a tank leaks 0.04 m³/hr of oil into a lake. the oil forms a semicircular disk with a thickness of 10⁻⁶…

Explanation:

Step1: Find the volume formula of the semi - circular disk

The volume $V$ of a semi - circular disk with radius $r$ and thickness $h$ is $V=\frac{1}{2}\pi r^{2}h$. Given $h = 10^{-6}$ m, so $V=\frac{1}{2}\pi r^{2}(10^{-6})=5\times10^{-7}\pi r^{2}$.

Step2: Differentiate the volume formula with respect to time $t$

Using the chain - rule, $\frac{dV}{dt}=5\times10^{-7}\pi\times2r\frac{dr}{dt}=10^{-6}\pi r\frac{dr}{dt}$.

Step3: Identify the value of $\frac{dV}{dt}$

The tank leaks oil at a rate of $\frac{dV}{dt}=0.04$ m³/hr.

Step4: Solve for $\frac{dr}{dt}$

We know $\frac{dV}{dt}=0.04$, $r = 145$ m. Substitute these values into $\frac{dV}{dt}=10^{-6}\pi r\frac{dr}{dt}$. Then $0.04=10^{-6}\pi\times145\times\frac{dr}{dt}$. First, we can rewrite the equation as $\frac{dr}{dt}=\frac{0.04}{10^{-6}\pi\times145}$. Calculate $\frac{0.04}{10^{-6}\pi\times145}=\frac{0.04\times10^{6}}{\pi\times145}=\frac{40000}{\pi\times145}\approx87.965$.

Answer:

$87.965$