Here's one we haven't worked with before: A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at the rate of 0.1 cm/hr as the slick spreads. (Note: 1 cm = 0.01 m.) At what rate is the radius of the slick increasing when the radius is 8 meters? (You can think of this oil slick as a very flat cylinder; its volume is given by V = pi(r^2)h, where r is the radius and h is the height of this cylinder.)
if the thickness is h, then
dh/dt = -0.1
v = πr^2 h = 10,000 cm^3 so r^2 = 10000/(πh)
dv/dt = 2πrh dr/dt + πr^2 dh/dt
since v is constant, dv/dt=0, and so
πr^2 dh/dt = -2πrh dr/dt
r dh/dt = -2h dr/dt
when r=8,
8(-0.1) = -2(100/√(8π)) dr/dt
dr/dt = -√(2π)/125 cm/hr
To find the rate at which the radius of the oil slick is increasing, we need to relate the variables given in the problem and differentiate with respect to time.
Let's first convert the thickness of the oil slick from centimeters to meters:
0.1 cm/hr = 0.1 * 0.01 m/hr = 0.001 m/hr
We are given that the volume of the oil slick is 1 cubic meter. Assuming the slick remains circular, the volume can be expressed in terms of the radius and thickness as follows:
V = πr^2h
We also know that the thickness, h, is decreasing at a constant rate of 0.001 m/hr.
Differentiating the volume equation with respect to time gives us:
dV/dt = (dV/dr)(dr/dt) + (dV/dh)(dh/dt)
Since the rate of change of volume, dV/dt, is zero (the volume remains constant at 1 cubic meter), and dh/dt is given as -0.001 m/hr, the equation simplifies to:
0 = (dV/dr)(dr/dt) - (dV/dh)(0.001)
Now, let's differentiate the volume equation with respect to the radius, r:
dV/dr = 2πrh
Substituting this expression into the simplified equation, we have:
0 = (2πrh)(dr/dt) - (dV/dh)(0.001)
Since the volume, V, is given as 1 cubic meter, we can find dV/dh as follows:
V = πr^2h
1 = π(8^2)h
1 = 64πh
h = 1/(64π)
Substituting this value of h into the equation, we have:
0 = (2πr(1/(64π)))(dr/dt) - (dV/dh)(0.001)
0 = (1/32)(dr/dt) - (dV/dh)(0.001)
Simplifying further, we get:
0 = (1/32)(dr/dt) - (0.001)(0.001)
Now, let's solve for dr/dt, the rate at which the radius is increasing:
(1/32)(dr/dt) = 0.001^2
(1/32)(dr/dt) = 0.000001
Multiplying both sides of the equation by 32, we get:
dr/dt = 32(0.000001)
dr/dt = 0.000032 meters per hour
Therefore, when the radius of the slick is 8 meters, it is increasing at a rate of 0.000032 meters per hour.
To find the rate at which the radius of the oil slick is increasing, we can use related rates and the formula for the volume of a cylinder.
Given information:
- The thickness of the oil slick is decreasing at a rate of 0.1 cm/hr.
- The volume of the oil slick is 1 cubic meter.
To start, let's convert the units to match the given rate of thickness:
1 cm = 0.01 m
Therefore, the rate of change of thickness is 0.1 cm/hr = (0.1 cm/hr) * (0.01 m/cm) = 0.001 m/hr.
Now, let's differentiate the volume formula with respect to time:
V = πr^2h
To differentiate, we need to write the volume equation in terms of variables we can differentiate. Since the height (h) is decreasing at a constant rate, we can express it in terms of the radius (r) using the given rate of thickness (0.001 m/hr):
h = 0.001 t
Now, substitute this expression for h in the volume equation:
V = πr^2(0.001 t)
Taking the derivative with respect to time (t) on both sides:
dV/dt = πr^2(0.001)
The left side of the equation represents the rate at which the volume is changing, which is equal to the rate at which the oil is being spilled (since the volume is given as 1 cubic meter).
dV/dt = 1
Therefore, by equating the rates:
1 = πr^2(0.001)
Now, let's solve for r in terms of t:
r^2 = 1 / (π(0.001))
r^2 = 1 / (0.00314)
r^2 ≈ 318.471
Taking the square root of both sides:
r ≈ √318.471
r ≈ 17.85 meters
Since we are asked to find the rate at which the radius is increasing when it is 8 meters, let's differentiate the equation:
r^2 = 318.471
2r(dr/dt) = 0
dr/dt = 0 / (2r)
dr/dt = 0
Therefore, when the radius is 8 meters, the rate at which the radius of the slick is increasing is 0 meters per hour.
Please note that the rate is 0 because the oil slick is not spreading at the given instant, and thus, its radius is not changing.
I hope this explanation helps! Let me know if you have any further questions.