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Parametric Equations, Vector Functions, Curvilinear Motion solved questions
Parametric Equations, Vector Functions, Curvilinear Motion solved questions
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CHAPTER 34Parametric Equations, Vector Functions,
Curvilinear Motion
PARAMETRIC EQUATIONS OF PLANE CURVES
34.1 Sketch the curve given by the parametric equations x = a cos 6, y = a sin 6.
Note that x2 + y2 = a2 cos2 0 + a2 sin2 0 — a2(cos2 6 + sin2 0) = a2. Thus, we have a circle of radius a with
center at the origin. As shown in Fig. 34-1, the parameter 6 can be thought of as the angle between the positive
jc-axis and the vector from the origin to the curve.
Fig. 34-1 Fig. 34-2
34.2 Sketch the curve with the parametric equations x = 2 cos 0, y = 3 sin 6.
x2 y 2
-T + -g - 1. Hence, the curve is an ellipse with semimajor axis of length 3 along the y-axis and semiminor
axis of length 2 along the x-axis (Fig. 34-2).
34.3 Sketch the curve with the parametric equations x = t, y = t2.
y = t2 = x2. Hence, the curve is a parabola with vertex at the origin and the y-axis as its axis of symmetry
(Fig. 34-3).
Fig. 34-3 Fig. 34-4
34.4 Sketch the curve with the parametric equations x = t, y = t2.
x = 1 + (3 — y)2, x — l = (y - 3)2. Hence, the curve is a parabola with vertex at (1,3) and axis of symmetry
y = 3 (Fig. 34-4).
34.5 Sketch the curve with the parametric equations x = sin t, y = —3 + 2 cos t.
x2 + = sin 2 1 + cos 2 1 = 1. Thus, we have an ellipse with center (0, —3), semimajor axis of length 2
along the y-axis, and semiminor axis of length 1 along the line y = —3 (Fig. 34-5).
274
, PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR MOTION 275
Fig. 34-5 Fig. 34-6
34.6 Sketch the curve with the parametric equations x = sec t, y = tan t.
X2 = y2 + l. Hence, x2 — y2 = l. Thus, the curve is a rectangular hyperbola with the perpendicular
asymptotes y = ±x. See Fig. 34-6.
34.7 Sketch the curve with the parametric equations x = sin t, y = cos 2t.
y = cos 2t = 1 — 2 sin 2 1 = 1 — 2x2, defined for \x\ ^ 1. Thus, the curve is an arc of a parabola, with vertex at
(0,1), opening downward, and with the _y-axis as axis of symmetry (Fig. 34-7).
Fig. 34-7 Fig. 34-8
34.8 Sketch the curve with the parametric equations x = t + 1 It, y = t - 1 It.
x2 = t2+ 2 + 1/12, y2 = t2-2+l/t2. Subtracting the second equation from the first, we obtain the
hyperbola Jt 2 -y 2 = 4 (Fig. 34-8).
34.9 Sketch the curve with the parametric equations * = 1 + t, y = l-t.
x + y = 2. Thus, we have a straight line, going through the point (1,1) and parallel to the vector (1, —1); see
Fig. 34-9.
Fig. 34-9 Fig. 34-10
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