An object is moving in the plane according to these parametric equations:
x(t) = πt + cos(4πt + π/2)
y(t) = sin(4πt + π/2)
A portion of the "cycloidal" path (on the time interval [0,1]) is shown in this figure:

At time t=0, the object is located at (0,1). Give EXACT ANSWERS to all questions below; use pi to denote the number π.
(a) The horizontal velocity is .
(b) The vertical velocity is .
(c) The slope of the tangent line to the path at time t is

(d) The equation of the tangent line to the path at time t=1/6 is
y = (x - ) +

(e) The first time the tangent line is vertical will be t=

(f) The second time the tangent line is vertical will be t=

(g) What is the maximum horizontal velocity?
(h) What is the minimum horizontal velocity?
(i) The speed of the object at time t is s(t)=

(j)Compute s'(t)=

(k) The maximum speed is

(l) The minimum speed is

a. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π

b. vy = 4π cos (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. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax

h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for 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. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.

obasola1 obasola1 · Answer 2 · 2025-04-02T03:22:57-04:00

Answer:

.a vx = -3π

b.vy = 0

c.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.f. t = -0.35

g.vx = π - 4πsin (4π(0.124) + π/2)

h.vmax=4π cos (4π(0.045) + π/2)

i.s(t) = [x(t)^2 + y(t)^2]^(1/2)

s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt

Explanation:

x(t) = πt + cos(4πt + π/2)

differentiating te orizontal distance wit respect to time t, will give horizontal velocity

change in displacement per change in time is velocity

vx = dx/dt = π - 4πsin (4πt + π/2)

vx = π - 4π sin (0 + π/2)

at t =0, substituting te value of t into the above

vx = π - 4π (1)

vx = -3π

b,

y(t) = sin(4πt + π/2)

differentiate wit respect to t

dy/dt=4π cos (4πt + π/2)

π/2=90

when t=0

b. vy=dy/dt = 4π cos (4πt + π/2)

vy = 0

c. slope of te tanent line

y(t)/x(t)

c. m = sin(4πt + π/2) / [πt + cos(4πt + π/2)]

d. at t=1/6, we substitute into answer gotten in c

m = sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]

e. t = -1.0

f. t = -0.35

g. Solve for t

vx = π - 4πsin (4πt + π/2) = 0

at maximum v=0

(4πt + π/2)=0.0043

t=-π/2+0.0043/(4π)

t=-0.124

vx = π - 4πsin (4π(0.124) + π/2)

h. Solve for t

vy = 4π cos (4πt + π/2) = 0

(4πt + π/2=1

t=0.57/4π

t=0.045

vmax=4π cos (4π(0.045) + π/2)

i. resultant of te displacement

s(t) = [x(t)^2 + y(t)^2]^(1/2)

h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt

s'(t)=

k and l. Solve for the values of t

d [x(t)^2 + y(t)^2]^(1/2) / dt = 0

And substitute to determine the maximum and minimum speeds.