Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 m/s, how fast is the area of the spill increasing when the radius is
23 m?
so you have
A = πr^2
dA/dt = 2πr dr/dt
Plugging in your numbers,
dA/dt = 2π(23m)(1m/s) = 46π m^2/s
To find the rate at which the area of the oil spill is increasing, we need to derive the equation for the area of a circle and then differentiate it with respect to time.
The area of a circle can be given by the formula: A = πr^2, where A is the area and r is the radius.
Given that the radius of the oil spill is increasing at a constant rate of 1 m/s, we can express it as: r = 23 m.
Now, let's differentiate the equation for the area of the circle with respect to time (t):
dA/dt = d/dt (πr^2)
To differentiate the equation, we use the chain rule:
dA/dt = 2πr(dr/dt)
Substituting the given value for r and the rate of change of the radius:
dA/dt = 2π(23)(1)
Calculating the above expression, we get:
dA/dt = 46π m^2/s
Therefore, when the radius is 23 m, the area of the oil spill is increasing at a rate of 46π m^2/s.
To find how fast the area of the spill is increasing when the radius is 23 m, 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 radius is increasing at a constant rate of 1 m/s. This means that the derivative of the radius with respect to time (dr/dt) is 1.
To find how fast the area is increasing with respect to time (dA/dt), we need to take the derivative of the area formula with respect to time:
dA/dt = d/dt(πr^2)
Using the power rule of differentiation, we can find the derivative of r^2 with respect to time:
d(r^2)/dt = 2r(dr/dt)
Here, dr/dt is given as 1 m/s, and we are interested in finding dA/dt when r = 23 m.
Plugging in the values:
dA/dt = 2(23)(1)π
Simplifying:
dA/dt = 46π
Therefore, the area of the oil spill is increasing at a rate of 46π square meters per second when the radius is 23 m.