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Applied Mathematics

  1. Mar 20, 2010 #1
    Given that a particle P in space would have position vector r(t)= xi + yj + zk,we can find its velocity v(t) by determining dy/dx of r(t) and even go further to computing the speed v given by the magnitude of v(t).

    With that being the case. If the magnitude of the speed v is defined, what would its magnitude be, and what would we call it ?
     
  2. jcsd
  3. Mar 20, 2010 #2
    v(t)=r'(t)=i+j+k so magnitude is square root of 3. You would call it what ever the units are. Kilometers per min, miles per hour, etc.
     
  4. Mar 20, 2010 #3
    I think you have to know the values of x, y and z as functions of time to determine the
    speed vectors sx, sy and sz. Then the speed of the particle is given by

    [tex]Sp\ =\ \sqrt{sx^{2}(t)+sy^{2}(t)+sz^{2}(t)}[/tex]

    Measure its magnitude with the Physical property (example: meter/second).
     
    Last edited: Mar 20, 2010
  5. Mar 20, 2010 #4

    Jmf

    User Avatar

    You don't determine [tex]\frac{dy}{dx}[/tex] - at least, not if y was defined as you orginally put. I.e. the if position is:

    [tex]\textbf{r}(t) = x\textbf{i} + y\textbf{j} + z\textbf{k}}[/tex]

    where x(t), y(t) and z(t) are functions of time, then the velocity is:

    [tex]\textbf{v}(t) = \textbf{r'}(t) = \frac{d}{dt\textbf{r}(t) = \frac{dx}{dt}\textbf{i} + \frac{dy}{dt}\textbf{j} + \frac{dz}{dt}\textbf{k}[/tex]

    The magnitude of this is usually what is termed the speed, the speed will be:

    [tex]|\textbf{v}(t)| = |\textbf{r'}(t)| = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2[/tex]

    If you take the next time-derivative, you get the acceleration:

    [tex]\textbf{a}(t) = \textbf{v'}(t) = \textbf{r''}(t) = \frac{d^2}{dt^2}}\textbf{r}(t) = \frac{d^2x}{dt^2}\textbf{i} + \frac{d^2y}{dt^2}\textbf{j} + \frac{d^2z}{dt^2}\textbf{k}[/tex]

    Does this make things any clearer?
     
    Last edited: Mar 20, 2010
  6. Aug 1, 2010 #5
    it makes clear sense!
     
  7. Apr 9, 2011 #6
    Iterated definite integrals

    I came across this problem, where I have to change a double Cartesian integral into an equivalent double Polar integral and then i am told to evaluate. The region R bounding the double Cartesian integral is R = {(x^2 + y^2): -1 <= x <= 1 and -(1-x^2)^(1/2) <= y <= (1-x^2)^(1/2) }. Then I found that -secz <= r <= secz, z=theta, which are the limits of integration for dx. Question is how do I find the limits of integration for dy? this is where my difficulty lies.
     
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