2}=4\sqrt{2}.
$$

Thus, when $$t=3$$,  
$$
y=4\sqrt{2}.
$$
**33)** The original population was approximately 33.33 flies. **34)** When $$ t=3 $$, $$ y=4\sqrt{2} $$.

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