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**Au101****Member**- Registered: 2010-12-01
- Posts: 351

Okay, so, having hopefully got myself re-acquainted with the very basics of differentiation, I now realise how much basic geometry I've forgotten (sigh - if only i still had my formula books ). 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 (2013-06-05 12:16:11)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,869

hi Au101,

Welcome back; nice to hear from you again.

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

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

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} (2013-06-05 19:39:52)*

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,869

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

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,918

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

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

Thanks

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,918

No problem.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**Au101****Member**- Registered: 2010-12-01
- Posts: 351

It's great to be back, bob bundy 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:

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**Au101****Member**- Registered: 2010-12-01
- Posts: 351

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,869

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

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**Au101****Member**- Registered: 2010-12-01
- Posts: 351

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 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?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,869

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

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**Au101****Member**- Registered: 2010-12-01
- Posts: 351

Okay, thanks bob bundy 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 It's good for me to understand as much as possible though

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