Help with 2 higher order derivative questions please

**coldfire** · October 19th 2008, 02:13 PM

Hi, can someone help me with these?

A particle moves along the x-axis, its position at time t is given by x(t) = 9t / 3+8t^2, t>0 where t is seconds and x is meters
The acceleration of the particle equals 0 at time t=_____ and t=_____ seconds

So I found the second derivative to be this big mess:

(-1296t^2 - 6912t^4 - 9216t^6 - 2592t + 18432t^5) / (9 +48t^2 +64t^4)^2
^ I hope this is right, but maybe it's where I went wrong

then I know to set the equation equal to 0, but then I don't know what to do next, any help?

also I need help with this question:

The function y = e^rx satisfies the equation y'' - 7.4y' + 12y = 0 if r=____ or if r=____

no idea what to do here.

thanks in advance!

**skeeter** · October 19th 2008, 03:43 PM

$\frac{d}{dt} \left(\frac{9t}{3+8t^2}\right) =$

$\frac{(3+8t^2)(9) - (9t)(16t)}{(3+8t^2)^2} = \frac{9[(3+8t^2)-16t^2]}{(3+8t^2)^2} = \frac{9(3-8t^2)}{(3+8t^2)^2}$

$\frac{d}{dt} \left[\frac{9(3-8t^2)}{(3+8t^2)^2}\right] =$

$\frac{9[(3+8t^2)^2(-16t) - (3-8t^2) \cdot 32t(3+8t^2)]}{(3+8t^2)^4}$

don't waste your time trying to simplify this mess, just sub in t = 0 ... you get a(0) = 0

$y = e^{rx}$

$y' = re^{rx}$

$y'' = r^2e^{rx}$

$y'' - 7.4y' + 12y = 0$

$r^2e^{rx} - 7.4re^{rx} + 12e^{rx} = e^{rx}(r^2 - 7.4r + 12) = 0$

solve the quadratic factor for r.

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