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

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 )

change tension

v=ir = 3x15 =45 v