Math / Calculus

QuestionBiology Because of an insufficient oxygen supply, the trout population in a lake is dying. The population's rate of change can be modeled by $\frac{d P}{d t}=-130 e^{-t / 30}$ where $t$ is the time in days. When $t=0$ , the population is 3,900 . (a) Find a model for the population. $P(t)=$ $\square$ (b) What is the population of fish after 14 days? (Round your answer to the nearest integer.) $\square$ fish (c) How long (in days) will it take for the entire trout population to die? (Assume the entire population has died off when the population is less than one. Round your answer to one decimal place.) $\square$ days Need Help? Readit Watchit

Studdy Solution

STEP 1

What is this asking? We're figuring out how a trout population changes over time, given how quickly it's decreasing, and we want to know the population at a specific time and when it'll practically disappear. Watch out! Don't forget that the given rate is negative, meaning the population is decreasing.
Also, make sure to round your answers correctly as requested!

STEP 2

1. Find the population model.
2. Calculate the population after 14 days.
3. Determine when the population dies off.

STEP 3

We're given the rate of change of the population, $\frac{dP}{dt} = -130e^{-t/30}$ .
To find the population $P(t)$ itself, we need to integrate the rate with respect to time $t$ .
This is because the integral of a rate of change gives us the original function, in this case, the population.
So, we have: $P(t) = \int -130e^{-t/30} dt$

STEP 4

Let's solve this integral: $P(t) = -130 \int e^{-t/30} dt$ We can use a simple u-substitution.
Let $u = -t/30$ .
Then $du = (-1/30)dt$ , so $dt = -30du$ .
Substituting, we get: $P(t) = -130 \int e^u (-30) du = 3900 \int e^u du = 3900e^u + C$ Substituting back for $t$ , we have: $P(t) = 3900e^{-t/30} + C$

STEP 5

We know that when $t = 0$ , the population is $P(0) = \textbf{3900}$ .
Let's use this initial condition to find the constant $C$ : $3900 = 3900e^{-0/30} + C$ $3900 = 3900(1) + C$ $C = 0$ So, our population model is: $P(t) = 3900e^{-t/30}$

STEP 6

We have our population model $P(t) = 3900e^{-t/30}$ .
We want to find the population after $\textbf{14}$ days, so we plug in $t = 14$ : $P(14) = 3900e^{-14/30}$

STEP 7

Now, let's calculate the result: $P(14) \approx 3900e^{-0.4667} \approx 3900(0.627) \approx 2445.3$ Rounding to the nearest integer, we get $\textbf{2445}$ fish.

STEP 8

We want to find when the population is less than one.
So, we set up the equation: $P(t) = 3900e^{-t/30} < 1$

STEP 9

Let's solve for $t$ : $e^{-t/30} < \frac{1}{3900}$ Taking the natural logarithm of both sides: $-\frac{t}{30} < \ln\left(\frac{1}{3900}\right)$ $-\frac{t}{30} < -\ln(3900)$ Multiplying both sides by $-30$ and flipping the inequality sign: $t > 30\ln(3900)$ $t > 30 \cdot 8.268 \approx 248.04$

STEP 10

Rounding to one decimal place, we get $t > \textbf{248.0}$ days.

STEP 11

(a) $P(t) = 3900e^{-t/30}$ (b) After 14 days, the population is approximately 2445 fish. (c) It will take approximately 248.0 days for the entire trout population to die.

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Studdy Solution

STEP 1

STEP 2

1. Find the population model.
2. Calculate the population after 14 days.
3. Determine when the population dies off.

STEP 3

STEP 4

STEP 5

We know that when $t = 0$ , the population is $P(0) = \textbf{3900}$ .
Let's use this initial condition to find the constant $C$ : $3900 = 3900e^{-0/30} + C$ $3900 = 3900(1) + C$ $C = 0$ So, our population model is: $P(t) = 3900e^{-t/30}$

STEP 6

We have our population model $P(t) = 3900e^{-t/30}$ .
We want to find the population after $\textbf{14}$ days, so we plug in $t = 14$ : $P(14) = 3900e^{-14/30}$

STEP 7

Now, let's calculate the result: $P(14) \approx 3900e^{-0.4667} \approx 3900(0.627) \approx 2445.3$ Rounding to the nearest integer, we get $\textbf{2445}$ fish.

STEP 8

We want to find when the population is less than one.
So, we set up the equation: $P(t) = 3900e^{-t/30} < 1$

STEP 9

Let's solve for $t$ : $e^{-t/30} < \frac{1}{3900}$ Taking the natural logarithm of both sides: $-\frac{t}{30} < \ln\left(\frac{1}{3900}\right)$ $-\frac{t}{30} < -\ln(3900)$ Multiplying both sides by $-30$ and flipping the inequality sign: $t > 30\ln(3900)$ $t > 30 \cdot 8.268 \approx 248.04$

STEP 10

Rounding to one decimal place, we get $t > \textbf{248.0}$ days.

STEP 11

(a) $P(t) = 3900e^{-t/30}$ (b) After 14 days, the population is approximately 2445 fish. (c) It will take approximately 248.0 days for the entire trout population to die.

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Studdy Solution

STEP 1

STEP 2

1. Find the population model.2. Calculate the population after 14 days.3. Determine when the population dies off.

STEP 3

STEP 4

STEP 5

STEP 6

We have our population model P(t)=3900e−t/30P(t) = 3900e^{-t/30}P(t)=3900e−t/30.We want to find the population after 14\textbf{14}14 days, so we **plug in** t=14t = 14t=14: P(14)=3900e−14/30P(14) = 3900e^{-14/30}P(14)=3900e−14/30

STEP 7

Now, let's **calculate** the result: P(14)≈3900e−0.4667≈3900(0.627)≈2445.3P(14) \approx 3900e^{-0.4667} \approx 3900(0.627) \approx 2445.3P(14)≈3900e−0.4667≈3900(0.627)≈2445.3 Rounding to the nearest integer, we get 2445\textbf{2445}2445 fish.

STEP 8

We want to find when the population is less than one.So, we set up the equation: P(t)=3900e−t/30<1P(t) = 3900e^{-t/30} < 1P(t)=3900e−t/30<1

STEP 9

STEP 10

Rounding to one decimal place, we get t>248.0t > \textbf{248.0}t>248.0 days.

STEP 11

(a) P(t)=3900e−t/30P(t) = 3900e^{-t/30}P(t)=3900e−t/30 (b) After 14 days, the population is approximately 2445 fish. (c) It will take approximately 248.0 days for the entire trout population to die.

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1. Find the population model.
2. Calculate the population after 14 days.
3. Determine when the population dies off.

We have our population model $P(t) = 3900e^{-t/30}$ .
We want to find the population after $\textbf{14}$ days, so we plug in $t = 14$ : $P(14) = 3900e^{-14/30}$

Now, let's calculate the result: $P(14) \approx 3900e^{-0.4667} \approx 3900(0.627) \approx 2445.3$ Rounding to the nearest integer, we get $\textbf{2445}$ fish.

We want to find when the population is less than one.
So, we set up the equation: $P(t) = 3900e^{-t/30} < 1$

Rounding to one decimal place, we get $t > \textbf{248.0}$ days.

(a) $P(t) = 3900e^{-t/30}$ (b) After 14 days, the population is approximately 2445 fish. (c) It will take approximately 248.0 days for the entire trout population to die.