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B Two ways of calculating speed?

  1. Feb 7, 2017 #1
    I know that the speed of an object can be calculated the following way:
    [tex] v = \frac {ds}{dt} = \sqrt{(v_x)^2 + (v_y)^2} [/tex]

    where ##s## is the position and ##v_x = \frac {dx}{dt}## and ditto for ##v_y##
    Based on its definition, one should be able to obtain the same result by calculating ##s = \sqrt{x^2 + y^2}## and then differentiating that. I squared s for simplicity and obtained the following:

    [tex] v = \frac {xv_x + yv_y}{\sqrt {x^2 + y^2}} [/tex]

    Or squaring again:

    [tex] v^2 = \frac {x^2v_x^2 + y^2v_y^2 + 2xyv_xv_y} {x^2 + y^2} [/tex]

    This expression should reduce to:

    [tex] v_x^2 + v_y^2 [/tex]

    But no matter what I try I can't simplify it to that.

    Furthermore, I tried this with an actual function: ##x = t^2## and ##y = t##. Their derivatives are ##2t## and ##1## respectively. Squaring those and adding them gives me ##\sqrt {4t^2 + 1} = v##. Trying the other way, ##s = t \sqrt {t^2 + 1}##, the derivative of which is ## \frac {2t^2 + 1}{\sqrt {t^2 + 1}}##. Squaring this gives me

    [tex] \frac {4t^4 + 4t^2 + 1}{t^2 + 1} [/tex]

    Now I can factor out ##4t^2## from the first two terms of the numerator, which leaves me with:

    [tex] \frac {4t^2(t^2 + 1) + 1}{t^2 + 1} [/tex]

    And I cannot reduce this to ##4t^2 + 1##

    Any help with this?

    Thank you
  2. jcsd
  3. Feb 7, 2017 #2
    Your expression for s is the expression for the distance of a point x,y from the origin. Taking the derivative of that will tell you how fast that distance changes with time. If the object moves in a circle of fixed radius about the origin, will this quantity describe what you think of as speed? Velocity is in general a vector quantity. You will want to take a time derivative of the vector (x,y) if you want a velocity. Speed is the magnitude of the velocity vector, which is what you wrote -- i.e. the square root of the sum of the squared components of the velocity.
  4. Feb 8, 2017 #3
    I got it.
    If I understood you well, I can't just differentiate s like that because then I just end up calculating the distance in a straight line from where I am, which is not descriptive of the path at all and is not the instant velocity. I have to take the limit in which both x and y are very small, and then calculate the very small change in s, which is:

    [tex] ds = \sqrt{(dx)^2 + (dy)^2} [/tex]

    Then divide by dt, to obtain the magnitude of the instant velocity. Thank you for your help
  5. Feb 8, 2017 #4
    Maybe you got it, but I just want to make clear: It's not the limit of the distance from the origin when x and y are small. The point is to take the limit of the distance between two points on the trajectory when the interval between them is small. If the path is given by x(t), y(t), then you can make a vector point to a point on the path at time t

    $$r = x(t)\hat{x}+y(t)\hat{y} $$

    The instantaneous velocity is the derivative of that "position vector":

    $$v = \dot x(t) \hat x + \dot y(t) \hat y$$

    Note that the velocity is a vector. The "speed" is a measure of how fast the thing is going along its trajectory in whatever direction it happens to be going. That is to say, the magnitude of the velocity vector at a given time, which is the scalar I think you are looking for.

    So to be precise you don't calculate a small change in s, as you've defined it, because that tells you how the distance from an arbitrary origin is changing -- which may be zero even for nonzero speed, as in the case of circular motion. You rather calculate the small change in in a quantity, my r for example, that can represent the distance between two points on the trajectory.
  6. Feb 8, 2017 #5
    Thank you for clarifying. This is what I meant by the small change in s. It's a bit like calculating how a quantity changes across the arc length of a circle, with directions, which is more or less what you do when you calculate ## \frac { d\vec r}{dt} ##
  7. Feb 8, 2017 #6
    It could be, but the vector r need not lie on a circle.
  8. Feb 8, 2017 #7
    Yes of course. That was just an example.
  9. Feb 8, 2017 #8


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    Science Advisor

    The mnemonics that I use for this sort of situation are, for instance:

    "The magnitude of the sum is not the sum of the magnitudes".
    "The mass of the sum is not the sum of the masses".
    "The square of the sum is not the sum of the squares".
  10. Feb 8, 2017 #9
    It is in classical physics right?
  11. Feb 8, 2017 #10


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    Science Advisor

    Yes. Mass is additive in classical physics.
  12. Feb 8, 2017 #11
    Right. Thank you
  13. Feb 8, 2017 #12


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    Homework Helper
    Gold Member

    Speed is defined as the magnitude of the derivative of the position vector. What you doing is calculating instead the derivative of the magnitude of the position vector which is not equal to speed because the derivative operator and the magnitude operator do not commute. It is
    ##|\frac{d\vec{r}}{dt}|\neq \frac{d|\vec{r}|}{dt}##
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