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Bad White [126]
3 years ago
9

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

ion 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
Physics
1 answer:
vovangra [49]3 years ago
7 0

Solve the differential equation:

d<em>V</em>/d<em>t</em> = <em>k V</em>   →   1/<em>V</em> d<em>V</em>/d<em>t</em> = <em>k</em>

→   d/d<em>t</em> [ln(<em>V</em>)] = <em>k</em>

→   ln(<em>V</em>) = <em>k t</em> + <em>C</em>

→   <em>V</em> (<em>t</em> )= exp(<em>k t</em> + <em>C </em>) = <em>C</em> exp(<em>k t</em> ) = <em>C e </em>ᵏᵗ

At <em>t</em> = 0, the glacier has volume 400 km³ of ice, so

<em>V</em> (0) = 400   →   <em>C</em> <em>e</em>⁰ = <em>C</em> = 400

Find when the glacier's volume is 300 km³:

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

→   <em>k t</em> = ln(3/4)

→   <em>t</em> = 1/<em>k</em> ln(3/4)

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

<em>V '</em> (<em>t </em>) = <em>C</em> <em>k</em> <em>e </em>ᵏᵗ   →   <em>V '</em> (1/<em>k</em> ln(3/4)) = 400 <em>k</em> exp(<em>k</em> × 1/<em>k</em> ln(3/4)) = -15

→   3/4 <em>k</em> = -3/80

→   <em>k</em> = -1/20

Then the volume <em>V</em> (<em>t</em> ) of the glacier at time <em>t</em> is

<em>V</em> (<em>t</em> ) = 400 exp(-1/20 <em>t </em>)

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3 years ago
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