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

650 Chapter 10 Parametric Equations and Polar Coordinates
Example 1 A curve C is defined by the parametric equations x − t 2 , y − t 3 2 3t.
(a) Show that C has two tangents at the point (3, 0) and find their equations.
(b) Find the points on C where the tangent is horizontal or vertical.
(c) Determine where the curve is concave upward or downward.
(d) Sketch the curve.
SOLUTION
(a) Notice that y − t 3 2 3t − tst 2 2 3d − 0 when t − 0 or t − 6s3 . Therefore the
point s3, 0d on C arises from two values of the parameter, t − s3 and t − 2s3 . This
indicates that C crosses itself at s3, 0d. Since
dy
dx − dyydt
dxydt − 3t 2 2 3
− 3 St 2 1 2t 2 tD
the slope of the tangent when t − 6s3 is dyydx − 66ys2s3 d − 6s3 , so the equations
of the tangents at s3, 0d are
y − s3 sx 2 3d and y − 2s3 sx 2 3d
y
t=_1
(1, 2)
0
t=1
(1, _2)
FIGURE 1
y=œ„3(x-3)
(3, 0)
x
y=_ œ„3(x-3)
(b) C has a horizontal tangent when dyydx − 0, that is, when dyydt − 0 and
dxydt ± 0. Since dyydt − 3t 2 2 3, this happens when t 2 − 1, that is, t − 61. The
corresponding points on C are s1, 22d and (1, 2). C has a vertical tangent when
dxydt − 2t − 0, that is, t − 0. (Note that dyydt ± 0 there.) The corresponding point
on C is (0, 0).
(c) To determine concavity we calculate the second derivative:
d
S
d 2 y
dx − dt dxD
2D
dy 3
S1 1 1 2 t
−
− 3st 2 1 1d
2 dx
2t
4t 3
dt
Thus the curve is concave upward when t . 0 and concave downward when t , 0.
(d) Using the information from parts (b) and (c), we sketch C in Figure 1.
Example 2
(a) Find the tangent to the cycloid x − rs 2 sin d, y − rs1 2 cos d at the point
where − y3. (See Example 10.1.7.)
(b) At what points is the tangent horizontal? When is it vertical?
SOLUTION
(a) The slope of the tangent line is
When − y3, we have
dy
dx − dyyd
dxyd −
r sin
rs1 2 cos d −
sin
1 2 cos
x − rS 3 2 sin 3D − rS 3 2 s3
2
D y − rS1 2 cos 3D − r 2
n
and
dy
dx −
sinsy3d
1 2 cossy3d − s3 y2
1 2 1 2
− s3
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