An object is moving in the plane according to these parametric equations: x ( t

Question

An object is moving in the plane according to these parametric equations:
x(t) = ?t + cos(4?t + ?/2)
y(t) = sin(4?t + ?/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 ?.

An object is moving in the plane according to these parametric equations: x(t) = ?t + cos(4?t + ?/2) y(t) = sin(4?t + ?/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 ?. The horizontal velocity is The vertical velocity is The slope of the tangent line to the path at time t is The equation of the tangent line to the path at time t=1/6 is The first time the tangent line is vertical will be t= The second time the tangent line is vertical will be t= What is the maximum horizontal velocity? What is the minimum horizontal velocity? The speed of the object at time t is s(t)= Compute s'(t)= The maximum speed is The minimum speed is

Explanation / Answer

x(t) = pi*t - sin( 4*pi*t )

y(t) = cos(4*pi*t)

a) horizontal velocity = x'(t) = pi - 4*pi cos(4*pi*t); x'(0) = -3*pi

b) Vertical velocity = y'(t) = -4*pi*sin(4*pi*t); y'(0) = 0

c) Slope= dy/dx = y'(t)/x'(t) = -4sin(4*pi*t)/( 1 - 4cos(4*pi*t))

d) y - y1 = m( x - x1)

m = -2/sqrt(3) = -1.155

y(1) = y(1/6) = -0.5

x(1) = x(1/6) = -0.342

so equation becomes: y+0.5 = -1.155 (x + 0.342)

e)Tangent line is vertical at x'(t) = 0

so, cos(4*pi*t) = 1/4

so, t = arccos(1/4)/(4*pi) = 0.1049 sec

f) t = (2*pi - arccos(1/4)) /(4*pi) = 0.3951 sec

g) max horizontal velocity when cosine term is -1

so max horizontal velocity = 5*pi

h) Minimum horizontal velocity = -3*pi

i) s(t) = sqrt( (x'(t))^2 + (y'(t))^2 )

s(t) = pi*sqrt(17 - 8 cos(4*pi*t))

j) s'(t) = 16*pi^2*sin(4*pi*t) / sqrt(17 - 8 cos(4*pi*t))

k) maximum speed is at s'(t) = 0

so sin(4*pi*t) = 0, so t = 0, 0.25 sec

so, maximum speed = 5*pi at t = 0.25 sec

l) minimum speed = 3*pi at t = 0

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