1. Let C(t) be the number of cougars on an island at time t years (where t > 0). The number of cougars is increasing at a rate directly proportional to 3500 - C(t). Also, C(0) = 1000, and C(5) = 2000.
1. Calculate C(10).
2. Find the limit as t tends to infinity of C(t) , and explain its meaning.
A rate directly proportional to 3500 - C(t) mean:
your answer should contain constant k.
dC / dt = k ∙ ( 3500 - C₍t₎ )
dC = k ∙ ( 3500 - C₍t₎ ) ∙ dt
dC / ( 3500 - C₍t₎ ) = k ∙ dt
∫ dC / ( 3500 - C₍t₎ ) = k ∙ ∫ dt
− ln | 3500 - C₍t₎ | = k⋅t + h
h = constant of integration
Multiply both sides by - 1
ln | 3500 - C₍t₎ | = - k⋅t - h
mark:
- h = m
ln | 3500 - C₍t₎ | = - k⋅t + m
3500 - C₍t₎ = e⁻ᵏᵗ⋅eᵐ
mark:
eᵐ = A
3500 - C₍t₎ = e⁻ᵏᵗ⋅A
- C₍t₎ = A⋅ e⁻ᵏᵗ - 3500
Multiply both sides by - 1
C₍t₎ = 3500 - A⋅ e⁻ᵏᵗ
Now:
t = 0
C(0) = 1000
1000 = 3500 - A⋅ e⁰
1000 = 3500 - A⋅ 1
1000 = 3500 - A
A = 3500 - 1000
A = 2500
t = 5
C(5) = 2000
2000 = 3500 - A⋅ e⁻ ⁵ᵏ
2000 = 3500 - 2500⋅e⁻ ⁵ᵏ
( 2000 - 3500 ) / - 2500 = e⁻ ⁵ᵏ
- 1500 / - 2500 = e⁻ ⁵ᵏ
- 3 ∙ 500 / - 5 ∙ 500 = e⁻ ⁵ᵏ
- 3 / - 5 = e⁻ ⁵ᵏ
3 / 5 = e⁻ ⁵ᵏ
e⁻ ⁵ᵏ = 3 / 5
ln ( e⁻ ⁵ᵏ ) = ln ( 3 / 5 )
- 5 k = ln ( 3 / 5 )
k = ln ( 3 / 5 ) / - 5
k = [ ln ( 3 ) - ln ( 5 ) ] / - 5
k = [ ln ( 5 ) - ln ( 3 ) ] / 5
k = 0.102165125
C₍t₎ = 3500 - 2500⋅e⁻⁰⋅¹⁰²¹⁶⁵¹²⁵ ᵗ
C₍10₎ = 3500 - 2500⋅e⁻¹⋅⁰²¹⁶⁵¹²⁵ = 2604.017 ≈ 2604
lim ( 3500 - 2500⋅e⁻¹⋅⁰²¹⁶⁵¹²⁵ ᵗ) = 3500 -
t->∞
lim (- 2500⋅e⁻¹⋅⁰²¹⁶⁵¹²⁵ ᵗ) =
t->∞
3500 - 2500 e⁻∞ = 3500 - 2500⋅0 = 3500 - 0 = 3500
The number of cougars on island can't be greater of 3500.
Correction:
C₍10₎ = 3500 - 2500⋅e⁻¹⋅⁰²¹⁶⁵¹²⁵ = 2600
thanks!!
1. Well, let's put on our math hats and solve this! We know that the rate of change of cougars, C'(t), is directly proportional to 3500 - C(t). So we can write this equation as C'(t) = k(3500 - C(t)), where k is the constant of proportionality.
To find C(10), we need to solve the differential equation. We can rearrange the equation as follows:
C'(t)/(3500 - C(t)) = k.
Now, we'll integrate both sides with respect to t:
∫C'(t)/(3500 - C(t)) dt = ∫k dt.
Integrating both sides gives us:
ln|3500 - C(t)| = kt + C.
We know that C(0) = 1000, so let's substitute that in:
ln|3500 - 1000| = k(0) + C.
ln|2500| = C.
Now, we can find the value of k using the information given for C(5):
ln|3500 - 2000| = k(5) + ln|2500|.
ln|1500| - ln|2500| = 5k.
Using logarithmic properties, we simplify:
ln(1500/2500) = 5k.
ln(3/5) = 5k.
Taking the inverse natural logarithm of both sides, we get:
3/5 = e^(5k).
Now we can find k:
k = ln(3/5)/5.
Using this k value, we can substitute it back into the equation to solve for C(10):
ln|3500 - C(10)| = (ln(3/5)/5)(10) + ln|2500|.
ln|3500 - C(10)| = 2ln(3/5) + ln|2500|.
Now we can exponentiate both sides:
3500 - C(10) = e^(2ln(3/5)) × 2500.
3500 - C(10) = (3/5)^2 × 2500.
3500 - C(10) = (9/25) × 2500.
Now, let's solve for C(10):
C(10) = 3500 - (9/25) × 2500.
I'll leave it to you to do the calculation and find the exact value of C(10).
2. Now, let's tackle the limit as t tends to infinity of C(t). As t gets bigger and bigger, what do you think happens to the number of cougars on the island?
Well, since the rate of change of cougars is directly proportional to 3500 - C(t), as C(t) approaches 3500, the rate of change gets smaller and smaller. This means that the number of cougars will approach a certain value, which in this case is 3500.
So, the limit as t tends to infinity of C(t) is 3500. In simpler terms, the number of cougars on the island will stabilize at 3500 as time goes on. It's like a cougar party that reaches maximum capacity!
To solve these questions, we will use the differential equation modeling the rate of change of cougars on the island:
dC/dt = k(3500 - C)
Where dC/dt represents the rate of change of cougars with respect to time, k is the constant of proportionality, and C represents the number of cougars on the island.
First, let's solve the differential equation to find the value of k. We have the initial condition C(0) = 1000.
dC/dt = k(3500 - C)
Integrating both sides:
∫(1/(3500 - C)) dC = ∫k dt
-ln|3500 - C| = kt + C1 (Where C1 is the constant of integration)
Applying the initial condition C(0) = 1000, we get:
-ln|3500 - 1000| = k(0) + C1
-ln(2500) = C1
Therefore, the particular solution is:
-ln|3500 - C| = kt - ln(2500)
Now, let's use the information given to solve the first question.
1. Calculate C(10):
To find out the value of C(10), we substitute t = 10 into our particular solution equation:
-ln|3500 - C| = k(10) - ln(2500)
We need to find the value of k to substitute it into this equation. Using the second piece of information given, C(5) = 2000, we can set up another equation:
-ln|3500 - 2000| = k(5) - ln(2500)
-k + ln(1500) = k(5) - ln(2500)
Simplifying this equation will give us the value of k. Once we have k, we can substitute t = 10 into our particular solution equation to find C(10).
2. Find the limit as t tends to infinity of C(t), and explain its meaning:
To find the limit of C(t) as t tends to infinity, we can analyze the behavior of the function as t becomes very large.
From the differential equation, as t increases, the term (3500 - C) becomes smaller, which means the rate of change of cougars tends to decrease. As time goes on, the increase in the number of cougars slows down and eventually reaches a point where the rate of change becomes zero. This point is known as the carrying capacity, which represents the maximum number of cougars the island can sustain in the long run.
Since the number of cougars cannot exceed the carrying capacity, the limit as t approaches infinity is equal to the carrying capacity. Therefore, finding the limit of C(t) as t approaches infinity will give us the carrying capacity of the island.