Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 7.5 mi2/hr. How rapidly is radius of the spill increasing when the area is 10 mi^2?
A = πr^2
dA/dt = 2πr dr/dt
when A = 10
πr^2 = 10
r^2 = 10/π
r = √(10/π)
you now have r and dr/dt, so just plug in and push buttons on your calculator
0.66905
YO stupidddd Reiny. did u even go to svchoolllll! LIL PEEEEEEEPP
To find how rapidly the radius of the spill is increasing, we can use 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 of the spill, A, is increasing at a constant rate of 7.5 mi^2/hr. We need to find how rapidly the radius, r, is changing when the area is 10 mi^2.
To do this, we can take the derivative of both sides of the equation with respect to time, t:
dA/dt = d(πr^2)/dt
The derivative of A with respect to t represents the rate of change of the area, dA/dt, and the derivative of πr^2 with respect to t represents the rate of change of the radius, dr/dt. Since the area is increasing at a constant rate, dA/dt can be replaced by the given value of 7.5 mi^2/hr.
So the equation becomes:
7.5 mi^2/hr = d(πr^2)/dt
We can simplify this equation by noticing that π is a constant and can be moved outside the derivative:
7.5 mi^2/hr = π * d(r^2)/dt
Next, we can evaluate the derivative d(r^2)/dt.
To do so, we can use the chain rule. The derivative of r^2 with respect to t can be expressed as:
d(r^2)/dt = 2r * dr/dt
Now, substituting this back into the equation:
7.5 mi^2/hr = π * (2r * dr/dt)
We know that we want to find dr/dt, which represents the rate of change of the radius. So we can rearrange the equation to solve for dr/dt:
dr/dt = (7.5 mi^2/hr) / (2πr)
Now, we can plug in the given value of the area, A = 10 mi^2, into this equation to find the rate at which the radius is changing:
dr/dt = (7.5 mi^2/hr) / (2π * r)
Let's calculate the rate at which the radius is increasing when the area is 10 mi^2. Assuming π ≈ 3.14, we substitute A = 10:
dr/dt = (7.5 mi^2/hr) / (2π * r)
= (7.5 mi^2/hr) / (2 * 3.14 * r)
≈ 1.19 / r
So, the rate at which the radius is increasing when the area is 10 mi^2 is approximately 1.19 / r mi/hr, where r represents the current radius of the spill.