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#1 2013-06-06 10:15:30

Au101
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Rates of change (conical vessel)

Okay, so, having hopefully got myself re-acquainted with the very basics of differentiation, I now realise how much basic geometry I've forgotten tongue (sigh - if only i still had my formula books tongue). Anyway, enough complaining, so I'm looking at the chain rule and rates of change and the first question I have is:



So, does anyone know where I've gone wrong?

Last edited by Au101 (2013-06-06 10:16:11)

#2 2013-06-06 16:47:38

bob bundy
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Re: Rates of change (conical vessel)

hi Au101,

Welcome back; nice to hear from you again.  smile

I've read and re-read this problem and I cannot find anything wrong with your answer.  Maybe the book answer is just a typo.  It would be easy to type two 1s rather than two zeros.  My brain to finger coordination does this to me all the time.  With the answer expressed as a multiple of pi it's hard to see why 110 (not divisible be 4) would be correct.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

#3 2013-06-06 17:38:18

{7/3}
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Re: Rates of change (conical vessel)

I am confused,since (i) wants rate depending on depth.so we should calculate dv/dh at h=5(which is 25π).correct me if i'm wrong.

Last edited by {7/3} (2013-06-06 17:39:52)


There are 10 kinds of people in the world,people who understand binary and people who don't.

#4 2013-06-06 20:14:56

bob bundy
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Re: Rates of change (conical vessel)

hi {7/3)

Rate is usually taken as wrt time. So asking for dV/dt

Also look at the units for the book answer.

This is what Au101 has done.



Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

#5 2013-06-06 21:35:55

{7/3}
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Re: Rates of change (conical vessel)

Ok.in that case shouldn't dV/dt be  a function in terms of t,not h?


There are 10 kinds of people in the world,people who understand binary and people who don't.

#6 2013-06-06 23:24:34

anonimnystefy
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Re: Rates of change (conical vessel)

Well, h is a function of t, so it comes down to the same thing.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#7 2013-06-07 00:05:08

{7/3}
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Re: Rates of change (conical vessel)

Thanks


There are 10 kinds of people in the world,people who understand binary and people who don't.

#8 2013-06-07 00:13:49

anonimnystefy
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Re: Rates of change (conical vessel)

No problem.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#9 2013-06-07 00:57:19

Au101
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Re: Rates of change (conical vessel)

It's great to be back, bob bundy smile Thanks so much. I have a feeling you're right, I just didn't have enough confidence in my answer, but the book agrees with me that the answer to part (ii) is:



{7/3} and anonimnystefy, you're absolutely right. We know that h varies with t, specifically, it varies at the uniform rate of:



(As the question tells us.)

And we know that V varies with h, specifically, it varies at a rate of:



(According to my calculations)

We know, then, that V also varies with t, since it varies with h, which varies with t. Specifically, by the chain rule, it varies at a rate of:

#10 2013-06-07 05:43:02

Au101
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Re: Rates of change (conical vessel)

Just a very very quick point of confusion I'd like to clear up, if I may: I have the question:



And after some mathematics, I come up with the solution that the length l of NT is:



Which agrees with the answer book, except the answer book does not have the modulus sign. I just wanted to clear-up why I can simply get rid of the modulus signs in this case, since my rustiness even extends to calculations of distance hmm tongue

#11 2013-06-07 18:19:51

bob bundy
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Re: Rates of change (conical vessel)

hi Au101,

Because it is measuring distance and that cannot be negative.  Without the || some values of t would give negative values for 4t^2 - 4t/3

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

#12 2013-06-07 21:58:43

Au101
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Re: Rates of change (conical vessel)

So, surely, I should keep them? The answer book simply has



I had thought that maybe the answer book was giving a simplified answer, it's an old A-level book, but the syllabus was very different back then, so I'm never sure what I'm expected to know tongue But looking at it, if I draw a graph, I don't think T can ever - on this graph - be above N on the y-axis, so - presumably - the distance can always be given by N - T, with no need to worry about what would happen if T were to occur above N, giving a negative distance?

#13 2013-06-07 22:34:51

bob bundy
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Re: Rates of change (conical vessel)

Maybe it depends on the way a question is worded.  I've met some where the distance along the x axis  of a particle from the origin is given as a function of t.  For some t the distance comes out negative and you're supposed to interpret that as meaning the particle is to the left of (0,0) .

I doubt it would loose you marks either way.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

#14 2013-06-07 22:40:03

Au101
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Re: Rates of change (conical vessel)

Okay, thanks bob bundy smile I don't think I'll worry about it too much, I mean, I just want to get my calculus back to a good enough level to start looking at some new maths and physics &c., so it's just some general practice smile It's good for me to understand as much as possible though smile

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