Acircular oil spill grows at a rate given by the differential equation dr/dt = k/r, where r represents the radius of the spill in feet, and time is measured in hours. if the radius of the spill is 400 feet 16 hours after the spill begins, what is the value of k? include units in your answer.

k = 5000 ft²/h

Step-by-step explanation:

r₀ = 0 ft

r₁ = 400 ft

t₀ = 0 h

t₁ = 16 h

k = ?

Knowing that

dr/dt = k/r    ⇒   k*dt = r*dr   ⇒  ∫ k*dt = ∫ r*dr

⇒   k ∫ dt = ∫ r*dr     ⇒   k*t + C₁ = (1/2)*r² + C₂

since  0 h  ≤ t ≤ 16 h   and    0 ft ≤ r ≤ 400 ft

we have

k*(16 h - 0 h) = (1/2)*((400 ft)² - (0 ft)²)

⇒  16 h*k = 80000 ft²

⇒   k = 5000 ft²/h

assuming you mean addition: addition and subtraction are reverse operations.

step-by-step explanation:

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step-by-step explanation:

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