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?
Pretty similar to the truck and the police car:
Area = A = pi r^2
dA/dt = 2 pi r dr/dt (which by the way is the circumference times the outward speed logically enough.)
part 2 is rate of volume change
Vol = V = pi r^2 h where h is thickness
dV/dt = pi r^2 (dh/dt) + h (2 pi r dr/dt)
but
dV/dt = 0 given so
pi r^2 dh/dt = - 2 pi r h dr/dt
we did dr/dt in part 1
dh/dt = - 2(h/r)dr/dt
-2.667* 10^-5
To find the answers to the given questions, we can use the formulas for the area of a circle and the volume of a cylinder.
a. To find the rate at which the area of the slick is expanding, we need to differentiate the formula for the area of a circle with respect to time:
A = πr^2
where A is the area of the slick and r is the radius. Differentiating both sides of the equation with respect to time (t), we get:
dA/dt = 2πr(dr/dt)
Given that dr/dt = 0.1 m/min and r = 150 m, we can plug in these values to find the rate of expansion of the area:
dA/dt = 2π(150)(0.1) = 94.2 m^2/min
Therefore, the area of the slick is expanding at a rate of 94.2 m^2/min.
b. To find the rate at which the thickness of the slick is decreasing, we need to differentiate the formula for the volume of a cylinder with respect to time:
V = πr^2h
where V is the volume of the slick, r is the radius, and h is the thickness. Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = 2πr(dr/dt)h + πr^2(dh/dt)
Given that dV/dt = 0 (since the volume of oil spilled remains fixed) and r = 150 m, we can solve for dh/dt:
0 = 2π(150)(0.1)h + π(150)^2(dh/dt)
Simplifying the equation, we get:
-30πh = 22500π(dh/dt)
Dividing both sides of the equation by -30πh, we find:
dh/dt = -750 m/min
Therefore, the thickness of the slick is decreasing at a rate of 750 m/min.
To answer these questions, we'll need to use some calculus and related rates principles. Let's break it down step by step:
a. How fast is the area of the slick expanding?
We're given that the radius of the slick is expanding at a rate of 0.1 m/min. Let's call the radius of the slick at any given time t as r(t). We want to find the rate at which the area of the slick is expanding, so we need to find dA/dt, where A represents the area of the slick.
The area of a circle is given by the formula: A = πr^2, where π is a constant.
To find dA/dt, we can differentiate both sides of the equation with respect to time (t):
dA/dt = d(πr^2)/dt
Using the chain rule, we can calculate the derivative of r^2 with respect to t:
dA/dt = 2πr(dr/dt)
Substituting the given values, we have:
dA/dt = 2π(150)(0.1) = 30π m^2/min
Therefore, the area of the slick is expanding at a rate of 30π square meters per minute.
b. If the circular slick has the same thickness everywhere, and the volume of oil spilled remains fixed, how fast is the thickness of the slick decreasing?
We're given that the thickness of the slick is 0.02 meters. Let's call the thickness of the slick at any given time t as h(t). We want to find the rate at which the thickness of the slick is decreasing, so we need to find dh/dt.
The volume of the slick can be calculated as V = Ah, where A is the area of the slick and h is the thickness. Since the volume of the oil spilled remains fixed, the volume V is constant.
Therefore, dV/dt = 0, which implies d(Ah)/dt = 0
Using the product rule, we can expand d(Ah)/dt as follows:
dV/dt = A(dh/dt) + h(dA/dt) = 0
Since dA/dt = 30π m^2/min (as calculated in part a), we can solve for dh/dt:
30π(0.02) + 150(dh/dt) = 0
dh/dt = -(30π(0.02))/150
= -0.02π/5
= -0.004π m/min
Therefore, the thickness of the slick is decreasing at a rate of 0.004π meters per minute.