A particle travels along a periodic orbit given as the parametric curve r(t) = isint +jcost(2t) + 2kcos(t)...?

Question

Find the minimum and maximum distance of the particle from the origin.

Answer 1

First, you need the function for the distance from the origin, but you can ignore the square root: d(t) = sin^2(t) + cos^2(2t) + 4cos^2(t).

Now use the first derivative test to find critical values and compare them to see which is farthest and which is nearest.

In other words, solve

2sin(t)cos(t) - 4cos(2t)sin(2t) - 8cos(t)sin(t) = 0

for t. Those are your critical values. Then test them by plugging them back into the distance formula.

Answer 2

A simpler way than finding the derivative is as follows: Let u = (cos t)^2. Recall that (sin t)^2 = 1-u and cos 2t = 2u-1. So D^2 (the square of the distance from the origin) is

1 - u + (2u - 1)^2 + 4u.

This simplifies to 4u^2 - u + 2 which equals 4(u - 1/8)^2 + 31/16, obtained by completing the square.

Now the answer is obvious. It has a minimum of 31/16 when (cos t)^2 = 1/8 and a maximum of 4(1-1/8)^2 +31/16 obtained when (cos t)^2 = 1. [Note that (cos t)^2 varies between 0 and 1.]

Answer 3

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