With the position function r(t)= (-2 cos 3t) i + (sin 3t) j . What is the maximum speed of the object?

Question

possible speeds of the object....

a. 0

b. 3

c. 6

d. 9

e. 12

f. 15

g. 18

or is it none of these?

Answer 1

maximum speed is where ||r'(t)||^2 is maximum, which is where d/dt ||r'(t)||^2 = 0 and d2/dt2 ||r'(t)||^2 < 0.

Since r'(t) = (6 sin 3t)i + (3 cos 3t)j,

||r'(t)||^2 = 36 (sin 3t)^2 + 9 (cos 3t)^2.

Differentiate this with respect to t and set it to zero, and solve for t. Then verify that the second derivative at this t is negative.

d/dt ||r'(t)||^2 = 72 sin 3t * 3 cos 3t - 18 cos 3t * 3 sin 3t
= 81 sin 6t = 0

This is zero when 6t = k*pi, or when t = k*pi/6. Now just take the second derivative, which is 486 cos 3t, and see which value of k will yield a negative value. Whichever value of k does this tells you that t is where speed is maximum.

Answer 2

dr /dt = (- 6sin 3t) i + (3cos 3t) j
|dr/dt| = [36 sin² (3t) + 9 cos² (3t)]^(1/2)
|dr/dt| = [36 sin² (3t) + 9.(1 - sin² (3t)]^(1/2)
|dr/dt| = [ 27 sin² (3t) + 9 ]^(1/2)
MAX value = 36^(1/2) = 6

OPTION c