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 use the relationship between the area of a circle and its radius.

The area of a circle is given by the formula A = πr^2, where A represents the area and r represents the radius.

We are given that the area of the spill increases at a constant rate of 6.5 mi^2/hr. This tells us that dA/dt = 6.5, where dA/dt represents the rate of change of the area with respect to time.

To find the rate at which the radius is increasing, we need to find dr/dt, where dr/dt represents the rate of change of the radius with respect to time.

We are asked to find dr/dt when the area is 10 mi^2. This means A = 10.

Differentiating both sides of the area formula with respect to time, we get:
dA/dt = 2πr(dr/dt)

Substituting the given values, we have:
6.5 = 2πr(dr/dt)

To solve for dr/dt, we need to find the value of r when A = 10 mi^2.

From the area formula, we have:
10 = πr^2

Solving for r, we get:
r^2 = 10/π
r = √(10/π) or approximately 1.78 mi

Substituting the value of r into the equation for the rate of change, we have:
6.5 = 2π(1.78)(dr/dt)

Simplifying the equation, we get:
dr/dt = 6.5 / (2π(1.78))
dr/dt ≈ 0.58 mi/hr

Therefore, the radius is increasing at a rate of approximately 0.58 mi/hr when the area is 10 mi^2.

To find out how rapidly the radius of the oil spill is increasing when the area is 10 mi^2, we can use the relationship between the area and the radius of a circle. The formula for the area of a circle is A = πr^2, where A represents the area and r represents the radius.

We know that the area increases at a constant rate of 6.5 mi^2/hr. This means that the derivative of the area with respect to time is constant and equal to 6.5 mi^2/hr.

Let's differentiate the area formula with respect to time to find the relationship between the rate of change of the area and the rate of change of the radius:

dA/dt = 2πr(dr/dt)

Here, dA/dt represents the rate of change of the area, and dr/dt represents the rate of change of the radius.

We are given that the area is 10 mi^2. Substituting A = 10 and dA/dt = 6.5 into the equation, we can solve for dr/dt:

6.5 = 2πr(dr/dt)
dr/dt = 6.5 / (2πr)

Now we can find the rate at which the radius is increasing by substituting the values into the equation.