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Derivative notation uniformality

  1. Jan 4, 2012 #1
    1. The problem statement, all variables and given/known data

    I have elected to put this in the "math" section because it is primarily a math question however, please read to problem knowing that it came from an engineering dynamics text book.

    Given:

    (as written)

    A particle moving along y= x - (x^2 / 400) where x and y are in ft. If the velocity component in the x direction remains constant at Vx= 2 ft/s , determine the magnitudes of velocity and acceleration when x = 20 ft.


    2. Relevant equations

    V = ds/dt
    A = dv/dt

    3. The attempt at a solution

    Ok, at first I became a little confused so I drew a Cartesian co-ordinate system and plotted the curve Y going up X horizontal. I then realized that its a path equation so it needs to be differentiated.

    The problem becomes I don't really understand the notation

    It should possible differentiate the path saying f(x,y)'= the velocity function

    What I don't get is how you can use Leibniz's notation for this problem to solve it. After being frustrated at the lack of clarity I looked through the book and at the solution and it appears the book's notes on how to solve this problem and the solution uses Newtonian notation.

    I apologize for not knowing how to make the dot character

    Saying that (first line of solution)

    ydot = xdot - (2x (xdot) / 400 )

    The problem is that I tend to drop terms like the underlined xdot above because I am not really sure where it comes from. It seems to me that there is some implicit magic going on. I would have no problem in Leibniz notation and the reason this problem gave me pause was because it didn't appear to be easy to express in Leibinz notation.

    Can the question be properly framed by saying given a curve S(x,y) such that

    S(x,y)= x-(x^2/400)-y

    then saying:

    S(x,y)ds/dt=(x-(x^2/400)-y)ds/dt
    V(x,y) = x ds/dt - (x^2 ds/dt / 400) - y ds/dt ?

    would it be even better to say

    S(x,y)= x(i) -(x^2(i)/400)-y(j) ?

    Is there a guide or reference to newtonian notation (other that wikipedia etc.) that makes it easy to understand?

    Does Newtonian notation in this example treat X and Y as functions themselves and not variables which is why the x dot term exists? And if so would X and Y Always notate a function and directions would be explicitly in i,j,k ? Furthermore why would you treat X and Y as a functions in this equation when it also corresponds directly to the inputs of that function?

    Also is there a guide to uber correct and uniform notation and/or an explanation of the mathematical and physical symbols.

    Thanks so much
     
  2. jcsd
  3. Jan 4, 2012 #2

    ehild

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    The problem says that the particle moves along a path with coordinates related according to y(t)=x(t) - (x(t)^2 / 400) (t is the time).
    The components of velocity are the time derivatives of the coordinates:
    [itex]v_x=\dot x[/itex], [itex] v_y=\dot y[/itex].
    vx is given, vx=2 ft/s, and it is constant.
    You get vy applying the chain rule for (y(x(t)). You did it already and got
    [itex]v_y=\dot x-2x \dot x/400=v_x-2xv_x/400[/itex].

    Substitute x=20 ft and vx=2 ft/s and you get vy at x=20 ft.

    Do you know how to determine the magnitude of velocity or of any vector from the components?

    The acceleration is the time derivative of the velocity. ax=0 as vx =constant. Differentiate [itex]v_y=v_x-2x v_x /400[/itex] to get ay.

    ehild
     
  4. Jan 4, 2012 #3
    Everything makes sense except for the problem I guess. I get the "root sum square" concept etc. but I think what is confusing me is that the problem states that the particle moves along the path but does not explicitly define the fact that there is a t parameter in the equation. You yourself added the y(t) (which makes everything much much clearer and was not in the problem statement). Should I always assume that when given a path there is always a parameter of t? Or is this problem just written poorly?

    thanks
     
  5. Jan 5, 2012 #4

    ehild

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    If the particle moves it does in time and space. There is no motion without passing time. Whenever there is some reference to velocity or speed, time is involved.
    If the path is given as a curve, relation between coordinates, time is not necessarily involved. You can find the distance between two points of the path, or the tangent of the path, without having any object moving along.

    Yes, you can say that the problem was written poorly. But it happens in Physics problems quite often. Read the text carefully, and try to read the mind of the problem writer.

    ehild
     
    Last edited: Jan 5, 2012
  6. Jan 5, 2012 #5
    Thanks for your replies its starting to make sense I guess I am just lost in a sea of math/physics/engineering notation.

    Would this be the correct way of expressing this problem in terms of leibniz notation?

    Its over rigorous but I really want to understand the notation and what meaning it carries.

    My attempt at reconciling the notation is below would it be

    d(y(x))/dt= (x(t) -(x(t)^2/400))dx/dt ?

    then how would you evaluate x(t) dx/dt and the other terms?

    would x(t)dx/dt turn into dx/dt via a chain rule? which could then be expressed as xdot or simply just v?

    then x(t)^2/400 would turn into 2x dx/dt /400 which simplifies to 2x dx/dt

    to give an overall expression of

    d(y(x)/dt = dx/dt -( 2x(dx/dt) /400) ?
     
  7. Jan 5, 2012 #6

    ehild

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    d(y(x))/dt=d( x - (x^2 / 400))/dt

    The derivative of a sum is the sum of the derivatives, d( x - (x^2 / 400))/dt=dx/dt-d(x^2/400)/dt.

    The derivative of x(t)^2/400 is (2x(t)/400) dx/dt, so

    dy/dt=dx/dt-(2x/400) dx/dt.

    Isolate dx/dt:

    so dy/dt=[1-(x/200)]dx/dt.

    You need to calculate dy/dt when dx/dt , the velocity in the x direction is 2 ft/s and x=20 ft.


    ehild
     
  8. Jan 5, 2012 #7
    Ok so what this is saying is that you are differentiating a function y(x) w.r.t to time

    But then it appears that you are treating x as a function of time and not as a variable so why wouldn't

    y(x) = x(t) -(x(t)^2/400)

    d(y(x))/dt= (x(t) -(x(t)^2/400))dx/dt

    be correct?

    Because at least to me d(y(x))/dt=d( x - (x^2 / 400))/dt seems like you are differentiating to some other variable not represented in the equation. Are we not using t as the parameter and therefore what we are differentiating with respect to?
     
  9. Jan 5, 2012 #8

    ehild

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    Why should it be correct?

    x is a function of the time. Otherwise dx/dt has no sense.

    y(t)=x(t)-x2(t)/400 is a composite function dependent on x, which is function of t.

    I think you have learnt about chain rule. What is the derivative of the function f(g(x)) = sin(x2), for example?

    In the problem in your post, t is the independent variable and y=y(x(t)).

    dy/dt=(dy/dx) (dx/dt).

    dy/dx=1-2x/400. So dy/dt=(1-2x/400)(dx/dt).

    ehild
     
    Last edited: Jan 5, 2012
  10. Jan 5, 2012 #9
    ok got ya thanks
     
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