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 )
The volume in terms of time will be "V(t) = 400 exp (-[tex]\frac{1}{20}[/tex] t)".
According to the question,
Glacier's volume, V(0) = 400 km³
→ [tex]\frac{dV}{dt}[/tex] = kV
[tex]\frac{1}{V}[/tex] [tex]\frac{dV}{dt}[/tex] = k
ln(V) = kt + C
Now,
V (t) = exp (kt + C)
= C exp (kt)
= C[tex]e^{kt}[/tex]
When volume of glacier be "300 km³",
V(t) = 400 [tex]e^{kt}[/tex] = 300
[tex]e^{kt}[/tex] = [tex]\frac{3}{4}[/tex]
By taking log,
kt = ln([tex]\frac{3}{4}[/tex])
t = [tex]\frac{1}{k}[/tex] ln([tex]\frac{3}{4}[/tex])
When volume decrease at 15 km³/yr, then
→ V'(t) = Ck[tex]e^{kt}[/tex]
= 400 k exp (k × [tex]\frac{1}{k}[/tex] kn ([tex]\frac{3}{4}[/tex]))
= -15
Now,
[tex]\frac{3}{4}[/tex] k = - [tex]\frac{3}{80}[/tex]
By applying cross-multiplication,
k = - [tex]\frac{1}{20}[/tex]
Thus the response above is correct.
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