The radius of a circular oil slick expands at a rate of 8 m/min.
a. How fast is the area of the oil slick increasing when the radius is 30 m? = 480pi
b.If the radius is 0 at time t=0 , how fast is the area increasing after 2 mins?
i cannot get part b!
At a rate of 8 m/min, after 2 minutes the radius will be 16 m.
Now sub r=16 into your dA/dt you had in a)
To solve part b, we need to find the rate at which the area of the oil slick is increasing after 2 minutes.
We are given that the radius of the oil slick expands at a rate of 8 m/min. This means that the rate at which the radius is increasing is constant.
We can use the formula for the area of a circle:
A = πr^2
where A is the area and r is the radius.
To find the rate at which the area is increasing, we need to differentiate both sides of the equation with respect to time, t, since we are looking for the rate of change with respect to time.
dA/dt = d/dt(πr^2)
First, let's find the derivative of r^2 with respect to t. Since the rate of change of the radius is constant, we can write it as dr/dt.
d/dr(r^2) = 2r * (dr/dt)
Now we can substitute the given values into the equation:
dA/dt = d/dr(r^2) * (dr/dt)
= 2r * (dr/dt)
At t = 2 minutes, the radius, r, would be equal to 0 + 8 * 2 (since the radius starts from 0 and is increasing at a rate of 8 m/min).
So, r = 16.
Substituting this value into the equation, we have:
dA/dt = 2 * (16) * (dr/dt)
= 32 * 8
= 256 m^2/min
Therefore, the area of the oil slick is increasing at a rate of 256 m^2/min after 2 minutes.
To find the rate at which the area of the oil slick is increasing, we can use the formula for the area of a circle, which is given by A = πr^2.
a. To find how fast the area of the oil slick is increasing when the radius is 30 m, we can differentiate the equation for the area with respect to time. This gives us:
dA/dt = 2πr(dr/dt)
where dA/dt is the rate at which the area is changing with respect to time and dr/dt is the rate at which the radius is changing with respect to time.
Given that dr/dt = 8 m/min when the radius is 30 m, we can substitute these values into the equation:
dA/dt = 2π(30)(8)
= 480π
So, when the radius is 30 m, the area of the oil slick is increasing at a rate of 480π square meters per minute.
b. In this case, we are given that the radius is 0 at time t=0. We want to find how fast the area of the oil slick is increasing after 2 minutes, so we need to calculate the radius at that time.
Since the radius is expanding at a rate of 8 m/min, after 2 minutes it will have expanded by:
Δr = (rate of expansion) x (time)
= 8 m/min x 2 min
= 16 m
The radius after 2 minutes would be the initial radius (0 m) plus the expansion:
r = 0 m + 16 m
= 16 m
Now that we have the radius, we can use the same formula as in part a to find how fast the area is increasing:
dA/dt = 2πr(dr/dt)
= 2π(16)(8)
= 256π
So, after 2 minutes, the area of the oil slick is increasing at a rate of 256π square meters per minute.