I have spilled 200cm^3 of vegetable oil on my kitchen floor. The resulting oil slick can be modeled as a cylinder with a very small height; the height is decreasing at a rate of 1/1000πcm/sec and the radius is currently 50cm.
How quickly is the radius of the oil slick increasing, in cm/sec?
V=PI*r^2 h
dv/dt=0=PIr^2 dh/dt + 2PI r h dr/dt
solve for dr/dt, you are given dh/dt.
To find out how quickly the radius of the oil slick is increasing, we can use the formula for the volume of a cylinder, which is V = πr^2h, where V is the volume, r is the radius, and h is the height.
Given that the height is decreasing at a rate of 1/1000π cm/sec, we can represent the rate of change of the height as dh/dt = -1/1000π cm/sec. Notice the negative sign because the height is decreasing.
We are interested in finding dr/dt, the rate at which the radius is changing. We can differentiate the volume formula with respect to time to relate dr/dt and dh/dt:
dV/dt = d(πr^2h)/dt
Now, using the product rule of differentiation, we can break down the formula:
dV/dt = 2πrh(dr/dt) + πr^2(dh/dt)
Since the height is very small, we can assume dh/dt ≈ 0. Therefore, the second term on the right side of the equation becomes negligible.
dV/dt ≈ 2πrh(dr/dt)
Now, we can substitute the known values into the equation:
200 cm^3/sec ≈ 2π(50 cm)(0.01π cm/sec)(dr/dt)
Simplifying, we can cancel out common factors:
1 ≈ π(50 cm)(0.01 cm/sec)(dr/dt)
Now we can solve for dr/dt:
dr/dt ≈ 1/(π(50 cm)(0.01 cm/sec))
Calculating this out, we find that:
dr/dt ≈ 0.002 cm/sec
Therefore, the radius of the oil slick is increasing at a rate of approximately 0.002 cm/sec.