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Thread: I do not understand Leibniz notation

  1. #1
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    I do not understand Leibniz notation

    So I have started a 3rd calculus class and my prof uses primarily leibniz notation. (Ex. $\displaystyle \frac{dy}{dx}$) and I don't really understand how it all works.

    Like I get that when you write dy/dx it's basically the difference between x and y as that difference gets arbitrarily small.

    I also understand that in an integral, the dx stands for a tiny verticle sliver and the integral means we are adding up an infinite number of those slivers.

    I get lost after that. For example, how does the chain rule use this notation? I've heard there are benefits to it in explaining the chain rule, but I don't quite follow it. An example/explanation would be great.

    Finally I am given equations such as:
    $\displaystyle \int_a^b \sqrt{(dx)^2 + (dy)^2} dx $
    for the arclength of a function and:

    $\displaystyle \int_{t_1}^{t_2} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$
    for the arclenth of a function given using a parametric equation.

    So what do these mean? In the first one, what would I replace dx and dy with. How about in the second one, what is (dx/dt)?

    Thanks in advance,
    -Keilan
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Keilan View Post
    So I have started a 3rd calculus class and my prof uses primarily leibniz notation. (Ex. $\displaystyle \frac{dy}{dx}$) and I don't really understand how it all works.

    Like I get that when you write dy/dx it's basically the difference between x and y as that difference gets arbitrarily small.

    I also understand that in an integral, the dx stands for a tiny verticle sliver and the integral means we are adding up an infinite number of those slivers.

    I get lost after that. For example, how does the chain rule use this notation? I've heard there are benefits to it in explaining the chain rule, but I don't quite follow it. An example/explanation would be great.

    Finally I am given equations such as:
    $\displaystyle \int_a^b \sqrt{(dx)^2 + (dy)^2} dx $
    for the arclength of a function and:

    $\displaystyle \int_{t_1}^{t_2} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$
    for the arclenth of a function given using a parametric equation.

    So what do these mean? In the first one, what would I replace dx and dy with. How about in the second one, what is (dx/dt)?

    Thanks in advance,
    -Keilan
    Just think of $\displaystyle dx$ and $\displaystyle dy$ as very small quantities, but still numbers. (Don't just treat them like symbols.) For the arc length one, you might see it as $\displaystyle \int_a^b\sqrt{(dx)^2+(dy)^2}$, but watch how it transforms to something more understandable.

    $\displaystyle \int_a^b\sqrt{(dx)^2+(dy)^2}=\int_a^b\sqrt{(dx)^2\ left(1+\frac{(dy)^2}{(dx)^2}\right)}$ $\displaystyle =\int_a^b dx\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \int_a^b\sqrt{1+[f'(x)]^2}\,dx$

    which is much easier to work with. The original formula comes from the Pythagorean Theorem, actually. The idea is that if you zoom in really close on any curve, a tiny part of it will look like a straight line, whose distance is given by $\displaystyle \sqrt{(\Delta x)^2+(\Delta y)^2}$.

    where the small change in $\displaystyle x$ is $\displaystyle \Delta x$ and the small change in $\displaystyle y$ is $\displaystyle \Delta y$. To find the total length of the curve, you just add up all those little parts:

    $\displaystyle \sum_{1}^{n}\sqrt{(\Delta x)^2+(\Delta y)^2}$

    Letting $\displaystyle n\to\infty$ changes this into an integral (recall Riemann sums).

    $\displaystyle \int_a^b\sqrt{(dx)^2+(dy)^2}$
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  3. #3
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    Alright, just some brief followup questions here.

    For an integral where I have (dx) or (dy) on their own, am I required to transform them into $\displaystyle \frac{dy}{dx}$ in order to be able to use the formula? Or would there be some way to use those numbers without putting them into that familiar form?

    As for this one:
    $\displaystyle
    $
    $\displaystyle \int_{t_1}^{t_2} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}
    $

    What is
    $\displaystyle \frac{dx}{dt}$?

    Like say for example we were given
    $\displaystyle x(t) = 2t^2 + 5t$.

    Would I just replace
    $\displaystyle \frac{dx}{dt}$ with $\displaystyle 4t + 5$?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Swampscott, MA
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    Yes.

    If $\displaystyle x(t)=2t^2+5t$ and $\displaystyle y(t)=t^3-x+3$ (for example), then

    $\displaystyle \frac{dx}{dt}=4t+5$ and $\displaystyle \frac{dy}{dt}=3t^2-1$

    Therefore the integral $\displaystyle \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$ becomes:

    $\displaystyle \int_{t_1}^{t_2}\sqrt{(4t+5)^2+(3x^2-1)^2}\,dt$

    From there it's just a regular integral w.r.t $\displaystyle t$.
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