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?
The radius is increasing at
To solve this problem, we can relate the radius of the spill to its area using the formula for the area of a circle:
A = πr^2
where A is the area and r is the radius. We are given that the area increases at a constant rate of 6.5 mi^2/hr, so we can write this as:
dA/dt = 6.5
where dA/dt represents the rate of change of the area with respect to time. We want to find the rate at which the radius is increasing, or dr/dt, when the area is 10 mi^2.
To find dr/dt, we can differentiate both sides of the area formula with respect to time:
dA/dt = d/dt(πr^2)
Using the chain rule, we get:
dA/dt = 2πr(dr/dt)
Now we can substitute the value of dA/dt and the given area A = 10 mi^2 into the equation:
6.5 = 2πr(dr/dt)
Rearranging the equation, we get:
dr/dt = 6.5 / (2πr)
Substituting the radius r = √(A/π) = √(10/π) mi into the equation, we can calculate the rate at which the radius is increasing:
dr/dt = 6.5 / (2π √(10/π))
Simplifying further:
dr/dt = 6.5 / (2√(10π))
Therefore, the radius is increasing at a rate of dr/dt = 6.5 / (2√(10π)) mi/hr when the area is 10 mi^2.