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 May 12th, 2017, 06:45 AM #1 Newbie   Joined: May 2017 From: alabama Posts: 6 Thanks: 0 calculus A tank of water is being drained through an outlet, the height H (m) of water in the time t (s) is given by dh/dt = -(3x10^-3) √H given t = 0 and H = 4 determine an expression for H in terms of t. Can anybody point me in the right direction with this question? I don't really understand what it's asking. Last edited by skipjack; May 16th, 2017 at 06:02 AM.
May 12th, 2017, 07:08 AM   #2
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Quote:
 Originally Posted by tommyturner1993 A tank of water is being drained through an outlet, the height H (m) of water in the time t (s) is given by dh/dt = -(3x10^-3) √H
You mean dH/dt= -(3x10^(-3))√H don't you?

Quote:
 given t = 0 and H = 4
When t = 0, H = 4 (or H(0) = 4).

Quote:
 determine an expression for H in terms of t. Can anybody point me in the right direction with this question? I don't really understand what it's asking.
Since you are given the derivative of H, you want to integrate to get H.

You can write this as $\displaystyle \frac{dH}{H^{1/2}}= H^{-1/2}dH= -(3x10^{-3})dt$. Now integrate both sides. That will give you a "constant of integration". Use H(0)= 4 to determine that constant.

Last edited by skipjack; May 16th, 2017 at 06:03 AM.

 May 12th, 2017, 10:43 AM #3 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 I'm assuming here that the x is the symbol for multiplier here. In my example, I will use less confusing notation as x is a common variable. $\displaystyle \frac{dH}{dt}=-(3\cdot 10^{-3})\sqrt{H(t)}$ $\displaystyle \frac{dH}{\sqrt{H(t)}}=-(3\cdot 10^{-3})dt$ $\displaystyle \int H(t)^{-\frac{1}{2}}dH=\int -3\cdot 10^{-3}dt$ $\displaystyle 2\sqrt{H(t)}+C_1=-3\cdot 10^{-3}t+C_2$ $\displaystyle \sqrt{H(t)}=\frac{-3\cdot 10^{-3}}{2}t+C$ $\displaystyle H(t)=\bigg (\frac{-3\cdot 10^{-3}}{2}t+C\bigg )^2$ $\displaystyle H(0)=4$ $\displaystyle 4=\bigg (\frac{-3\cdot 10^{-3}}{2}(0)+C\bigg )^2$ $\displaystyle 2=C$ $\displaystyle H(t)=\bigg (\frac{-3\cdot 10^{-3}}{2}t+2\bigg )^2$ $\displaystyle H(t)=(.0015t+2)^2$ Hopefully, I didn't make a careless mistake. Hope this helps. Last edited by skipjack; May 16th, 2017 at 06:05 AM.
 May 16th, 2017, 05:06 AM #4 Newbie   Joined: May 2017 From: alabama Posts: 6 Thanks: 0 I think I may have noted the question down wrong, by 3x10^-3 I meant as in the metrix prefix (0.003) rather than 3 multiplied by ten to the power of minus 3 if that makes sense?
May 16th, 2017, 05:56 AM   #5
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Quote:
 Originally Posted by tommyturner1993 I think I may have noted the question down wrong, by 3x10^-3 I meant as in the metrix prefix (0.003) rather than 3 multiplied by ten to the power of minus 3 if that makes sense?
No, it doesn't! "3 multiplied by ten to the power of minus 3" is 0.003.

 May 16th, 2017, 08:22 AM #6 Newbie   Joined: May 2017 From: alabama Posts: 6 Thanks: 0 So if the question read A tank of water is being drained through an outlet, the height H (m) of water in the time t (s) is given by dh/dt = - (0.003) √H given t = 0 and H = 4 determine an expression for H in terms of t. That would give the same answer?

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