Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

|
Options

Au101
2013-06-07 22:40:03

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

bob bundy
2013-06-07 22:34:51

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

Au101
2013-06-07 21:58:43

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?

bob bundy
2013-06-07 18:19:51

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

Au101
2013-06-07 05:43:02

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

Au101
2013-06-07 00:57:19

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:

anonimnystefy
2013-06-07 00:13:49

No problem.

{7/3}
2013-06-07 00:05:08

Thanks

anonimnystefy
2013-06-06 23:24:34

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

{7/3}
2013-06-06 21:35:55

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

bob bundy
2013-06-06 20:14:56

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

{7/3}
2013-06-06 17:38:18

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.

bob bundy
2013-06-06 16:47:38

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

Au101
2013-06-06 10:15:30

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?