$12.50 Full detailed A+ answer
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- Posted on Nov 18, 2008 at 5:24:45PM
A:
Preview: ... city vector v(t), we take the derivative of r(t) with respect to t, as shown:<br>v(t) <br>= d r(t) / dt<br>= d(2t)/dt i + d(3t^2)/dt j + d[(sin t)e^(-t)]/dt k<br>= 2i + 6t j + (cos t - sin t)e^(-t) k<br><br>To find the acceleration vector a(t), we take the d ...
The full tutorial is about 279 words long .
$7.00 Answer and Explnation
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- Posted on Nov 18, 2008 at 5:49:41PM
A:
Preview: ... 3cost k<br><br>r'xr''= i j k<br> 2cost 0 -3sint<br> -2sint 0 -3cost<br> =oi-(-6(cost)^2-6(sint)^2)j+ok<br> = 6((cost)^2+(sint)^2)j<br> = 6j<br>abs(r'xr'')= sqrt(6^2)<br> = 6<br>abs(r')= sqrt( (2cost)^2+(-3sint)^2 ...
The full tutorial is about 256 words long .
$10.00 detailed solution
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- Posted on Nov. 19, 2008 at 12:58:25AM
A:
Preview: ... is the rate of change of path with respect to time t<br>so d/dt(r(t))=d/dt(2ti+3t^2j+(sint)e^(-t)k)<br>dr/dt=2i+6tj+(coste^(-t)-sint e^(-t))k,which is the velocity<br>rate of change of velocity is the ac ...
The full tutorial is about 115 words long .