Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 9 mi2/hr. How rapidly is radius of the spill increasing when the area is 4 mi2?
To find the rate at which the radius of the oil spill is increasing, we can use the formula for the area of a circle:
A = πr²
Where A is the area of the spill and r is the radius. We are told that the area is increasing at a constant rate of 9 mi²/hr.
First, let's differentiate both sides of the equation with respect to time (t):
dA/dt = 2πr(dr/dt)
Now, we need to solve for dr/dt, which represents the rate at which the radius is changing.
Since we know that dA/dt is 9 mi²/hr and we want to find the value of dr/dt when the area is 4 mi², we can substitute these values into the equation:
9 = 2πr(dr/dt)
We can simplify this by dividing both sides by 2πr:
9/(2πr) = dr/dt
Now, we substitute the given value of the area (A = 4 mi²) into the equation:
dr/dt = 9/(2πr)
Since we want to find dr/dt when the area is 4 mi², we substitute r = √(A/π) = √(4/π) into the equation:
dr/dt = 9/(2π * √(4/π))
dr/dt = 9/(2π * √(4/π))
dr/dt = 9/(2 * √(π/4))
dr/dt = 9/(2 * √(π/4))
dr/dt = 9/(2 * (1/2))
dr/dt = 9/1
dr/dt = 9 mi/hr
Therefore, when the area of the spill is 4 mi², the radius is increasing at a rate of 9 mi/hr.
a = πr^2
when a=4, r=2/√π
da/dt = 2πr dr/dt
9 = 2π(2/√π) dr/dt
dr/dt = 9/(4√π) mi/hr