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Math Help - 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. \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:
    \int_a^b \sqrt{(dx)^2 + (dy)^2} dx
    for the arclength of a function and:

    \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. \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:
    \int_a^b \sqrt{(dx)^2 + (dy)^2} dx
    for the arclength of a function and:

    \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 dx and 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 \int_a^b\sqrt{(dx)^2+(dy)^2}, but watch how it transforms to something more understandable.

    \int_a^b\sqrt{(dx)^2+(dy)^2}=\int_a^b\sqrt{(dx)^2\  left(1+\frac{(dy)^2}{(dx)^2}\right)} =\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 \sqrt{(\Delta x)^2+(\Delta y)^2}.

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

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

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

    \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 \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:
    <br />
\int_{t_1}^{t_2} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} <br />

    What is
    \frac{dx}{dt}?

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

    Would I just replace
    \frac{dx}{dt} with 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 x(t)=2t^2+5t and y(t)=t^3-x+3 (for example), then

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

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

    \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 t.
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