Question

a tank leaks $0.01 \frac{\text{m}^3}{\text{hr}}$ of oil into a lake. the oil forms a semicircular disk with a thickness of $10^{-6}$ meters. how rapidly is the radius of the disk increasing 5 hours after the tank begins leaking?
the radius of the disk is increasing by \boxed{} m/hr (round your answer to 3 decimal places)

Explanation:

Step1: Define Volume of Semicircular Disk

The volume \( V \) of a semicircular disk (a half - cylinder, essentially) with radius \( r \) and thickness (height) \( h \) is given by \( V=\frac{1}{2}\pi r^{2}h \). Here, \( h = 10^{-6}\) meters (constant thickness) and the rate of change of volume with respect to time \( t \) is \( \frac{dV}{dt}=0.01\space m^{3}/hr \).

Step2: Differentiate Volume with Respect to Time

Differentiate \( V=\frac{1}{2}\pi r^{2}h \) with respect to \( t \) using the chain rule. Since \( h \) is constant, we have:
\( \frac{dV}{dt}=\frac{1}{2}\pi h\times2r\frac{dr}{dt}=\pi r h\frac{dr}{dt} \)
We can solve for \( \frac{dr}{dt} \): \( \frac{dr}{dt}=\frac{1}{\pi r h}\frac{dV}{dt} \)

Step3: Find Volume at \( t = 5 \) hours

The volume leaked after \( t = 5 \) hours is \( V=\frac{dV}{dt}\times t \). Substituting \( \frac{dV}{dt}=0.01\space m^{3}/hr \) and \( t = 5 \) hours, we get \( V=0.01\times5 = 0.05\space m^{3} \)

Step4: Find Radius at \( t = 5 \) hours

Using the volume formula \( V=\frac{1}{2}\pi r^{2}h \), we can solve for \( r \). Rearranging for \( r \):
\( r^{2}=\frac{2V}{\pi h} \)
Substitute \( V = 0.05\space m^{3} \), \( h=10^{-6}\space m \) and \( \pi\approx3.1416 \):
\( r^{2}=\frac{2\times0.05}{\pi\times10^{-6}}=\frac{0.1}{\pi\times10^{-6}}\approx\frac{0.1}{3.1416\times10^{-6}}\approx\frac{0.1}{3.1416}\times10^{6}\approx0.03183\times10^{6}=31830 \)
\( r=\sqrt{31830}\approx178.41\space m \)

Step5: Substitute Values to Find \( \frac{dr}{dt} \)

Now substitute \( \frac{dV}{dt}=0.01\space m^{3}/hr \), \( r\approx178.41\space m \), \( h = 10^{-6}\space m \) and \( \pi\approx3.1416 \) into the formula for \( \frac{dr}{dt} \):
\( \frac{dr}{dt}=\frac{1}{\pi r h}\frac{dV}{dt} \)
\( \frac{dr}{dt}=\frac{0.01}{\pi\times178.41\times10^{-6}} \)
First, calculate the denominator: \( \pi\times178.41\times10^{-6}\approx3.1416\times178.41\times10^{-6}\approx560.5\times10^{-6}=5.605\times10^{-4} \)
Then, \( \frac{dr}{dt}=\frac{0.01}{5.605\times10^{-4}}\approx\frac{10^{-2}}{5.605\times10^{-4}}=\frac{10^{2}}{5.605}\approx\frac{100}{5.605}\approx17.84 \)

Answer:

\( 17.840 \) (rounded to 3 decimal places)