A circular oil slick floats on the surface of still water. Its area is increasing at a rate of 10m^2/min. At what rate is the radius (r metres) increasing?
a = pi r^2
da/dt = 2pi r dr/dt
Now plug in your numbers
To find the rate at which the radius is increasing, we can use the relationship between the area of a circle and its radius.
The formula for the area of a circle is A = πr^2, where A is the area and r is the radius.
To find the rate at which the radius is changing with respect to time, we can differentiate both sides of the equation with respect to time (t):
dA/dt = d(πr^2)/dt
Given that the area is increasing at a rate of 10m^2/min, we can substitute dA/dt with 10:
10 = d/dt (πr^2)
Using the chain rule, we can differentiate πr^2 with respect to r and multiply by dr/dt (the rate at which the radius is changing):
10 = 2πr(dr/dt)
Now we need to solve for dr/dt, the rate at which the radius is increasing. Rearranging the equation, we get:
dr/dt = 10 / (2πr)
So, the rate at which the radius is increasing is 10 / (2πr) meters per minute.
To determine the rate at which the radius is increasing, we need to use the formula for the area of a circle, which is A = πr^2, where A represents the area and r represents the radius.
Given that the area is increasing at a rate of 10 m^2/min, we can differentiate the equation with respect to time (t) to find the relation between the rate of change of the area (dA/dt) and the rate of change of the radius (dr/dt).
dA/dt = 2πr * dr/dt
Here, we are looking for the rate of change of the radius (dr/dt). We know the rate of change of the area (dA/dt) is 10 m^2/min.
Substituting the values into the equation, we have:
10 = 2πr * dr/dt
Now, we can solve for dr/dt:
dr/dt = 10 / (2πr)
Therefore, the rate at which the radius is increasing is 10 / (2πr) m/min.