Rhys_M
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For part B below, how can I get the differential equation into a normal equation?

I've tried lots of times and got h=1/9t - ht/300
But i think this is really wrong.
Thanks
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mqb2766
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h is not a constant, its a function of t, so you have to integrate it like the examples you were doing (yesterday), so integrate wrt h on one side and integrate wrt t on the other. The solution should involve logs / exponentials.
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Rhys_M
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(Original post by mqb2766)
h is not a constant, its a function of t, so you have to integrate it like the examples you were doing (yesterday), so integrate wrt h on one side and integrate wrt t on the other. The solution should involve logs / exponentials.
I'm strugglying to manipulate it to get that, as it involves adding and minusing terms so i cant get anything where there isn't an h without a dt or vice versa
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mqb2766
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You did a harder problem yesterday and you've ended up with a
h dt
integral, which is still a problem.

If I wrote a simplified version as
dh/dt = (1+h)
how would you do it? Note the brackets on the right. Then a simple generalization to the actual numbers in the question.
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Rhys_M
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Ohh so do you mean treat the derivative as a fraction
So
Dh/dt = 1/9 - h/300
Would become
Dt/dh = 9 - 300/h
Which is a valid integral?
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the bear
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you need to separate the variables. this involves putting the right hand expression 100 -3h into brackets so it can be moved correctly.
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Rhys_M
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(Original post by the bear)
you need to separate the variables. this involves putting the right hand expression 100 -3h into brackets so it can be moved correctly.
I don't understand how brackets will help - how can I separate the variables in this case?
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the bear
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(Original post by Rhys_M)
I don't understand how brackets will help - how can I separate the variables in this case?
you need to get all of the bits relating to h onto the LHS, and all the bits relating to t onto the RHS. that is why putting brackets round 100 - 3h will help.
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mqb2766
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For the example I gave
1/(1 + h) dh = dt
and integrate. The brackets don't help by themselves, but dividing by the complete right hand side does (and then "multiply" by dt), rather than rearranging it by adding and subtracting terms.
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Rhys_M
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(Original post by mqb2766)
For the example I gave
1/(1 + h) dh = dt
and integrate. The brackets don't help by themselves, but dividing by the complete right hand side does (and then "multiply" by dt), rather than rearranging it by adding and subtracting terms.
Oh ok thank you - so is this getting close?
I don't think it's fully right as the answer for part b is 300ln4 seconds - which I can't get to
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mqb2766
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No, about a third of the way down, the reciprocal of
1/a + 1/b
is not
a + b

Without any simplification, but divide by the right hand side (and left) you have
1/(100 - 3h) dh = 1/900 dt
and integrate ...

The left hand side is a log, the right is "t". So if you write h = ..., the right hand side will basically be an exponential in t, which you should expect if the differential equation is dh/dt ~ h.
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Rhys_M
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(Original post by mqb2766)
No, about a third of the way down, the reciprocal of
1/a + 1/b
is not
a + b

Without any simplification, but divide by the right hand side (and left) you have
1/(100 - 3h) dh = 1/900 dt
and integrate ...

The left hand side is a log, the right is "t". So if you write h = ..., the right hand side will basically be an exponential in t, which you should expect if the differential equation is dh/dt ~ h.
I've got the answer now - thank you so much !
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