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Hi, I'm having a little difficulty understanding the presentation of the chain rule in my textbook. Here's what it says:
Now, I'm having a few problems understanding this, my first problem is with the following sentence:
I'm not sure I understand why "there will be a functional relationship between x and y."
Say I try a couple of examples - just in order to see it in practise. Let's say - to take some random examples from the top of my head - we say:
Now, admittedly, I could have made my life easier for myself here, I just wanted a couple of curves, but if I look at these two curves, I really can't see any reason why there must be a relationship between x and y.
So that's my first problem which I was hoping for some help with - I was wondering if somebody might be able to unpack and explain this sentence a little for me.
The second problem I have is I'm afraid I simply can't see where:
comes from. I just don't follow and I was wondering - again - if somebody could just add a little bit more explanation so I can see why:
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hi Au101,
Yes, i'll explain. It's a bit past my bedtime, but I'll do it now if you stay online.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Firstly, the functions of t.
Choose a value of t and work out x and y, using your formulas. Plot the point (x,y)
Now choose another value of t and repeat.
You'll get a series of points on the graph. Those points display that x is a function of y and vice versa.
I'll post this and see if you're still with me.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I'm still with you if you're here, but I wouldn't want you to stay up on my account!
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Oh drat, you're not.
Never mind I'm warmed up to it now so I'll carry on.
delta x etc is used for a small but not yet zero amount in the x direction.
So you can start with delta y over delta x, and divide top and bottom by delta t to get
If you let delta t tend to zero then you'd be dividing by zero so you need the small but not yet zero amounts.
Then, is it valid to let the amounts all tend to zero so you get
The rigorous proof of that step is the one that is left out. You could probably convince yourself it is OK by making up an example, but I suggest easier formulas for x and y in terms of t.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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But for now I shall let t = 1 and 2, if that works nicely?
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but I wouldn't want you to stay up on my account!
I've just driven 70 miles round the M25 so my brain is wide awake, even though I'm theoretically tired having spent the day doing jobs for my Mum. So this is probably a good way to unwind.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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The points, by the way, would then be (13, -5) and (64, 7). I think, sadly, we ended up talking past each other just a little bit, but it doesn't seem to have mattered too much because I got the last message
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but I wouldn't want you to stay up on my account!
I've just driven 70 miles round the M25 so my brain is wide awake, even though I'm theoretically tired having spent the day doing jobs for my Mum. So this is probably a good way to unwind.
Bob
Ooooh I have mostly bad memories of the M25
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Just read your last post. What I had in mind was something like:
x = 2t
y = t^2
Take t = 1 for one point and t = 1.1 for a second.
Compute delta t, delta x and delta y.
Then change t = 1.1 into t = 1.01 so you're letting delta t tend to zero.
It's easy enough to eliminate the t from the above so you have y as a function of x. That way you can calculate lim (delta y / delta x) and compare results.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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So we're saying that we can definitely see that y may be expressed as a function of x, although quite what that function would be I don't know - a function in terms of t, it seems.
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With my simple functions
x = 2t so t = (x/2)
Therefore y = (x/2)^2
It would be somewhat harder with the functions you chose.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I see - I think, but I feel like I'm missing something deeper. It's like I'm almost there with it, but not quite. I can see that where you have the points:
You can say that the gradient (say m) is given by the change in y over the change in x. That is to say:
And that δx and δy must both be functions of t, but I guess I'm not quite sure I see why the gradient of the graph y = φ(x) is equal to gradient of the graph y = g(t) divided by the gradient of the graph x = f(t). It feels kind of right and yet, I don't see it somehow.
Edit: obviously that should really be the gradient of the curves at an equivalent point - it's actually not very easy to express!
Last edited by Au101 (2015-01-24 11:55:02)
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With my simple functions
x = 2t so t = (x/2)
Therefore y = (x/2)^2
It would be somewhat harder with the functions you chose.
Bob
Yes I see that, so we can see that since both x and y are functions of t, there will be some way of relating x and y. They are both expressed in terms of t so what we can do is express t as some new function of x and thereby express y in terms of x (since y is a function of t and we have just expressed t as a function of x, so y may also be expressed as a function of x.) Clearly, in some cases y will have to be expressed as a function of x and t.
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That's why I suggested a simple example first. When I find a bit of maths hard, I look for simple cases first so I can slide into the hard stuff in small steps.
Try it with simple functions and just two points, letting the second move gradually closer to the first. If you set up the formulas on a spreadsheet it will be easy to change 1.1 into 1.01 into 1.001 ...........
Then change the fixed pint t=1 into t=2 and repeat. Or change the formulas into harder ones.
Think I'd better sign off now and wash away the literal garden mud and metaphorical middle lane hogging driver anguish. Good night. See you tomorrow.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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What happens if neither f(t) nor g(t) is invertible, though? Surely then we could not express t as a function of x or as a function of y and then we could not express y as a function of x by this method?
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That's why I suggested a simple example first. When I find a bit of maths hard, I look for simple cases first so I can slide into the hard stuff in small steps.
Try it with simple functions and just two points, letting the second move gradually closer to the first. If you set up the formulas on a spreadsheet it will be easy to change 1.1 into 1.01 into 1.001 ...........
Then change the fixed pint t=1 into t=2 and repeat. Or change the formulas into harder ones.
Think I'd better sign off now and wash away the literal garden mud and metaphorical middle lane hogging driver anguish. Good night. See you tomorrow.
Bob
Sleep well, and thanks for delaying your well-earned sleep for me!
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