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.

Question

1. Calculate C(10).
2. Find the limit as t tends to infinity of C(t) , and explain its meaning.

Answer 1

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.

Answer 2

Correction:

C₍10₎ = 3500 - 2500⋅e⁻¹⋅⁰²¹⁶⁵¹²⁵ = 2600

Answer 3

thanks!!