If the fish population, p, in a lake can be modelled by the function p(t)=15(t^2+30)(t+8), where t is the time, in years , from now.
How do I determine the rate of change when the population when there are 5000 fish in the lake?
That's what I have done, I have expanded it to get the derivative, but I am having problems solving for t.
To determine the rate of change of the fish population when there are 5000 fish in the lake, we need to find the derivative of the function p(t) with respect to t and evaluate it when p(t) = 5000.
Let's start by finding the derivative of p(t). The function p(t) can be rewritten as p(t) = 15t^3 + 480t^2 + 240t.
To find the derivative, differentiate each term in the function with respect to t:
d/dt (15t^3) = 45t^2
d/dt (480t^2) = 960t
d/dt (240t) = 240
The derivative of p(t), denoted by p'(t), is given by:
p'(t) = 45t^2 + 960t + 240
Next, we substitute p(t) = 5000 into p'(t) and solve for t:
5000 = 45t^2 + 960t + 240
Now, we have a quadratic equation that can be solved in order to find the values of t when the population is 5000.
However, the equation cannot be factored easily, so we will solve it using the quadratic formula:
t = (-b +/- sqrt(b^2 - 4ac)) / 2a
For our equation, a = 45, b = 960, and c = -4760.
Substituting these values, we get:
t = (-960 +/- sqrt(960^2 - 4 * 45 * -4760)) / (2 * 45)
Now, we can calculate the values of t:
t = (-960 +/- sqrt(921600 - (-855600))) / 90
t = (-960 +/- sqrt(1777200)) / 90
t = (-960 +/- 1332.437) / 90
There are two possible values for t:
1. t = (372.437) / 90 ≈ 4.16
2. t = (-2292.437) / 90 ≈ -25.47
Since time cannot be negative in this context, we can ignore the second value of t.
Therefore, when there are 5000 fish in the lake, the rate of change of the population is given by p'(t) at t = 4.16.
plugging in t = 4.16 in p'(t) we get:
p'(4.16) = 45(4.16^2) + 960(4.16) + 240
≈ 1224.94
Therefore, the rate of change of the population when there are 5000 fish in the lake is approximately 1224.94 fish per year.
To determine the rate of change of the fish population at a specific point, you need to find the derivative of the population function with respect to time, t, and then substitute the value of time when the population is 5000.
Step 1: Find the derivative of the population function
Differentiate the function p(t) with respect to t using the power rule and the chain rule:
p'(t) = d/dt [15(t^2+30)(t+8)]
= 15 [(2t)(t+8) + (t^2+30)(1)]
= 15 [2t(t+8) + t^2+30]
= 15 [2t^2 + 16t + t^2 + 30]
= 15 [3t^2 + 16t + 30]
= 45t^2 + 240t + 450
Step 2: Substitute the value of time when the population is 5000
Replace t with the appropriate value in the derivative equation:
p'(t) = 45t^2 + 240t + 450
p'(t) = 45(t^2 + 5.33t + 10)
Now, we need to find the value of t when the population is 5000.
15(t^2+30)(t+8) = 5000
(t^2+30)(t+8) = 5000/15
(t^2+30)(t+8) = 333.33
You can solve this equation algebraically or use numerical methods like graphing the function and finding the intersection point where the population is approximately 5000.
Once you have the value of t, you substitute it back into the derivative equation:
p'(t) = 45(t^2 + 5.33t + 10) = 45(5000^2 + 5.33(5000) + 10)
Evaluating p'(t) will give you the rate of change of the fish population when there are 5000 fish in the lake.
You take the derivative of p
expand p
p=15(t^3+240+30t+8t^2)
p'=15(3t^2+30+16t)
in the first equation,p=5000, so you have to solve for t. A little algebra is required. Then, for the real t, put that in the p' equation