Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 6.5 {\rm mi}^2{\rm /hr}. How rapidly is radius of the spill increasing when the area is 10 {\rm mi}^2?
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same question, just change some of the numbers.
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To find the rate at which the radius of the oil spill is increasing, we need to use some calculus and related rates.
Let's denote the radius of the oil spill as r (in miles) and the area of the oil spill as A (in square miles).
We are given that the area of the spill is increasing at a constant rate of 6.5 mi^2/hr. This means that dA/dt = 6.5 mi^2/hr.
We are asked to find how rapidly the radius is increasing when the area is 10 mi^2. So we need to find dr/dt when A = 10 mi^2.
The relationship between the radius and the area of a circle is given by the formula: A = πr^2.
To solve for the rate at which the radius is changing, we'll need to differentiate both sides of the equation with respect to time (t).
dA/dt = d(πr^2)/dt
6.5 mi^2/hr = 2πr (dr/dt)
Now we can plug in the given area A = 10 mi^2 and solve for dr/dt.
6.5 mi^2/hr = 2πr (dr/dt)
6.5 mi^2/hr = 2π(√(A/π)) (dr/dt) (Substitute r = √(A/π))
6.5 mi^2/hr = 2√(10/π)π (dr/dt) (Substitute A = 10 mi^2)
Simplifying the equation further:
6.5 mi^2/hr = 2√(10π) (dr/dt)
dr/dt = 6.5 mi^2/hr / (2√(10π))
Finally, we can calculate the value of dr/dt:
dr/dt ≈ 0.323 mi/hr
So, when the area of the oil spill is 10 mi^2, the radius is increasing at a rate of approximately 0.323 miles per hour.