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timeforchg · Feb 15, 2012, 06:45 PM

The position of a particle at time, t, during the time interval t = 0 and t = 1, is given by the co-ordinates

$X = t - \frac{t^{3}}{3}, Y = t^{2}, Z = t + \frac{t^{3}}{3}$

Find the following at time, t:

1) Velocity of the particle
2) speed
3)distance traveled from t=0 to t=1

corrigan · Feb 15, 2012, 10:32 PM

okay, we have

$x(t)=t-\frac{t^3}{3} \Rightarrow \frac{dx}{dt}=1-t^2$

$y(t)=t^2 \Rightarrow \frac{dy}{dt}=2t$

$z(t)=t+\frac{t^3}{3} \Rightarrow \frac{dz}{dt} = 1 +t^2$

1) velocity is just the change in each of the coordinates over the change in time, so we have

$v(t) = \frac{dx}{dt}i+\frac{dy}{dt}j+\frac{dz}{dt}k = (1-t^2)i + 2tk + (1+t^2)j$

2) speed is just the change in distance traveled over the change in time. If you notice it's just the pythagorean theorem with our derivatives.

$s(t)=\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2}$

$= \sqrt{(1-t^2)^2 + (2t)^2 + (1+t^2)^2}$

$= \sqrt{1-2t^2 +t^4 +4t^2 + 1 +2t^2 +t^4}$

$= \sqrt{2+4t^2 +2t^4}$

$= \sqrt{2(1+t^2)^2} = (1+t^2)\sqrt{2}$

3) total distance traveled is just the integration of our speed over the time in question, in this case, from 0 to 1.

$\int_0^1 s(t) dt = \int_0^1 (1+t^2)\sqrt{2} dt$

$= \sqrt{2} \int_0^1 (1+t^2) dt = \sqrt{2}(t+\frac{t^3}{3}) |_0^1$

$= \sqrt{2}(1+\frac{1}{3} - 0) = \frac{4 \sqrt{2}}{3}$

I hope this helps.

timeforchg · Feb 15, 2012, 10:59 PM

Wow. That was fast :)

What if the question asked for I)the acceleration, ii) unit tangent, iii)radius of curvature and principal normal ?

*what do they mean by principal normal?

Thanks appreciate your prompt reply.

corrigan · Feb 16, 2012, 06:06 PM

A few things we will need:

$x(t)=t-\frac{t^3}{3} \Rightarrow \frac{dx}{dt}=1-t^2 \Rightarrow \frac{d^2x}{dt^2}= -2t$

$y(t)=t^2 \Rightarrow \frac{dy}{dt}=2t \Rightarrow \frac{d^2y}{dt^2}= 2$

$z(t)=t+\frac{t^3}{3} \Rightarrow \frac{dz}{dt} = 1 +t^2 \Rightarrow \frac{d^2z}{dt^2}= 2t$

i) acceleration (and velocity for that matter) is a lot like what it would be for single variable calculus, it's just the double derivative of the position function.

$a(t) = \frac{d^2x}{dt^2}i+\frac{d^2y}{dt^2}j+\frac{d^2z}{ dt^2}k = -2ti + 2k + 2tj$

ii) unit tangent. Since velocity is really just speed and a direction, think of the unit tangent as the velocity with the speed part "taken out".

$T(t)= \frac{v(t)}{s(t)}= \frac{\frac{dx}{dt}i+\frac{dy}{dt}j+\frac{dz}{dt}k }{\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2}}$

$=\frac{(1-t^2)i + 2tj + (1+t^2)k }{(1+t^2)\sqrt{2}}$

$=\frac{1}{(1+t^2)\sqrt{2}}((1-t^2)i + 2tj + (1+t^2)k)$

iii) Principal normal. Since the tangent is a line in a three dimensional space, there are an infinite number of lines that are perpendicular to it ( the tangent). The principal normal is the particular line that is perpendicular to the tangent and is pointing towards the inside of the curve.

All it is is the derivative of $T(t)$ , which is

$T^{\prime}(t)= \frac{d}{dt}(\frac{(1-t^2)i + 2tj + (1+t^2)k }{(1+t^2)\sqrt{2}})$

$= \frac{((1-t^2)i + 2tk + (1+t^2)j)(2t\sqrt{2})-(-2ti+jk+2tk)((1+t^2)\sqrt{2})}{((1+t^2)\sqrt{2})^2}$ (quotient rule)

$= \frac{(2t\sqrt{2}-2t^3\sqrt{2}+2t\sqrt{2}+2t^3\sqrt{2})i +(4t^3\sqrt{2}-2\sqrt{2}-2t^2\sqrt{2})j+(2t\sqrt{2}+2t^3\sqrt{2}-2t\sqrt{2}-2t^3\sqrt{2})k}{2(1+2t^2+t^4)}$

$= \frac{(4t\sqrt{2})i +(4t^3\sqrt{2}-2t^2\sqrt{2}-2\sqrt{2})j+0k}{2(1+2t^2+t^4)}$

$= \frac{sqrt{2}}{1+2t^2+t^4}(2ti+(2t^3-t^2-1)j +0k)$

Curvature is how much the function curves at a given spot. (i know you didn't ask, but I'm telling you anyway) Curvature is just the magnitude of the principal normal. Which is:

$C(t)=\frac{sqrt{2}}{1+2t^2+t^4}\sqrt{(2t)^2+(2t^3-t^2-1)^2 +0^2}$

And radius of convergence is just the radius of a circle with the same curvature at a given point. That's computed by just taking the reciprocal of the curvature.

I hope this helps :)

timeforchg · Feb 18, 2012, 07:25 AM

Your steps are very clear.. Thanks for explaining! :)