Results 1 to 15 of 15

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

  1. #1
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member anonimnystefy's Avatar
    Joined
    Jul 2011
    Posts
    157
    Thanks
    3

    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))
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    Re: Two quick questions on the 'differential' notation

    Thank you! Would you also know the answer to part (2)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member anonimnystefy's Avatar
    Joined
    Jul 2011
    Posts
    157
    Thanks
    3

    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    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]
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    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? ...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Two quick questions on the 'differential' notation

    Ok, Sorry, forget my answer, I was confused.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    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.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Jul 2011
    Posts
    27

    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.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,724
    Thanks
    1476

    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}
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    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.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    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}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A Quick Question on Valid Set Builder Notation
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: July 7th 2011, 12:10 PM
  2. Replies: 1
    Last Post: January 23rd 2011, 12:07 AM
  3. Quick Notation Question
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: October 19th 2010, 10:35 PM
  4. Quick Question about Vector Notation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 22nd 2009, 07:45 AM
  5. Quick question about notation
    Posted in the Statistics Forum
    Replies: 3
    Last Post: December 16th 2007, 10:24 PM

Search Tags


/mathhelpforum @mathhelpforum