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Math Help - Two quick questions on the 'differential' notation

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    Question Two quick questions on the 'differential' notation

    (1) Why is it OK to change the integrand's step like this: \int \frac{dx}{x\log x}=\int\frac{d(\log x)}{\log x}=\log\log x + C
    What is the (geometrical or some other) intuition behind this?
    This is from a (somewhat) rigorous analysis book ( \epsilon- \delta proofs galore), and such integration-step-substitutions were being made without explanation.

    (2) Let's say that \alpha=f\left(x(t),y(t)\right). Please help me write \frac{d\alpha}{dt} with differentials:
    \frac{d\alpha}{dt}=\frac{d}{dt}f(x(t),y(t))=f'(x(t  ),y(t))\cdot\text{?}
    How do you continue?
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    Re: Two quick questions on the 'differential' notation

    1) i don't think it is intuition.it is just integration by substitution.

    if you say that some t=log(x)
    then dt/dx=1/x

    or dt=dx/x

    d(\log x)=\frac{dx}{x}

    now you can substitute dx/x with d(log(x))
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    Re: Two quick questions on the 'differential' notation

    Thank you! Would you also know the answer to part (2)?
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    Re: Two quick questions on the 'differential' notation

    sure!

    2)  \frac{d}{dt}f(x(t),y(t))=x'(t) \cdot y'(t) \cdot f'(x(t),y(t))

    You get that from the chain rule. Chain rule - Wikipedia, the free encyclopedia

    edit: i fixed the whole thing right up.
    Last edited by anonimnystefy; August 2nd 2011 at 01:51 PM.
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    MHF Contributor Siron's Avatar
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    Re: Two quick questions on the 'differential' notation

    Quote Originally Posted by anonimnystefy View Post
    sure!

    2)  \frac{d}{dt}f(x(t),y(t))=x'(t) \cdot y'(t) \cdot f'(x(t),y(t))

    You get that from the chain rule. Chain rule - Wikipedia, the free encyclopedia

    my latex isn't working.
    You have to use [ tex ] ... [ /tex ], not [ math] ... [ /math]
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    Re: Two quick questions on the 'differential' notation

    Quote Originally Posted by anonimnystefy View Post
    2)  \frac{d}{dt}f(x(t),y(t))=x'(t) \cdot y'(t) \cdot f'(x(t),y(t))
    Only now did I remember to quote and change the tags to [ TEX ] myself.
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    Re: Two quick questions on the 'differential' notation

    Quote Originally Posted by anonimnystefy View Post
    2)  \frac{d}{dt}f(x(t),y(t))=x'(t) \cdot y'(t) \cdot f'(x(t),y(t))
    How would one apply this to, say, the slope \alpha of a parameterized curve \alpha=\arctan\left(\frac{\dot{y}}{\dot{x}}\right) at the point (x,y)?

    \frac{d}{dt} \Big{(}\alpha\Big{)} = \frac{d}{dt} \left(\arctan(\frac{ \dot{y} }{ \dot{x} })\right)=\frac{1}{\dot{x}^2+\dot{y}^2} \cdot \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)} = \hdots

    Now, what is \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}? The final answer should be d\alpha=\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\dot{x}^2+\dot{y}^2}dt, so the \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)} apparently equals \dot{x}\ddot{y}-\ddot{x}\dot{y}. How do you get this?
    Last edited by courteous; August 3rd 2011 at 06:33 AM. Reason: added emphasis to "slope of a parameterized curve", the germ of my question
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    MHF Contributor Siron's Avatar
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    Re: Two quick questions on the 'differential' notation

    Quote Originally Posted by courteous View Post
    How would one apply this to, say, the slope of a parameterized curve \alpha=\arctan\left(\frac{\dot{y}}{\dot{x}}\right) at the point (x,y)?

    \frac{d}{dt} \Big{(}\alpha\Big{)} = \frac{d}{dt} \left(\arctan(\frac{ \dot{y} }{ \dot{x} })\right)=\frac{1}{\dot{x}^2+\dot{y}^2} \cdot \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)} = \hdots

    Now, what is \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}? The final answer should be d\alpha=\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\dot{x}^2+\dot{y}^2}dt, so the \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)} apparently equals \dot{x}\ddot{y}-\ddot{x}\dot{y}. How do you get this?
    There's something strange I think, if you want to calculate:
    \frac{d}{dt}\left(\alpha\right)=\frac{d}{dt}\left[\arctan\left(\frac{y}{x}\right)\right]=0
    Because there's no variable t in the function, that means \alpha is a constant.

    My other question, I'm not famimiar with the dots on the letters. What do they mean? ...
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    Re: Two quick questions on the 'differential' notation

    x and y are both functions of the parameter t. Therefore, \dot{x}=\frac{dx}{dt}. Oh, and the dot is Newton's way of writing the derivative.
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    Re: Two quick questions on the 'differential' notation

    Ok, Sorry, forget my answer, I was confused.
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    Re: Two quick questions on the 'differential' notation

    Got slightly further: since \frac{\dot{y}}{\dot{x}}=\frac{dy}{dx}, we can rewrite \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big)= \frac{d}{dt}\Big(\frac{dy}{dx}\Big)=\frac{d}{dx} \Big(\frac{dy}{dx}\Big)\cdot\frac{dx}{dt}=y''(x) \cdot \dot{x}

    To put everything in one place:
    \frac{d}{dt} \left(\arctan(\frac{ \dot{y} }{ \dot{x} })\right)=\frac{1}{\dot{x}^2+\dot{y}^2} \cdot \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)} = \frac{1}{\dot{x}^2+\dot{y}^2} \cdot y''(x) \cdot \dot{x} = \hdots

    Is this going in the right direction? If so, then what is y''(x) if you have x(t) and y(t)?

    PS. Also, when x and y are functions of t, does there always exist y(x)?
    Last edited by courteous; August 3rd 2011 at 03:58 AM.
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    Re: Two quick questions on the 'differential' notation

    Quote Originally Posted by anonimnystefy View Post
    sure!

    2)  \frac{d}{dt}f(x(t),y(t))=x'(t) \cdot y'(t) \cdot f'(x(t),y(t))
    I might mistake, but it doesn't seem right to me. What does f'(a) mean ? If it's the differential of f in a, then it's a linear application and your equality isn't homogeneous.

    \alpha' (t) = \mathrm d_{(x(t),y(t))}f(x'(t),y'(t)) by a simple composition of f and t \mapsto (x(t),y(t)).

    If f : \mathbb R^2 \rightarrow \mathbb R, then we express it like that :

    \alpha' (t) = x'(t)\frac {\partial f}{\partial x}(x(t),y(t)) + y'(t)\frac{\partial f}{\partial y}(x(t),y(t))


    @courteous : can you precise your goal, because your function \alpha = \arctan \left( \frac {\. x} {\.y} \right) is not of the form t \mapsto f(x(t),y(t)).
    Last edited by pece; August 3rd 2011 at 06:49 AM.
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    Re: Two quick questions on the 'differential' notation

    Yes, what anonimnystefy wrote was wrong. If x and y are both functions of t and f is a function of x and y then
    \frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}
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    Re: Two quick questions on the 'differential' notation

    Quote Originally Posted by pece View Post
    @courteous : can you precise your goal, because your function \alpha = \arctan \left( \frac {\. x} {\.y} \right) is not of the form t \mapsto f(x(t),y(t)).
    Hello pece! I want to differentiate the slope/angle \alpha of a parameterized curve with regards to parameter t ... that is, I'd like to find \frac{d\alpha}{dt}=\frac{d}{dt}\arctan\Big(\frac{ \dot{y} }{ \dot{x} }\Big) at a point (x(t),y(t)).

    EDIT: And so that everything is summed-up in this one post, this is what I've figured out so far (not necessarily correct though):
    \frac{d}{dt} \left(\arctan(\frac{ \dot{y} }{ \dot{x} })\right)=\frac{1}{\dot{x}^2+\dot{y}^2} \cdot \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)} = \frac{1}{\dot{x}^2+\dot{y}^2} \cdot y''(x) \cdot \dot{x} = \hdots

    Now, I don't know what to do with the y''(x)=\frac{d^2y}{dx^2} term.
    Last edited by courteous; August 3rd 2011 at 11:20 AM.
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    Re: Two quick questions on the 'differential' notation

    I highly doubt that no one knows the answer ... HallsofIvy certainly knows it. Please, help ... what is \frac{d}{dt}\Big(\alpha\Big)=\frac{d}{dt}\Big( \arctan(\frac{ \dot{y} }{ \dot{x} }) \Big)? The complete question is in previous (#14) post.



    PS. The final answer should be \frac{d\alpha}{dt}=\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\dot{x}^2+\dot{y}^2}.
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