As a glacier melts, the volume V of the ice, measured in cubic kilometers, decreases at a rate modeled by the differential equation dVdt=kV

Question

As a glacier melts, the volume V of the ice, measured in cubic kilometers, decreases at a rate modeled by the differential equation dVdt=kV , where t is measured in years. The volume of the glacier is 400km3 at time t=0 . At the moment when the volume of the glacier is 300km3 , the volume is decreasing at the rate of 15km3 per year. What is the volume V in terms of time

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Orla Orla 5 years 2021-08-30T04:00:07+00:00 1 Answers 904 views 0

Answers ( )

  1. Xavia
    0
    2021-08-30T04:01:12+00:00
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    Solve the differential equation:

    dV/dt = k V   →   1/V dV/dt = k

    →   d/dt [ln(V)] = k

    →   ln(V) = k t + C

    →   V (t )= exp(k t + C ) = C exp(k t ) = C e ᵏᵗ

    At t = 0, the glacier has volume 400 km³ of ice, so

    V (0) = 400   →   C e⁰ = C = 400

    Find when the glacier’s volume is 300 km³:

    V (t ) = 400 e ᵏᵗ = 300   →   e ᵏᵗ = 3/4

    →   k t = ln(3/4)

    →   t = 1/k ln(3/4)

    At this time, the volume is decreasing at a rate of 15 km³/yr, so

    V ‘ (t ) = C k e ᵏᵗ   →   V ‘ (1/k ln(3/4)) = 400 k exp(k × 1/k ln(3/4)) = -15

    →   3/4 k = -3/80

    →   k = -1/20

    Then the volume V (t ) of the glacier at time t is

    V (t ) = 400 exp(-1/20 t )

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