Question

heres an optimization problem, much thanks to anyone who could shed some light...
"a ruptured oil tanker causes a circular oil slik on the surface of the ocean. when it's radius is 200meters, the radius of the slick is expanding to 0.8m/min it's thickness is 0.08 meters. at the moment, how fast is the area of the slick expanding?

Answer 1

i dont understand the thickness is not important here
just the radius

Answer 2

well i think the thickness implies that you are dealing with a cylinder here
so somehow i was thinking about relating the volume equation to the area and find the rate

Answer 3

the answer is (16/3)pi for what it's worth, still no clue how it's done

Answer 4

could u draw what u understand about the problem

Answer 5

the radius being 200 and the sides or height being .8

Answer 6

@ganeshie8
@mathstudent55
@whpalmer4

Answer 7

Are you sure you've copied all the numbers and so on correctly?

at t = 0, the radius is 200 m, and at that point the area is \(\pi *200^2 \approx 125664 \text{ m}^2\)
at t = 1 min, if the radius is increasing by 0.8 m/min, the new radius is 200.8 m and the new area is \(\pi * 200.8^2 \approx 126671\) and the difference between those is \(\approx 320\pi\)

Answer 8

right, but the question is asking at that moment
so its looking for the instant rate? i think?

Answer 9

right, but the instant rate will be approximately the same as what happens in 1 minute. I'm just linearizing about the point of interest. that's not going to give a factor of 60 error :-)

now, are there any units on that rate? perhaps it is in m^2/s, in which case 60 would be exactly the factor we would expect...

Answer 10

oh yeah, sorry, the units are in meters^2/sec

Answer 11

okay. then it is easy.

the area of the slick is just the area of a circle with radius \(r\):
\[A = \pi r^2\]the rate of change of area with respect to the radius is\[\frac{dA}{dr} = 2\pi r\]Agreed?

Answer 12

yes.

Answer 13

the rate of change of the radius with respect to time is \[\frac{dr}{dt} = 0.8 \text{ m/min} * 1 \text{ min}/60\text{ s} = \frac{0.8}{60} \text{ m/s}\]
\[\frac{dA}{dt} = \frac{dA}{dr}*\frac{dr}{dt}\]
Plug in the numbers and you've got your answer

Answer 14

so thickness was useless

Answer 15

It would seem so...

Answer 16

wow, literally spent almost 2 hours on that
thanks so much!

Answer 17

that's a fair amount of oil that makes up that slick, however — about 10 million liters, or 2.6 million gallons, about 4x the volume of an Olympic swimming pool

Answer 18

the second part of the question states that the thickness is uniform and volume of oil spilled remains fixed. how fast is the thickness of the slick decreasing when the radius is 200 meters?

Answer 19

is this where the thickness comes into play?

Answer 20

so our instantaneous rate of change is \(16\pi/3\text{ m}^2\text{/s}\approx 16.7552\text{ m}^2\text{/s}\)

Let's do the same calculation I did originally (with the right units this time!) and see how close it is.

at t = 0, r = 200 m, A = \(40000\pi\approx 125664\)
at t = 1s, r = 200 + 0.8/60, A = \(200^2+2*0.8/60 + (0.8/60)^2)\pi \approx 125680\)
\[\Delta A = 125680-125664 = 16\]\[\frac{16}{3}\pi \approx 16.7552 \text{ or about } 4.5\% \text{ error}\]

So, it's a useful technique for sanity checking your answer. We pretend the instantaneous rate of change is constant for the next short interval of time and compute the change, and compare with what the problem gave us.

Answer 21

right! ok thanks! =-)