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Math Help - confusion with differentiation

  1. #1
    j55
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    confusion with differentiation

    hi, im going over differentiation atm but reading from a book, and it states that v=ds/dt

    now could someone explain what d stands for? is it delta or what :S



    because theres another equation a=dv/dt = d(squared)s/dt(squared) so does d stand for something that represents a single value if its being squared or what?

    thanks in advance.
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by j55 View Post
    hi, im going over differentiation atm but reading from a book, and it states that v=ds/dt

    now could someone explain what d stands for? is it delta or what :S



    because theres another equation a=dv/dt = d(squared)s/dt(squared) so does d stand for something that represents a single value if its being squared or what?

    thanks in advance.
    \frac{dy}{dx} is a different way of saying f'(x). It means the derivative of y=f(x) with respect to x.

    f''(x)= \frac{d^2y}{dx^2} means the second derivative of y=f(x) with respect to x.
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    Junior Member eddie2042's Avatar
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    Quote Originally Posted by j55 View Post
    hi, im going over differentiation atm but reading from a book, and it states that v=ds/dt

    now could someone explain what d stands for? is it delta or what :S



    because theres another equation a=dv/dt = d(squared)s/dt(squared) so does d stand for something that represents a single value if its being squared or what?

    thanks in advance.
    \frac{d}{dx} is simply Leibniz notation for saying 'The derivative of ___ with respect to x." In your case, the dx is a dt, so you're differentiating with respect to the variable 't', which stands for time.

    Usually in mathematics and entry-level physics, 'S' stands for the Position function. It's a mathematical function that describes an object location relative to a fixed point with respect to time (i.e. Time is on the x-axis).

    V=\frac{ds}{dt} simply means that the derivative of the Position function is 'V', the Velocity function. The rate of change of the position function, with respect to time, will tell us it's velocity. Easy to understand right?

    A=\frac{dV}{dt} = \frac{d^2S}{dt^2} simply means that Acceleration is the first derivative of the Velocity function and/or the second derivative of the Position function. Why is there a squared over the d and t in the second derivative, you ask? That simply means it's the second derivative of the position function.

    \frac{d}{dt}\cdot\frac{dS}{dt} = \frac{d^2S}{dt^2}

    Cheers!
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    In plain English...

    Here is a quote from Calculus Made Easy by Silvanus P. Thompson & Marvin Gardner.

    " d merely means "a little bit of." Thus dx means a little bit of x;or du means a little bit of u. Ordinary mathematicians think it more polite to say "an element of," instead

    of "a little bit of". Just as you please. But you will find that these little bits (or elements) may be considered infinitely small."
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    Super Member Anonymous1's Avatar
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    Quote Originally Posted by dbakeg00 View Post
    In plain English...

    Here is a quote from Calculus Made Easy by Silvanus P. Thompson & Marvin Gardner.

    " d merely means "a little bit of." Thus dx means a little bit of x;or du means a little bit of u. Ordinary mathematicians think it more polite to say "an element of," instead

    of "a little bit of". Just as you please. But you will find that these little bits (or elements) may be considered infinitely small."
    When you said "in plain english" I though it wasn't going to be confusing.

    I'm gonna go ahead and not read that book.
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  6. #6
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by Anonymous1 View Post
    When you said "in plain english" I though it wasn't going to be confusing.

    I'm gonna go ahead and not read that book.
    LOOOL exactly what i though.

    confusing as hell...

    maybe calculus isn't meant to be understood?
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    Super Member Anonymous1's Avatar
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    Quote Originally Posted by eddie2042 View Post
    LOOOL exactly what i though.

    confusing as hell...

    maybe calculus isn't meant to be understood?
    Calculus is actually rather intuitive. Like everything else it just takes a little practice. The derivative is related to the slope of a function, but it is a rate of change.

    Stop for a second.

    Think about what "rate of change" means just with your knowledge of the english language, post what you think, and we will work through it together.

    []EDIT[] Sorry this was aimed at j55
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    I guess we all learn in different ways...that little paragraph was pretty clear to me
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  9. #9
    j55
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    re

    im getting to grips with it, but naturally its still slightly confusing, especially when you mentioned about it being a little bit of, that makes me think of it as a little bit of this divided by a little bit of that, but im assuming thats not the case and theres no actual division included? its just a representation, am i correct?

    and as to your post anonayous1, rate of change simply means how fast or slow something is changing, or is it alot more complicated.

    thanks for your posts though, great help!
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  10. #10
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by j55 View Post
    a little bit of this divided by a little bit of that
    Haha!

    Quote Originally Posted by j55 View Post
    and as to your post anonayous1, rate of change simply means how fast or slow something is changing, or is it alot more complicated.
    Okay, so we are changing at some rate. What does this mean?

    Say we have a function f(x), the derivate of function f'(x), describes the slope at a point. This enables us to "move along the function" f(x) because we can map out its slope as its inputs change

    f(x)
    |
    |
    | /*
    |/__________x (Sorry when I extend the function further it comes out jumbled. You should draw it out)
    0 1..

    This is some function f(x). say we take the derivative and obtain f'(x). Then f'(1) describes the slope at *. Or, the rate f(x) is changing when x=1.

    So we are answering the question: at what rate does f(x) change when its inputs change?

    Is this clear?
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  11. #11
    j55
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    re

    why the 'haha!' ?

    and yeah it does make sense but i am still reading up on it when i can. The 1 above the f is that the same as a suffix? to represent the first value.

    also i dont see how the actual question is answered. how do you show the rate at which f(x) changes?

    and finally ' we are changing at some rate. What does this mean? ' i dont know what this means
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    Super Member Anonymous1's Avatar
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    Quote Originally Posted by j55 View Post
    why the 'haha!' ?
    Because I thought \frac{little-bit-of-this}{little-bit-of-that} was funny of course!

    Quote Originally Posted by j55 View Post
    and yeah it does make sense but i am still reading up on it when i can. The 1 above the f is that the same as a suffix? to represent the first value.
    The 1 is the point x=1 on the x-axis, the * is the the location of the function when x=1, (1,f(1)). You know like (x,y)...

    Quote Originally Posted by j55 View Post
    also i dont see how the actual question is answered. how do you show the rate at which f(x) changes?
    For nonlinear functions the rate at which f(x) changes depends on the point you are referring to. Mapping these instantaneous rates of change is what allows us to move along the function. The derivative is the tool that accomplishes this. For instance f'(1)= (a number) is the rate at which we are changing on the function at (1,f(1)).

    Quote Originally Posted by j55 View Post
    and finally ' we are changing at some rate. What does this mean? ' i dont know what this means
    For instance f'(1)= (a number) is the rate at which we are changing on the function at (1,f(1)).
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  13. #13
    j55
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    re

    right its really coming together now, might sound naive but the ds/dt isnt an actual equation but a notation showing derivative of what with respect to what? :S .
    Last edited by j55; April 8th 2010 at 05:41 AM.
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  14. #14
    j55
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    re

    anyone?
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