related rates: An oil spill in the shape of a circle in the

[thegirl]

I have a problem that I need help with (otherwise I wouldnt be here) mainly B. I am just unsure how to set it up

The question is:

An oil spill in the shape of a circle in the ocean with a radius of 150 meters is expanding 0.1 meters/minute and its thickness is 0.02 meters. At that moment:

a) how fast is the area expanding?

I think I came up with 94.2 m^2 / min

b) The circular slick has the same thickness everywhere, and the volume remains fixed. How fast is the thickness of the slick decreasing?

 

[skeeter]

Re: related rates

I have a problem that I need help with (otherwise I wouldnt be here) mainly B. I am just unsure how to set it up

The question is....an oil spill in the shape of a circle in the ocean with a radius of 150 meters is expanding 0.1 meters/miunte and its thickness is 0.02 meters. At that moment:

a) how fast is the area expanding I think I came up with 94.2 m^2 / min

correct

b) the circular slick has the same thickness everywhere, and the volume remains fixed, how fast is the thickness of the slick decreasing?

let h = thickness of the slick
V = pi*r²*h
dV/dt = pi*r²*(dh/dt) + h*2pi*r*(dr/dt)
since the volume remains fixed, dV/dt = 0. you have all the necessary info to
solve for dh/dt ... its value should be negative, correct?

 

[Unco]

The question is....an oil spill in the shape of a circle in the ocean with a radius of 150 meters is expanding 0.1 meters/miunte and its thickness is 0.02 meters. At that moment:

a) how fast is the area expanding I think I came up with 94.2 m^2 / min

Agreed, well done.

b) the circular slick has the same thickness everywhere, and the volume remains fixed, how fast is the thickness of the slick decreasing?

If V is the volume of the spill, r the radius, h the thickness, then by the chain rule \(\displaystyle \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} \, + \, \frac{dV}{dh} \cdot \frac{dh}{dt}\).