A tank leaks 0.11 m3hr of oil into a lake. The oil forms a semicircular disk with a thickness of 10−6 meters.

Question

A tank leaks  of oil into a lake. The oil forms a semicircular disk with a thickness of  meters. How rapidly is the radius of the disk increasing  hours after the tank begins leaking?The radius of the disk is increasing by ____ m/hr (round your answer to 3 decimal places)

Answer 1

V=(h)(πr²)/2

take derivative of both sides

DV/dt=(dh/dt)(πr²)/2+(h)(πr)(dr/dt)

Now let's use what we know

DV/dt = .11 meters cubed per hour

Dh/dt = 0. Height isn't changing so rate is 0

Now we need to figure out what value we need for r. To do so, we will use the original formula for volume. Remember it has been leaking for 5 hours so

5 hours(.11 meters cubed per hour) = .55 meters cubed

so

V=(h)(πr^2)/2

.55 m³ = (10^-6 m)(πr²)/2 m²

Solve for r

591.727 meters

Now let's plug everything in for the differential.

DV/dt=(dh/dt)(πr²)/2+(h)(πr)(dr/dt)

.11=(0)(π)(591.727)²+(10^-6)(π)(591.727)(dr/dt)

.11/[(10^-6)(π)(591.727)=dr/dt = 59.173 m/hr