The parametric equations for the path of the chalk mark are x ( t ) = 70 ? t + 3

Question

The parametric equations for the path of the chalk mark are
x(t) = 70?t + 35cos(–2?t – ?/2)
y(t) = 35 + 35sin(–2?t – ?/2)
A portion of the path (on the time interval [0,2]) is shown in this figure:

Give EXACT ANSWERS to all questions below; use pi to denote the number ?.

(a) The horizontal velocity is

(b) The vertical velocity is

(c) Compute x'(t) at these times:

t = 0sec

t = 0.25sec

t = 0.5sec

t = 0.75sec

t = 1 sec

(d) Compute y'(t) at these times:

t = 0sec

t = 0.25sec

t = 0.5sec

t = 0.75sec

t = 1 sec

(e) When is the first time the horizontal velocity will be a maximum?

(f) When is the first time the horizontal velocity will be a minimum?

(g) What is the maximum horizontal velocity?

(h) What is the minimum horizontal velocity?

(i) When is the first time the vertical velocity will be a maximum?

(j) When is the first time the vertical velocity be a minimum?

(k) What is the maximum vertical velocity?

(l) What is the minimum vertical velocity?

(m) The speed of the moving chalk mark at time t is defined by the equation
s(t) = ?{(x'(t))2 + (y'(t))2}.

Compute s(t).

(n) During the first 2 seconds, how many times will the speed be equal to 50 cm/sec?

Explanation / Answer

x(t) = 70t + 35cos(–2t – /2)
y(t) = 35 + 35sin(–2t – /2)

(a) The horizontal velocity is

x'(t) = 70 + 35(-sin(–2t – /2))(–2)

x'(t) = 70 + 70(sin(–2t – /2))

x'(t) = 70 + 70(sin(–2t - /2))

(b) The vertical velocity is

y'(t) = 0 + 35(cos(–2t – /2))*(–2)

y'(t) = -70(cos(–2t – /2))

(c) Compute x'(t) at these times:

t = 0sec

x'(0) = 70 + 70(sin(0 - /2)) =0

t = 0.25sec

x'(0.25) = 70 + 70(sin(–2*0.25 - /2)) =70

t = 0.5sec

x'(0.5) = 70 + 70(sin(–2*0.5 - /2)) =140

t = 0.75sec

x'(0.75) = 70 + 70(sin(–2*0.75 - /2)) =70

t = 1 sec

x'(1) = 70 + 70(sin(–2 - /2)) =0

(d) Compute y'(t) at these times:

t = 0sec

y'(0) = -70(cos(0 – /2)) =0

t = 0.25sec

y'(0.25) = -70(cos(–2*0.25 – /2)) =70

t = 0.5sec

y'(0.5) = -70(cos(–2*0.5 – /2)) =0

t = 0.75sec

y'(0.75) = -70(cos(–2*0.75 – /2)) = -70

t = 1 sec

y'(1) = -70(cos(–2 – /2)) =0

(e)first time the horizontal velocity will be a maximum at 0.75sec

(f) first time the horizontal velocity will be a minimum at 0.25sec

(g) maximum horizontal velocity =140

(h) minimum horizontal velocity=0

(i) first time the vertical velocity will be a maximum=0.5sec

(j) first time the vertical velocity be a minimum=0.75

(k) maximum vertical velocity=70

(l) minimum vertical velocity=-70

(m)s(t) = {(x'(t))2 + (y'(t))2}

s(t) =  {(70 + 70(sin(–2t - /2)) )2 + (-70(cos(–2t – /2)) )2}

s(t) = 70 {(1+ (sin(–2t - /2)) )2 + ((cos(–2t – /2)) )2}

s(t) = 70 {(1+ (sin2(–2t - /2)) +2(sin(–2t - /2)) ) + ((cos(–2t – /2)) )2}

s(t) = 70 {(1+ 1+2(sin(–2t - /2)) )}

s(t) = (702){(1+(sin(–2t - /2)) )}

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