Question

15.) suppose a population of snails is introduced to an aquarium. the population of snails grows according to the function $p(t) = \frac{352t + 55}{t + 5}$, where $t$ is time in months since the introduction of the snails.

a.) what was the initial population?

b.) what is the carrying capacity of the aquarium? (the carrying capacity is the maximum population supported indefinitely by the environment.)

c.) when will the population reach half the carrying capacity?

16.) the amount of miligrams of a certain medicine in a patients bloodstream $t$ hours after administration is given by $m(t) = 300e^{-2t}$.

a.) how many miligrams of the medicine were initially administered?

b.) how many miligrams will be in the patients bloodstream after 2 hours? (round to the nearest tenth.)

c.) find $\lim\limits_{t \to \infty} m(t)$. interpret the meaning of this limit.

Explanation:

For Problem 15:

Step1: Initial population at $t=0$

Substitute $t=0$ into $p(t)$:
$p(0) = \frac{352(0) + 55}{0 + 5} = \frac{55}{5}$

Step2: Calculate initial population

$p(0) = 11$

Step3: Find carrying capacity (limit as $t\to\infty$)

Divide numerator/denominator by $t$:
$\lim_{t\to\infty} p(t) = \lim_{t\to\infty} \frac{352 + \frac{55}{t}}{1 + \frac{5}{t}}$

Step4: Evaluate the limit

As $t\to\infty$, $\frac{55}{t}\to0$, $\frac{5}{t}\to0$, so $\lim_{t\to\infty} p(t) = 352$

Step5: Set $p(t)$ to half carrying capacity

Half of 352 is 176. Solve $\frac{352t + 55}{t + 5} = 176$

Step6: Cross-multiply and solve for $t$

$352t + 55 = 176(t + 5)$
$352t + 55 = 176t + 880$
$352t - 176t = 880 - 55$
$176t = 825$
$t = \frac{825}{176} = 4.6875$

Step1: Initial medicine at $t=0$

Substitute $t=0$ into $m(t)$:
$m(0) = 300e^{-2(0)} = 300e^0$

Step2: Calculate initial dosage

$e^0=1$, so $m(0)=300$

Step3: Medicine at $t=2$ hours

Substitute $t=2$ into $m(t)$:
$m(2) = 300e^{-2(2)} = 300e^{-4}$

Step4: Compute and round the value

$e^{-4}\approx0.0183156$, so $m(2)\approx300\times0.0183156\approx5.5$

Step5: Find $\lim_{t\to\infty} m(t)$

Evaluate $\lim_{t\to\infty} 300e^{-2t} = 300\lim_{t\to\infty} \frac{1}{e^{2t}}$

Step6: Evaluate the limit

As $t\to\infty$, $e^{2t}\to\infty$, so $\frac{1}{e^{2t}}\to0$, thus $\lim_{t\to\infty} m(t)=0$

Answer:

a.) 11
b.) 352
c.) 4.6875 months

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For Problem 16: