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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.

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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