Solutions to the Second Homework Set (page 22) Parametrization r

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Solutions to the Second Homework Set (page 22)
1. (a) Parametrization: Let r(t) = hcos t; sin ti and t 2 (0; 2). The t
1
1
value for the point P = ( 2 ; 2 ) can be found from the parametrization: We need to nd t so that r(t) = h 12 ; 12 i so t must satisfy the
two equations
cos t = p1 ;
sin t = p1
2
2
from which we conclude that t = 4 .
Tangent vector: r (t) = h, sin t; cos ti. Evaluating r at t = 4
yields
r ( 4 ) = h, p12 ; p12 i
for the tangent vector.
(b) Parametrization: Let r(t) = h1+cos t; ,1+sin ti, t 2 (0; 2). Find
t so that r(t) = h1 + 12 ; ,1 + 12 i. So
1 + cos t = 1 + p1 ;
,1 + sin t = ,1 + p1
2
2
. So t = 4 .
Tangent vector: r (t) = h, sin t; cos ti and r ( 4 ) = h, 12 ; 12 i.
2
(c) Parametrization: First divide the equation by 6 to get x22 + 2y3 = 1.
2
Now, let x22 = cos2 t and 2y3 = sin2 t, from which we get
r3
p
x(t) = 2 cos t;
y (t) =
2 sin t;
and, therefore,
r3
p
r(t) = h 2 cos t; 2 sin ti:
p
p
p
p
0
0
0
p
p
0
0
p
p
Next, we nd t so that r(t) = h 12 ; 2 32 i or
p
p
p
Hence, t satises
2 cos t = p1 ;
2
cos t = 12 ;
p
r3
p
p3 :
2 sin t =
2 2
p
sin t = 23 ;
which implies that t = 3 .
q
p
Tangent vector: r (t) = h, q2 sin t; 32 cos ti which when evaluated at t = 3 yields r ( 3 ) = h, 32 ; 2 32 i.
(d) Parametrization:
0
p
0
p
1
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