calculus

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.

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  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.

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  2. Correction:

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

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  3. thanks!!

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