1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook