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May 12th, 2017, 06:45 AM   #1
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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.
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May 12th, 2017, 07:08 AM   #2
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Quote:
Originally Posted by tommyturner1993 View Post
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.
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May 12th, 2017, 10:43 AM   #3
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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.
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May 16th, 2017, 05:06 AM   #4
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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?
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May 16th, 2017, 05:56 AM   #5
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Quote:
Originally Posted by tommyturner1993 View Post
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.
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May 16th, 2017, 08:22 AM   #6
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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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