When a reservoir is created by a dam,50 fish are introduced into the reservoir ,which has an estimated carrying capacity of 8000 fish.
A logistic model of the population is
P(t)= 400,000 / 50 + 795e^-0.5t , where t is measured in years.
How long does it take for the population to reach 5000 fish and 90% of the carrying capacity?
Should I use the derivatives to calculate it?
no use for derivatives here.
90% of capacity is 7200 fish
So, just solve for t when P(t) = 5000 or 7200
Oo..I see..but what to do with e?
Sorry,totally forgot about that..
Thanks^^
Well, I could give you a serious answer using derivatives, but where's the fun in that? Let's take a different approach!
To find out how long it takes for the population to reach 5000 fish, we can simply substitute that value into the population equation and solve for t. It's like a math mystery, where the solution is the culprit!
Plugging in P(t) = 5000 into the equation, we have:
5000 = 400,000 / (50 + 795e^(-0.5t))
Now, don't be scared. We just need to get rid of that denominator to solve for t. You see, numbers can be like irrational exes, always causing division! So, we'll multiply both sides of the equation by (50 + 795e^(-0.5t)):
(50 + 795e^(-0.5t)) * 5000 = 400,000
Now comes the tricky part. We need to isolate t, our mathematical slippery suspect. So, we'll divide both sides of the equation by 5000:
50 + 795e^(-0.5t) = 80
Subtracting 50 from both sides:
795e^(-0.5t) = 30
And now, we're ready to employ the powerful weapon of mathematics: the natural logarithm (ln). We'll apply the natural logarithm to both sides:
ln(795e^(-0.5t)) = ln(30)
Next, we'll use the logarithmic property to bring that pesky exponent down:
ln(795) + ln(e^(-0.5t)) = ln(30)
But wait, there's more! Thanks to the power of logarithms, e and ln are like two peas in a pod. The natural logarithm of e cancels out, leaving us with:
ln(795) - 0.5t = ln(30)
Now, to find out what t is up to, we'll subtract ln(795) from both sides:
-0.5t = ln(30) - ln(795)
Last steps, I promise! We just divide both sides by -0.5:
t = (ln(30) - ln(795)) / -0.5
Now, grab your calculator and get cracking! And remember, solving equations is like solving a mystery. Just follow the clues and try not to get lost in the logarithmic labyrinth!
Yes, you can use derivatives to calculate the time it takes for the population to reach certain levels. In this case, the population function is given by:
P(t) = 400,000 / (50 + 795e^(-0.5t))
To find the time it takes for the population to reach 5000 fish, you need to solve the equation:
5000 = 400,000 / (50 + 795e^(-0.5t))
To find the time it takes for the population to reach 90% of the carrying capacity, you need to solve the equation:
0.9 * 8000 = 400,000 / (50 + 795e^(-0.5t))
To solve these equations, you can rearrange them to isolate the exponential term and then use logarithms or other numerical methods to solve for t. However, using derivatives can simplify the process.
Taking the derivative of the population function with respect to t, you have:
dP(t) / dt = 400,000 * (0.5)(795e^(-0.5t)) / (50 + 795e^(-0.5t))^2
To find the time it takes for the population to reach a certain level, you can set dP(t)/dt equal to 0 and solve for t. This is because the population growth rate is zero when the population reaches a certain level.
For example, to find the time it takes for the population to reach 5000 fish, you can set dP(t)/dt = 0 and solve for t:
0 = 400,000 * (0.5)(795e^(-0.5t)) / (50 + 795e^(-0.5t))^2
By solving this equation, you would find the time it takes for the population to reach 5000 fish.
Similarly, you can set dP(t)/dt equal to 0.9 * 8000 to find the time it takes for the population to reach 90% of the carrying capacity.
So, yes, derivatives are useful for calculating the time it takes for the population to reach certain levels.
come on, guy. You must have solved exponential problems before, no?
5000 = 400,000 / (50 + 795e^-0.5t)
50 + 795e^-.5t = 80
795e^-.5t = 30
e^-.5t = 30/795
-.5t = log(30/795)
t = -2log(30/795) = 6.55