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http://wims.unice.fr/wims/wims.cgi?session=PU545FE4E4.6&+lang=en&+module=tool/geometry/animtrace.en&+cmd=help&+special_parm=demo&+demo_num=6#demo6http://wims.unice.fr/wims/wims.cgi?session=PU545FE4E4.6&+lang=en&+module=tool/geometry/animtrace.en&+cmd=help&+special_parm=demo&+demo_num=6#demo6


http://wims.unice.fr/wims/wims.cgi?session=PU545FE4E4.5&+lang=en&+module=tool/geometry/animtrace.en&+cmd=help&+special_parm=demo&+demo_num=8#demo8http://wims.unice.fr/wims/wims.cgi?session=PU545FE4E4.5&+lang=en&+module=tool/geometry/animtrace.en&+cmd=help&+special_parm=demo&+demo_num=8#demo8


http://www2.scc-fl.edu/lvosbury/images/HeatSeekanim.gifhttp://wims.unice.fr/wims/wims.cgi?session=PU545FE4E4.5&+lang=en&+module=tool/geometry/animtrace.en&+cmd=help&+special_parm=demo&+demo_num=8#demo8

From grad T = r'(t) we equate vector components to get dx/dt = -2x and dy/dt = -4y.  Let 0 be the value of t at the starting point so that x(0) = 2 and y(0) = 4. 


http://www2.scc-fl.edu/lvosbury/images/P884No48b.gifhttp://wims.unice.fr/wims/wims.cgi?session=PU545FE4E4.5&+lang=en&+module=tool/geometry/animtrace.en&+cmd=help&+special_parm=demo&+demo_num=8#demo8

From grad T = r'(t) we equate vector components to get 

dx/dt = -3, dy/dt = -1, and dz/dt = -2z.  

Let 0 be the value of t at the starting point so that 

x(0) = 2, y(0) = 2, and z(0) = 5.  

We have three differential equations to solve and their solutions 

will yield x = -3t + 2, y = -t + 2, and z = 5e-2t.

Thus r(t) = < -3t + 2 , -t + 2 , 5e-2t >


Koordinat Polarhttp://wims.unice.fr/wims/wims.cgi?session=PU545FE4E4.5&+lang=en&+module=tool/geometry/animtrace.en&+cmd=help&+special_parm=demo&+demo_num=8#demo8


Koordinat Polarhttp://wims.unice.fr/wims/wims.cgi?session=PU545FE4E4.5&+lang=en&+module=tool/geometry/animtrace.en&+cmd=help&+special_parm=demo&+demo_num=8#demo8


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