A ruptured oil tanker causes a circular oil slick on the surface of the ocean. When its radius is 150 meters, the radius of the slick is expanding at 0.1 m/min, and its thickness is .02 m. At that moment:
a. How fast is the area of the slick expanding?
b. If the circular slick has the same thickness as everywhere, and the volume of oil spilled remains fixed, how fast is the thickness of the slick decreasing?
Area = πr^2
dA/dt = 2πr(dr/dt)
for the given data
dA/dt = 2π(150)(.1) m^2/min
= 94.25 m^2/min
Volume = Areax height = Ah
dV/dt = A(dh/dt) + h(dA/dt)
so when r = 150, A = π(150)^2 = 22500π
dV/dt = 0 (it remains constant) and
h = .02
0 = 22500π(dh/dt) + .02(94.25)
dh/dt = - .02(94.25)/22500π
= - .000026667 m/min
check my arithmetic
To solve this problem, we can use the formulas for the area of a circle and the volume of a cylinder.
a. To find how fast the area of the slick is expanding, we need to find the rate of change of the area with respect to time.
The formula for the area of a circle is A = πr^2, where A is the area and r is the radius. Taking the derivative with respect to time (t), we get:
dA/dt = 2πr(dr/dt)
Given that dr/dt = 0.1 m/min and r = 150 meters, we can plug in the values and solve for dA/dt:
dA/dt = 2π(150)(0.1) = 30π ≈ 94.25 m^2/min
Therefore, the area of the slick is expanding at a rate of approximately 94.25 m^2/min.
b. To find how fast the thickness of the slick is decreasing, we need to find the rate of change of the thickness with respect to time.
The formula for the volume of a cylinder is V = πr^2h, where V is the volume, r is the radius, and h is the height (thickness).
Given that the volume of oil spilled remains fixed, it means the volume (V) stays constant. Therefore, we have dV/dt = 0.
Taking the derivative of the volume equation with respect to time, we get:
0 = 2πr(dr/dt)h + πr^2(dh/dt)
We are given that dr/dt = 0.1 m/min and h = 0.02 m. We need to solve for dh/dt, the rate at which thickness is decreasing:
0 = 2π(150)(0.1)(0.02) + π(150)^2(dh/dt)
0 = 0.6π + 22500π(dh/dt)
22500π(dh/dt) = - 0.6π
dh/dt = (-0.6π)/(22500π)
Simplifying this expression, we find:
dh/dt = - 0.000027 ≈ - 2.7 x 10^-5 m/min
Therefore, the thickness of the slick is decreasing at a rate of approximately 2.7 x 10^-5 m/min.
To find the answers to these questions, we need to use the formulas for the area of a circle and the volume of a cylinder.
a. To find how fast the area of the slick is expanding, we can use the formula for the area of a circle, A = πr^2, where A is the area and r is the radius.
Given that the radius is expanding at a rate of 0.1 m/min, we can use the chain rule of differentiation to find the rate of change of the area with respect to time.
Let's differentiate the equation A = πr^2 with respect to time:
dA/dt = d(πr^2)/dt
= 2πr(dr/dt)
Here, dA/dt represents the rate of change of the area, dr/dt represents the rate at which the radius is changing, and r is the radius at that moment.
Substituting the values we have:
dr/dt = 0.1 m/min (given)
r = 150 m (given)
dA/dt = 2π(150)(0.1)
= 30π m^2/min
≈ 94.25 m^2/min (approximately)
Thus, the area of the slick is expanding at a rate of approximately 94.25 m^2/min.
b. To find how fast the thickness of the slick is decreasing, we can use the formula for the volume of a cylinder, V = πr^2h, where V is the volume, r is the radius, and h is the thickness.
Given that the volume of the oil spilled remains fixed, we can differentiate the equation V = πr^2h with respect to time, and then solve for dh/dt.
Let's differentiate the equation V = πr^2h with respect to time:
dV/dt = d(πr^2h)/dt
= 2πrh(dr/dt) + πr^2(dh/dt)
Since the volume remains fixed, dV/dt is zero.
Thus, 0 = 2π(150)(0.1)h + π(150)^2(dh/dt)
Simplifying the equation:
0 = 30πh + 22500π(dh/dt)
Now we can solve for dh/dt:
22500π(dh/dt) = -30πh
dh/dt = -30h/22500
dh/dt = -h/750
Substituting h = 0.02 m (thickness given):
dh/dt ≈ -(0.02)/750
dh/dt ≈ -0.0000267 m/min (approximately)
Thus, the thickness of the slick is decreasing at a rate of approximately 0.0000267 m/min.