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Applied Mathematics

  • Thread starter paile
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  • #1
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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 ?
 

Answers and Replies

  • #2
699
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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.
 
  • #3
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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 ?
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).
 
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  • #4
Jmf
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,we can find its velocity v(t) by determining dy/dx of r(t)
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?
 
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  • #5
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it makes clear sense!
 
  • #6
3
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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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