oil is leaking from a pipeline on the surface of a lake and forms an oil slick whose volume increases at a constant rate of 2000 cubic centimeters per minute. The oil slick takes the form of a right circular cylinder with both its radius r and height changing h is given by V= pie times r to the second power times h)
a) at the instant when the radius of oil slick is 100 cenimetters and the hieght is 0.5 cm the radius is increasing at the rate of 2.5 cm per min. at this instant what is the rate of change of the height of oil slick with respect to time in cm per min
let the thickness of the oil slick be h cm
let the radius be r cm
let the volume be V cm^3
given:
dV/dt = 2000cm^3/min
when r=100 cm, dr/dt = 2.5 cm/min and
V=pi(r^2)h
dV/dt = pi(r^2)dh/dt + 2pi(r)(h)dr/dt
2000 = pi)100^2)dh/dt + 2pi(100)(.5)(2.5)
solve for dh/dt
To find the rate of change of the height of the oil slick with respect to time, we can use the given equation for the volume of the oil slick:
V = π * r^2 * h
We are given the following values:
Radius, r = 100 cm (constant)
Initial height, h = 0.5 cm (constant)
Rate of change of the radius, dr/dt = 2.5 cm/min (given)
To find the rate of change of the height, we need to differentiate the volume equation with respect to time:
dV/dt = d(π * r^2 * h)/dt
Since r and h are constants in this case, the equation simplifies to:
dV/dt = π * (2r * dr/dt * h)
Substituting the given values:
dV/dt = π * (2 * 100 cm * 2.5 cm/min * 0.5 cm)
Calculating this expression:
dV/dt = π * (500 cm^2 * 2.5 cm/min)
Simplifying:
dV/dt = 1250π cm^3/min
Therefore, the rate of change of the height of the oil slick with respect to time is 1250π cm^3/min.
To find the rate of change of the height of the oil slick with respect to time, we need to differentiate the volume function V = πr^2h with respect to time t. This will give us an equation relating the rate of change of height with respect to time, dh/dt, to the known rates of change of volume and radius.
Given:
- Volume rate of change: dV/dt = 2000 cm^3/min
- Radius rate of change: dr/dt = 2.5 cm/min
- Radius: r = 100 cm
- Height: h = 0.5 cm
Differentiating the volume function, we get:
dV/dt = π(2rh)(dr/dt) + πr^2(dh/dt)
2000 = π(2(100)(0.5))(2.5) + π(100^2)(dh/dt)
2000 = π(100)(2.5) + π(10000)(dh/dt)
2000 = 250π + 10000π(dh/dt)
Simplifying the equation, we have:
2000 - 250π = 10000π(dh/dt)
Now we can solve for dh/dt, the rate of change of the height with respect to time:
dh/dt = (2000 - 250π) / (10000π)
Using a calculator to approximate the value of π as 3.14:
dh/dt ≈ (2000 - 250(3.14)) / (10000(3.14))
dh/dt ≈ 1.18 cm/min
Therefore, at the given instant, the rate of change of the height of the oil slick with respect to time is approximately 1.18 cm/min.