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In other words, at time t , the particle is located at the point. Sketch the curve with parametric equations. CONCEPTUAL INSIGHT The graph of a function y = f ( x ) can always be parametrized in a simple way as c ( t ) = ( t , f ( t ))
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In other words, at time t, the particle is located at the point Sketch the curve with parametric equations
CONCEPTUAL INSIGHT The graph of a function y = f (x) can always be parametrized in a simple way as c(t) = (t, f (t)) For example, the parabola y = x2 is parametrized by c(t) = (t, t2) and the curve y = et by c (t) = (t, et). An advantage of parametric equations is that they enable us to describe curves that are not graphs of functions. For example, the curve below is not of the form y = f (x) but it can be expressed parametrically.
Eliminating the Parameter Describe the parametric curve c (t) = (2t − 4, 3 + t2) of the previous example in the form y = f (x). Solve for y as a function of x: Substitute:
A bullet follows the trajectory c (t) = (80t, 200t − 4.9t2) until it hits the ground, with t in seconds and distance in meters. Find: (a) The bullet’s height at t = 5s. (b) Its maximum height. The height of the bullet at time t is y (t) = 200t − 4.9t2 The maximum height occurs at the critical point of y (t):
Solution THEOREM 1 Parametrizationof a Line (a) The line through P = (a, b) of slope m is parametrized by for any r and s (with r 0) such that m = s/r. (b) The line through P = (a, b) and Q = (c, d) has parametrization The segment from P to Q corresponds to 0 ≤ 1 ≤ t.
(a) Usex = a + rt, to writetin terms of x… implies t= (x − a)/r: This is the equation of the line through P = (a, b) of slope m. The choice r = 1 and s = m yields the parametrizationbelow. (b) This parametrization defines a line that satisfies (x (0), y (0)) = (a, b) and (x (1), y (1)) = (c, d). Thus, it parametrizes the line through P and Q and traces the segment from P to Q as t varies from 0 to 1. THM 1
Parametrization of a Line Find parametric equations for the line through P = (3, −1) of slope m = 4.
The circle of radius R with center (a, b) has parametrization Let’s verify that a point (x, y) given by the above equation, satisfies the equation of the circle of radius R centered at (a, b): In general, to translatea parametric curve horizontally a units and vertically b units, replace c (t) = (x (t), y (t)) by c(t) = (a + x (t), b + y (t)).
Suppose we have a parametrizationc (t) = (x (t), y (t)) where x (t) is an even function and y (t) is an odd function, that is, x (−t) = x (t) and y (−t) = −y (t). In this case, c (−t) is the reflection of c (t) across the x-axis: c (−t) = (x (−t), y (−t)) = (x (t), −y (t)) The curve, therefore, is symmetric with respect to the x-axis.
Parametrization of an Ellipse Verify that the ellipse with equation is parametrized by Show that the equation of the ellipse is satisfied with x= acost, y = b sin t: Plot the case a = 4, b = 2.
To plot the case a = 4, b = 2, we connect the points corresponding to the t-values in the table. This gives us the top half of the ellipse corresponding to 0 ≤ t ≤ π. Then we observe that x(t) = 4 cost is even and y(t) = 2 sin t is odd. As noted earlier, this tells us that the bottom half of the ellipse is obtained by symmetry with respect to the x-axis. c (−t) = (x (−t), y (−t)) = (x (t), −y (t)) The curve, therefore, is symmetric with respect to the x-axis.
Different Parametrizations of the Same Curve Describe the motion of a particle moving along each of the following paths. (a)c1(t) = (t3, t6) (b)c2(t) = (t2, t4) (c)c3(t) = (cos t, cos2t) Each of these parametrizations satisfies y = x2, so all three parametrize portions of the parabola y = x2. c (t) = (t, f (t))
Different Parametrizations of the Same Curve Describe the motion of a particle moving along each of the following paths. (a)c1(t) = (t3, t6) (b)c2(t) = (t2, t4) (c)c3(t) = (cos t, cos2t) Each of these parametrizations satisfies y = x2, so all three parametrize portions of the parabola y = x2.
Different Parametrizations of the Same Curve Describe the motion of a particle moving along each of the following paths. (a)c1(t) = (t3, t6) (b)c2(t) = (t2, t4) (c)c3(t) = (cos t, cos2t) Each of these parametrizations satisfies y = x2, so all three parametrize portions of the parabola y = x2.