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a) Assuming even distribution of oil, calculate the volume in cubic meter oil slick when the radius is 1 km and the height is .23 meters

B) AT the exact instant in part a, the radius is increasing at the rate of 1.5 meters per minute. how quickly is the volume of the oil slick increasing at the same moment? How quickly is the height increasing at the same moment?

It is right cylinder

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  • $\begingroup$ I did part a and found the volume. r=1000m h=.25m v= 250,000 pi m^3 im stuck with part B. @RobinGoodfellow $\endgroup$ – Ace Acgel Nov 2 '14 at 19:34
  • $\begingroup$ @RobinGoodfellow $\endgroup$ – Ace Acgel Nov 2 '14 at 19:40
  • $\begingroup$ For you to answer the second part rate of change of volume, you need both rate of change of radius and rate of change of height. In the same part, if you know the rate of change of Volume, and rate of change of radius, you can find the rate of change of height. Are you missing any detail? $\endgroup$ – Satish Ramanathan Nov 2 '14 at 19:44
  • $\begingroup$ @satishramanathan thats all the details $\endgroup$ – Ace Acgel Nov 2 '14 at 19:48
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Answer:

$r_o-1000$ m

$h_o = .25$ m

Since it is even distribution,$ \frac{r_o}{h_o} = 4000 $

Thus $$r = 4000 h$$

Differentiating with respect to t,

$$\frac{dr}{dt} = 4000\frac{dh}{dt} => \frac{dh}{dt}= \dfrac{\frac{dr}{dt}}{4000}= \frac{1.5}{4000} = 0.000375\text{m/min}$$

$$V = \pi r^2 h$$

$$\frac{dV}{dt} = 2\pi r_o h_0\frac{dr}{dt}+\pi r_o^2\frac{dh}{dt} = \pi\left(2*1000*.25*1.5+1000^2*.000375\right) = 1125\pi \text{ }m^3\text{/min}$$

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