James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 13.2 Derivatives and Integrals of Vector functions 857
Example 1
(a) Find the derivative of rstd − s1 1 t 3 d i 1 te 2t j 1 sin 2t k.
(b) Find the unit tangent vector at the point where t − 0.
SOLUTION
(a) According to Theorem 2, we differentiate each component of r:
r9std − 3t 2 i 1 s1 2 tde 2t j 1 2 cos 2t k
(b) Since rs0d − i and r9s0d − j 1 2k, the unit tangent vector at the point s1, 0, 0d is
Ts0d −
r9s0d
| r9s0d | − j 1 2k − 1 j 1 2 k
s1 1 4 s5 s5
■
y
2
0
FIGURE 2
r(1)
(1, 1)
rª(1)
1
Notice from Figure 2 that the tangent
vector points in the direction of
increasing t. (See Exercise 58.)
x
Example 2 For the curve rstd − st i 1 s2 2 td j, find r9std and sketch the position
vector rs1d and the tangent vector r9s1d.
SOLUTION We have
r9std − 1
2st
i 2 j and r9s1d − 1 2 i 2 j
The curve is a plane curve and elimination of the parameter from the equations
x − st , y − 2 2 t gives y − 2 2 x 2 , x > 0. In Figure 2 we draw the position vector
rs1d − i 1 j starting at the origin and the tangent vector r9s1d starting at the corresponding
point s1, 1d.
■
Example 3 Find parametric equations for the tangent line to the helix with parametric
equations
x − 2 cos t y − sin t z − t
at the point s0, 1, y2d.
SOLUTION The vector equation of the helix is rstd − k2 cos t, sin t, tl, so
r9std − k22 sin t, cos t, 1l
The parameter value corresponding to the point s0, 1, y2d is t − y2, so the tangent
vector there is r9sy2d − k22, 0, 1l. The tangent line is the line through s0, 1, y2d
parallel to the vector k22, 0, 1l, so by Equations 12.5.2 its parametric equations are
x − 22t y − 1 z − 2 1 t ■
12
The helix and the tangent line in
Example 3 are shown in Figure 3.
8
z
4
FIGURE 3
0
_1
_0.5
y
0
0.5
1
2
0
x
_2
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