geometry/animtrace.en&+cmd=help&+special_parm=demo&+demo_num=6#demo6

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geometry/animtrace.en&+cmd=help&+special_parm=demo&+demo_num=6#demo6
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http://www2.scc-fl.edu/lvosbury/images/HeatSeekanim.gif
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.gif
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 Polar
Koordinat Polar
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