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An open can of oil is accidently dropped into a lake; assume the oil spreads over the surface as a circular disc of uniform thickness whose radius increases steadily at the rate of 10 cm/sec. At the moment when the radius is 1 m, the thickness of the oil slick is decreasing at the rate of 4 mm/sec. How fast is it decreasing when the radius is 2 m?


As the volume of the open can is fixed (let us say $V$), let the radius of the circular disc in lake be $r$ and the thickness of the circular disc be $h$.

$$V=\pi r^2h$$ $$\frac{dr}{dt}=\frac{10}{100}$$ $$\frac{dV}{dt}=0=\pi(r^2\frac{dh}{dt}+2rh\frac{dr}{dt}).........(1)$$ $$r^2\frac{dh}{dt}+2rh\frac{dr}{dt}=0$$

When $r=1,\frac{dh}{dt}=\frac{-4}{1000}$ and when $r=2,\frac{dh}{dt}=?$ $$(1)^2\times \frac{-4}{1000}+2(1)h\times \frac{10}{100}=0$$ $$h=\frac{2}{100}$$

When $r=2,$ $$(2)^2\frac{dh}{dt}+2(2)(\frac{2}{100})(\frac{10}{100})=0$$ $$\frac{dh}{dt}=-0.002 \mathrm{m/sec}$$

But the answer given in my book is $-0.0005$ m/sec.

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  • $\begingroup$ How did I manage to misspell does as dis? $\endgroup$ – TRiG Jun 20 '17 at 14:37
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The error in your method is that you found the height at the moment when $r=1$ and used it to find the change in height when $r=2$, however since the height is changing this doesn't work. It's easier to find a fixed value, i.e. the volume, as I've done below. We have

$$ h = \frac{V}{\pi r^2} \implies \frac{dh}{dt} = -\frac{2V}{\pi r^3} \frac{dr}{dt} $$

substitute in $r=1$ and $\frac{dh}{dt} = -0.004$

$$ -0.004 = -\frac{2V}{\pi (1)^3} 0.1 \implies \frac{2V}{\pi} = 0.04 $$

now use this volume when $r=2$

$$ \frac{dh}{dt} = -\frac{2V}{8 \pi } 0.1 = -\frac{2( 0.02\pi )}{8\pi}0.1 = -0.0005 $$

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The problem is that you took the thickness at r=1m and considered it the thickness at r=2m. You need to calculate the new thickness at r=2m from the volume. enter image description here

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  • $\begingroup$ I have to say this is marvelous handwriting, at least compared to my scrawl. $\endgroup$ – Dando18 Jun 20 '17 at 13:10
  • $\begingroup$ xD thanks you I guess. Are you interested in a question about directional derivatives? I posted it a couple of days ago but got nothing. $\endgroup$ – Khalid T. Salem Jun 20 '17 at 13:14
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    $\begingroup$ Why make everyone turn their head/computer/tablet/phone sideways to read what you wrote when there is a perfectly functional equation formatting system (MathJax) available? $\endgroup$ – Kyle Kanos Jun 20 '17 at 17:15
  • $\begingroup$ I'm using my phone and I couldn't use MathJax before but I did today after this post, sorry for the picture being sideways, I'll edit it. $\endgroup$ – Khalid T. Salem Jun 20 '17 at 17:44
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After I fixed a mistake in my computations, I got $-0.0005$m/sec.

Since $r'(t)=\frac1{10}$ and $r(0)=1$ we get $r(t)=\frac1{10}t+1$. Using $V=\pi r^2(t)h(t)$ we get $$ h(t)=\frac{V}{\pi r^2(t)}\text{ and }h'(t)=-\frac{2Vr'(t)}{\pi r^3(t)}=-\frac{V}{5\pi \left(\frac1{10}t+1\right)^3}. $$ Since $$ -\frac4{1000}=h'(0)=-\frac{V}{5\pi \left(\frac1{10}\cdot 0+1\right)^3}=-\frac{V}{5\pi} $$ we get $V=\frac{2\pi}{100}$ and therefore $$ h'(t)=-\frac{4}{1000\left(\frac1{10}t+1\right)^3}. $$ Now consider $2=r(t)$ iff $t=10$ and we get $$ h'(10)=-\frac{4}{1000\left(\frac1{10}\cdot 10+1\right)^3}=-\frac{4}{8000}=-0.0005. $$

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You have that $h$ changes so $h$ at $r=1$ is not the same as $h$ at $r=2$.

You have that at $r=1$ that $$V=\pi\frac{2}{100}=\frac{\pi}{50}$$ And since $V$ doesn't change for $r=2$ $$\frac{\pi}{50}=2^2h\pi=4h\pi\\\frac{1}{200}=h$$ Now plugging in your formula $$2^2\frac{dh}{dt}+2(2)\left(\frac1{200}\right)\frac{10}{100}=0$$ From this follows that $$\frac{dh}{dt}=-0.0005\frac{m}{\sec}$$

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