The area of a floating circular oil spill in the ocean is increasing at a rate of 96pi square meters per second. When the area of the spill is 1000pi how fast is the radius increasing in meters per second.
Area = πr^2
d(area)/dt = 2πr dr/dt
when area = 1000π
πr^2 = 1000π
r^2=1000
r = √1000 = 10√10
96π = 2π(10√10)dr/dt
dr/dt = 96/(20√10) or 24/(5√10) or 12√10/25
= appr. 1.52
Well, isn't that a slippery situation! Let's tackle this oil spill mathematically!
We can relate the area of a circular spill to its radius using the formula A = πr², where A is the area and r is the radius. If we differentiate both sides with respect to time, we get dA/dt = 2πr(dr/dt) using the chain rule.
Given that dA/dt = 96π square meters per second and A = 1000π square meters, we can substitute these values into our equation to solve for dr/dt:
96π = 2πr(dr/dt)
48 = r(dr/dt)
Now, if we plug in the value of A, we get 1000π = πr². By canceling out the πs on both sides, we're left with 1000 = r².
Substituting this value into our previous equation, we get:
48 = r(dr/dt)
48 = √1000(dr/dt)
48 = 10√10(dr/dt)
Solving for dr/dt, we find that:
dr/dt = 48 / (10√10)
dr/dt ≈ 4.8 / √10
So, the radius is increasing at approximately 4.8 / √10 meters per second. Just be careful not to slip on any oil, or your calculations might get oily too!
To find the rate at which the radius is increasing, we can use the derivative of the area with respect to time.
Given:
Area of the oil spill, A = 1000π m^2
Rate of change of the area, dA/dt = 96π m^2/s
We know that the formula for the area of a circle is A = πr^2, where r is the radius.
Taking the derivative of both sides with respect to time (t):
dA/dt = 2πr(dr/dt)
Now we can solve for dr/dt, the rate at which the radius is increasing.
96π = 2πr(dr/dt)
76 = r(dr/dt)
dr/dt = 76/r
Substituting the given area, the radius can be calculated:
A = πr^2
1000π = πr^2
1000 = r^2
r = √1000 = 10√10
Now we can substitute the radius value into the equation for dr/dt:
dr/dt = 76/(10√10)
dr/dt = 7.6/√10 meters per second
Therefore, the radius is increasing at a rate of approximately 7.6/√10 meters per second.
To find the rate at which the radius is increasing, we can use the formula for the area of a circle: A = πr^2, where A represents the area and r represents the radius.
Given that the area of the oil spill is increasing at a rate of 96π square meters per second, we can set up the following equation:
dA/dt = 96π
where dA/dt represents the rate of change of the area with respect to time.
We are also given that the area of the oil spill is 1000π, so we can substitute this value into the equation:
1000π = 96π
Now, let's solve for dt/dt, the rate at which the radius is increasing:
dA/dt = (dA/dr) * (dr/dt)
We are looking for dr/dt, so we need to find dA/dr, the derivative of the area with respect to the radius.
Taking the derivative of A = πr^2 with respect to r gives us:
dA/dr = 2πr
Substituting the values we have, we get:
96π = (2πr) * (dr/dt)
Now we can solve for dr/dt:
dr/dt = (96π) / (2πr)
dr/dt = 48 / r
Since we want to find the rate at which the radius is increasing when the area is 1000π, we can substitute A = 1000π into the equation:
dr/dt = 48 / r
dr/dt = 48 / (sqrt(1000π))
dr/dt ≈ 4.81 meters per second (rounded to two decimal places)
Therefore, when the area of the oil spill is 1000π square meters, the radius is increasing at a rate of approximately 4.81 meters per second.