An ocean oil well is leaking oil into the ocean at a rate of 1 m3/hr resulting in an oil slick in the form of a circular

disk (like a quarter or dime). If the slick remains in the shape of a disk with a constant thickness of 1/100 m as the disk expands, determine the rate at which the radius of the slick increases when the radius is 10 meters

To determine the rate at which the radius of the slick increases, we can use the related rates formula:

\(A = πr^2\)

where A is the area of the circular slick, and r is the radius.

We are given that the oil well is leaking at a rate of 1 m3/hr. Since the slick has a constant thickness of 1/100 m, the volume of oil spread over time is given by:

\(V = 1 \, \text{m}^3/\text{hr}\)

The volume of the slick is equal to the product of its area and thickness:

\(V = A \cdot \text{thickness}\)

Substituting the value of the thickness (1/100 m) and A = πr^2, we have:

\(1 \, \text{m}^3/\text{hr} = πr^2 \cdot \frac{1}{100}\)

Rearranging the equation to solve for A:

\(A = \frac{100}{π} \, r^2\)

Now, let's differentiate both sides of the equation with respect to time:

\(\frac{dA}{dt} = \frac{d}{dt} \left(\frac{100}{π} \, r^2\right)\)

The left side represents the rate at which the area of the slick is changing, and the right side represents the derivative of the expression \(\frac{100}{π} \, r^2\) with respect to time.

Using the power rule for differentiation, we can find the derivative:

\(\frac{dA}{dt} = \frac{100}{π} \cdot 2r \cdot \frac{dr}{dt}\)

Now we can substitute the given values, r = 10 meters and V = 1 m3/hr, into the equation:

\(1 \frac{\text{m}^3}{\text{hr}} = \frac{100}{π} \cdot 2(10) \cdot \frac{dr}{dt}\)

Simplifying further:

\(1 \frac{\text{m}^3}{\text{hr}} = \frac{2000}{π} \cdot \frac{dr}{dt}\)

Now we can solve for \(\frac{dr}{dt}\):

\(\frac{dr}{dt} = \frac{π}{2000}\cdot \frac{1}{r} \frac{\text{m}}{\text{hr}}\)

Therefore, when the radius of the slick is 10 meters, the rate at which the radius increases is \(\frac{π}{2000}\) meters per hour.