A. The horizontal component of velocity is
vx = dx/dt = π - 4πsin(4πt + π/2)
vx = π - 4πsin(0 + π/2)
vx = π - 4π(1)
vx = -3π
b. vy = 4πcos(4πt + π/2)
vy = 0
c. m = sin(4πt + π/2) / [πt + cos(4πt + π/2)]
d. m = sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f. t = -0.35
g. To find t, set
vx = π - 4πsin(4πt + π/2) = 0
Then use this to calculate vxmax
h. To determine t, set
vy = 4πcos(4πt + π/2) = 0
Then use this to find vymax
i. s(t) = [x(t)^2 + y(t)^2]^(1/2)
h. s'(t) = d[x(t)^2 + y(t)^2]^(1/2) / dt
k and l. Determine the values for t
d[x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to find both the maximum and minimum speeds.
