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# Differentiation watch

1. The first part of question is to find dy/dx when given x=t-sin(t) y=1-cos(t)

dy/dx= sin(t)/ (1-cos(t))

Using identities sin(t)=s*sin(t/2)*cos(t/2)
1-cos(t)=2*sin(t/2)*sin(t/2)

therefore dy/dx= cot(t/2)

Second part, reads Find d^2y/dx^2 at t=pi/2
I know that dy/dx(cot(x))=-csc^2(x)

But i think there is something else i have to do which i am missing, any help will be much appreciated, Thank you.
2. (Original post by OmegaKaos)
The first part of question is to find dy/dx when given x=t-sin(t) y=1-cos(t)

dy/dx= sin(t)/ (1-cos(t))

Using identities sin(t)=s*sin(t/2)*cos(t/2)
1-cos(t)=2*sin(t/2)*sin(t/2)

therefore dy/dx= cot(t/2)

Second part, reads Find d^2y/dx^2 at t=pi/2
I know that dy/dx(cot(x))=-csc^2(x)

But i think there is something else i have to do which i am missing, any help will be much appreciated, Thank you.
You need to use the chain rule - write

du/dx = (du/dt)(dt/dx)

and use u = dy/dx in the form you found earlier.
3. (Original post by davros)
You need to use the chain rule - write

du/dx = (du/dt)(dt/dx)

and use u = dy/dx in the form you found earlier.
I am sorry, it may just be me. But i am just not getting that.
u=cot(t/2) du/dt=-(1/2)*cosec^2(t/2)

What function do i use as dt/dx..... does du/dx represent d^2y/dx^2?
Again i know this should be really simple, but it just isn't happening at all for me today. Sorry for being a pain. I really do appreciate the help.
4. (Original post by OmegaKaos)
I am sorry, it may just be me. But i am just not getting that.
u=cot(t/2) du/dt=-(1/2)*cosec^2(t/2)

What function do i use as dt/dx..... does du/dx represent d^2y/dx^2?
Again i know this should be really simple, but it just isn't happening at all for me today. Sorry for being a pain. I really do appreciate the help.
Since you want
and you have as a function of t, you must use imp0licit differentiation
i.e. if
5. (Original post by brianeverit)
Since you want
and you have as a function of t, you must use imp0licit differentiation
i.e. if
so is that d^2y/dx^2=(d/dt(dy/dx)/dx/dt)

can anyone please confirm or deny that this is indeed the correct answer?
6. (Original post by OmegaKaos)
so is that d^2y/dx^2=(d/dt(dy/dx)/dx/dt)

can anyone please confirm or deny that this is indeed the correct answer?
Only done this very very quickly on a piece of rough paper, but assuming your cot(t/2) was correct for dy/dx I get the final answer to be -1.

Happy to be corrected if wrong
7. (Original post by davros)
Only done this very very quickly on a piece of rough paper, but assuming your cot(t/2) was correct for dy/dx I get the final answer to be -1.

Happy to be corrected if wrong
I get either -1/2 or -2 but not -1.

So confused.
8. (Original post by OmegaKaos)
I get either -1/2 or -2 but not -1.

So confused.
So we agree we want [d/dt(dy/dx)] / (dx/dt)

You claim (I haven't checked!) that dy/dx = cot(t/2), so d/dt(dy/dx) = -(1/2)cosec^2(t/2)
and dx/dt = 1 - cos t.

So at t = pi/2, dx/dt = 1 - 0 = 1 and
(-1/2)cosec^2(t/2) = (-1/2)cosec^2(pi/4)
cosec^2(pi/4) = 1/(sin^2(pi/4)) = 1/(1/2) = 2 so (-1/2)cosec^2(pi/4) = (-1/2)(2) = -1

and the final answer = -1/1 = -1

(if I haven't screwed up somewhere!)
9. (Original post by davros)
So we agree we want [d/dt(dy/dx)] / (dx/dt)

You claim (I haven't checked!) that dy/dx = cot(t/2), so d/dt(dy/dx) = -(1/2)cosec^2(t/2)
and dx/dt = 1 - cos t.

So at t = pi/2, dx/dt = 1 - 0 = 1 and
(-1/2)cosec^2(t/2) = (-1/2)cosec^2(pi/4)
cosec^2(pi/4) = 1/(sin^2(pi/4)) = 1/(1/2) = 2 so (-1/2)cosec^2(pi/4) = (-1/2)(2) = -1

and the final answer = -1/1 = -1

(if I haven't screwed up somewhere!)
Oh yeah of course, i forgot about my scale factor 1/2.... thats it, you are spot on mate. Thank you very much foor your help.

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