Calculus IV, Spring 10
Grinshpan
Quiz 3 Answers
A) Examine components: x√= cos(πt) is defined for all t, y = − ln t is
defined for t > 0, and z = 2 − t is defined for t ≤ 2. The domain of
→
−
r (t) is therefore 0 < t ≤ 2.
−
B) The graph of →
r (t) is a helix. Starting at (2, 0, 0) it winds around
the surface of the elliptic cylinder ( x2 )2 + ( y3 )2 = 1. At t increases,
→
−
r (t) gains elevation rotating clockwise.
C) Observe that (t sin t)2 + (t cos t)2 = t2 (sin2 t + cos2 t) = t2 . Hence,
for every t, the point (t sin t, t cos t, t2 ) lies on the surface of the
elliptic paraboloid x2 + y 2 = z.
√
D) The tangent line passes through the point r( 31 ) = (1, 3, 1) and is
√
−
parallel to the instantaneous velocity vector →
r 0 ( 31 ) = h−π 3, π, 3i.
Its vector equation is
√
√
→
−
r (s) = h1, 3, 1i + sh−π 3, π, 3i, −∞ < s < ∞,
tan
and its parametric equations are:
√
√
x = 1 − π 3s, y = 3 + πs,
z = 1 + 3s.
E)
Z
−c
h1, 0, tidt = ht + c1 , c2 , 21 t2 + c3 i = ht, 0, 21 t2 i + →
Z 1
h1, 0, tidt = h1, 0, 12 i
0
F) Letting x = t, y = t2 , z = −3t, we infer that the point r(t) lies in
the plane 2x − y + z + 2 = 0 if and only if 2t − t2 − 3t + 2 = 0. Thus
t2 + t − 2 = 0 and so t = 1, −2. This gives two points of intersection:
(1, 1, −3) and (−2, 4, 6). For each point of intersection we need to
−
determine the acute angle between the normal →
n = h2, −1, 1i and
−
tangent →
r 0 (t) = h1, 2t, −3i directions. We have:
q
|2·1+(−1)·2+1·(−3)|
1
√ √
θ(1) = arccos
= arccos 2 37
6 14
and
√ √
θ(−2) = arccos |2·1+(−1)·(−4)+1·(−3)|
= arccos 12
6 26
Approximations are θ(1) ≈ 71o , θ(−2) ≈ 76o .
q
3
13
.