Results 1 to 15 of 15

Thread: Two quick questions on the 'differential' notation

  1. #1
    Member
    Joined
    Aug 2008
    From
    nowhere
    Posts
    206

    Question Two quick questions on the 'differential' notation

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

    (2) Let's say that $\displaystyle \alpha=f\left(x(t),y(t)\right)$. Please help me write $\displaystyle \frac{d\alpha}{dt}$ with differentials:
    $\displaystyle \frac{d\alpha}{dt}=\frac{d}{dt}f(x(t),y(t))=f'(x(t ),y(t))\cdot\text{?}$
    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
    Joined
    Aug 2008
    From
    nowhere
    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)$\displaystyle \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; Aug 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
    Belgium
    Posts
    1,254
    Thanks
    24

    Re: Two quick questions on the 'differential' notation

    Quote Originally Posted by anonimnystefy View Post
    sure!

    2)$\displaystyle \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
    Joined
    Aug 2008
    From
    nowhere
    Posts
    206

    Re: Two quick questions on the 'differential' notation

    Quote Originally Posted by anonimnystefy View Post
    2)$\displaystyle \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
    Joined
    Aug 2008
    From
    nowhere
    Posts
    206

    Re: Two quick questions on the 'differential' notation

    Quote Originally Posted by anonimnystefy View Post
    2)$\displaystyle \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 $\displaystyle \alpha$ of a parameterized curve $\displaystyle \alpha=\arctan\left(\frac{\dot{y}}{\dot{x}}\right)$ at the point $\displaystyle (x,y)$?

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

    Now, what is $\displaystyle \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}$? The final answer should be $\displaystyle d\alpha=\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\dot{x}^2+\dot{y}^2}dt$, so the $\displaystyle \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}$ apparently equals $\displaystyle \dot{x}\ddot{y}-\ddot{x}\dot{y}$. How do you get this?
    Last edited by courteous; Aug 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
    Belgium
    Posts
    1,254
    Thanks
    24

    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 $\displaystyle \alpha=\arctan\left(\frac{\dot{y}}{\dot{x}}\right)$ at the point $\displaystyle (x,y)$?

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

    Now, what is $\displaystyle \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}$? The final answer should be $\displaystyle d\alpha=\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\dot{x}^2+\dot{y}^2}dt$, so the $\displaystyle \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}$ apparently equals $\displaystyle \dot{x}\ddot{y}-\ddot{x}\dot{y}$. How do you get this?
    There's something strange I think, if you want to calculate:
    $\displaystyle \frac{d}{dt}\left(\alpha\right)=\frac{d}{dt}\left[\arctan\left(\frac{y}{x}\right)\right]=0$
    Because there's no variable $\displaystyle t$ in the function, that means $\displaystyle \alpha$ 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
    Joined
    Aug 2008
    From
    nowhere
    Posts
    206

    Re: Two quick questions on the 'differential' notation

    $\displaystyle x$ and $\displaystyle y$ are both functions of the parameter $\displaystyle t$. Therefore, $\displaystyle \dot{x}=\frac{dx}{dt}$. 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
    Belgium
    Posts
    1,254
    Thanks
    24

    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
    Joined
    Aug 2008
    From
    nowhere
    Posts
    206

    Re: Two quick questions on the 'differential' notation

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

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

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

    PS. Also, when $\displaystyle x$ and $\displaystyle y$ are functions of $\displaystyle t$, does there always exist $\displaystyle y(x)$?
    Last edited by courteous; Aug 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)$\displaystyle \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 $\displaystyle f'(a)$ mean ? If it's the differential of $\displaystyle f$ in $\displaystyle a$, then it's a linear application and your equality isn't homogeneous.

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

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

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


    @courteous : can you precise your goal, because your function $\displaystyle \alpha = \arctan \left( \frac {\. x} {\.y} \right)$ is not of the form $\displaystyle t \mapsto f(x(t),y(t))$.
    Last edited by pece; Aug 3rd 2011 at 06:49 AM.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,779
    Thanks
    3028

    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
    $\displaystyle \frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}$
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Aug 2008
    From
    nowhere
    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 $\displaystyle \alpha = \arctan \left( \frac {\. x} {\.y} \right)$ is not of the form $\displaystyle t \mapsto f(x(t),y(t))$.
    Hello pece! I want to differentiate the slope/angle $\displaystyle \alpha$ of a parameterized curve with regards to parameter $\displaystyle t$ ... that is, I'd like to find $\displaystyle \frac{d\alpha}{dt}=\frac{d}{dt}\arctan\Big(\frac{ \dot{y} }{ \dot{x} }\Big)$ at a point $\displaystyle (x(t),y(t))$.

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

    Now, I don't know what to do with the $\displaystyle y''(x)=\frac{d^2y}{dx^2}$ term.
    Last edited by courteous; Aug 3rd 2011 at 11:20 AM.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    Aug 2008
    From
    nowhere
    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 $\displaystyle \frac{d}{dt}\Big(\alpha\Big)=\frac{d}{dt}\Big( \arctan(\frac{ \dot{y} }{ \dot{x} }) \Big)$? The complete question is in previous (#14) post.



    PS. The final answer should be $\displaystyle \frac{d\alpha}{dt}=\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\dot{x}^2+\dot{y}^2}$.
    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: Jul 7th 2011, 12:10 PM
  2. Replies: 1
    Last Post: Jan 23rd 2011, 12:07 AM
  3. Quick Notation Question
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Oct 19th 2010, 10:35 PM
  4. Quick Question about Vector Notation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 22nd 2009, 07:45 AM
  5. Quick question about notation
    Posted in the Statistics Forum
    Replies: 3
    Last Post: Dec 16th 2007, 10:24 PM

Search Tags


/mathhelpforum @mathhelpforum