The " is not a symbol that LaTeX recognises. More importantly, you should have a space between your \cos and your A, because it reads \cosA all as one string and \cosA is not a command it recognises - that's why you're getting an error. To be honest, you only really needed
h = c\cos A
is it important for you to have quotation marks around your \cos A. If it is you could do that, but I'm not sure I really understand why you would want to.
Also, did you mean to get Delta or were you hoping for Δ (\Delta)?
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.
Sleep well, and thanks for delaying your well-earned sleep for me!
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.
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.
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!
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.
Ooooh I have mostly bad memories of the M25
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:
Yes I puzzled over that. I am sure it is true, because, well I reasoned this out in a few ways I think. First, I thought:
Second I thought:
Which implies, looking at things from the other point of view, that:
And I was about to write that I wasn't sure how to rigorously derive the statement, it just seemed intuitively true to me, but I've now realised that surely my last two lines would do the trick?
God it's been a while, but I can do the trig, I just don't seem to be able to draw this diagram correctly. I've had a number of gos, where I've thought I've had it and yet, I always end up a long way out - an order of magnitude usually.
53. From A, the bearing of X, the foot of a tower, is 041° (N 41° E) and the angle of elevation of Y, the top of the tower, is 5°. The distance AX is 1000 m. From a point B, due E of A, the bearing of X is 313° (N 47° W). A, B and X lie in the same horizontal plane. Calculate
(a) the height of the tower,
(b) the distance BX,
(c) the angle of elevation of Y from B.
I've tried all sorts, I've looked at it every which way I know how, I was sure I'd be able to figure this one out, but - finally - I decided it was time to move on. So, what am I overlooking here?
...And I run out of steam. Anything else I try to do from there always leads me back to a restatement of something I've already worked out. So I thought maybe it's not about simple vector arithmetic, maybe I have to try and use the fact that it's given me a ratio to work out some other ratios, but I've not been able to make any breakthroughs. At one point I tried constructing a rectangle around the whole shape with vertices O and A as well as one directly above O at the same height as B and D and another at that height and directly above A. Finally I tried resolving b and expressing it as some vertical component x + some fraction of a, but I couldn't make any of these approaches work.
Thanks for all the help so far.
Okay, so - sorry, I've only just got a chance to work on this - I tried applying S1 to an F shape and then to a simple right-angled triangle and, well, all it did was distort my shapes! My nice right-angled triangle has been moved and the angles have changed dramatically I'm not really sure how to proceed and I'm also not really sure what the vector:
looks like, either :S