Differential equations Watch

Hashmi
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#1
Report Thread starter 9 years ago
#1

In part (i) I can verify that h satisfies the equation by substituting the value of dh/dt:

0.2 = 0.1 (9 - h)^(1/3)

2 = (9-h)^(1/3)

8 = 9 - h

h = 1

But how can I verify t ?

Also a hint for part (ii) will be appreciated.
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darkness9999
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#2
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I think for part (i) you already showed them that because  \frac{dh}{dt} proves both since t is a function of h

hint for part (ii) :=====> Chainrule
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Hashmi
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(Original post by darkness9999)
I think for part (i) you already showed them that because proves both since t is a function of h
Ok thanks.

hint for part (ii) :=====> Chainrule
I have tried this but the answer is wrong:

dh/dt x dt/dh = 0.1 (9 - h)^(1/3)

0.2 x dt/dh = 0.1 (9 - h)^(1/3)

dt/dh = 0.1/0.2 x (9-h)^(1/3)
dt/dh = 0.5 x (9 - h)^(1/3)

Integrating both sides


What have I done wrong?
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Mathematician!
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#4
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#4
Just re-arrange the differential equation. This gives:

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
{(9-h)^{-\frac{1}{3}} \, dh = 0.1 dt


Then integrate!

\int(9-h)^{-\frac{1}{3}} \, dh = \int 0.1\, dt

EDIT: Sorry about the difficulties I was having with the LaTeX there... It's all fixed now!
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Hashmi
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(Original post by Mathematician!)
Just re-arrange the differential equation. This gives:



Then integrate!



EDIT: Sorry about the difficulties I was having with the LaTeX there... It's all fixed now!
Thanks a lot, I am getting the correct answer now.
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Mathematician!
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#6
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(Original post by Hashmi)
Thanks a lot, I am getting the correct answer now.
No problem. :smile:

EDIT: And thanks for the rep!
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