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Published byLewis Mills Modified over 6 years ago
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In other words, at time t, the particle is located at the point
Sketch the curve with parametric equations
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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.
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Eliminating the Parameter Describe the parametric curve
c (t) = (2t − 4, 3 + t2) of the previous example in the form y = f (x).
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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 maximum height occurs at the critical point of y (t):
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THEOREM 1 Parametrization of a Line
Solution THEOREM 1 Parametrization of 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.
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This is the equation of the line through P = (a, b) of slope m
This is the equation of the line through P = (a, b) of slope m. The choice r = 1 and s = m yields the parametrization above. (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
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Parametrization of a Line Find parametric equations for the line through P = (3, −1) of slope m = 4.
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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 translate a 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)).
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Suppose we have a parametrization c(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.
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Parametrization of an Ellipse Verify that the ellipse with equation
is parametrized by Plot the case a = 4, b = 2.
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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 cos t 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.
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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, cos2 t) c (t) = (t, t2)
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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, cos2 t)
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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, cos2 t)
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Using Symmetry to Sketch a Loop Sketch the curve
c (t) = (t2 + 1, t3 − 4t) Label the points corresponding to t = 0, ±1, ±2, ±2.5. Step 1. Can we use symmetry?
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Using Symmetry to Sketch a Loop Sketch the curve
c (t) = (t2 + 1, t3 − 4t) Label the points corresponding to t = 0, ±1, ±2, ±2.5. Step 2. Analyze x (t), y (t) as functions of t.
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Using Symmetry to Sketch a Loop Sketch the curve
c (t) = (t2 + 1, t3 − 4t) Label the points corresponding to t = 0, ±1, ±2, ±2.5. Step 3. Plot points and join by an arc. The points c (0), c (1), c (2), c (2.5) are plotted and joined by an arc to create the sketch for t ≥ 0. The sketch is completed by reflecting across the x-axis.
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A cycloid is a curve traced by a point on the circumference of a rolling wheel. Cycloids are famous for their “brachistochrone property”. A cycloid. A stellar cast of mathematicians (including Galileo, Pascal, Newton, Leibniz, Huygens, and Bernoulli) studied the cycloid and discovered many of its remarkable properties. A slide designed so that an object sliding down (without friction) reaches the bottom in the least time must have the shape of an inverted cycloid. This is the brachistochrone property, a term derived from the Greek brachistos, “shortest,” and chronos, “time.”
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Parametrizing the Cycloid Find parametric equations for the cycloid generated by a point P on the unit circle. The point P is located at the origin at t = 0. At time t, the circle has rolled t radians along the x axis and the center C of the circle then has coordinates (t, 1). Figure (B) shows that we get from C to P by moving down cos t units and to the left sin t units, giving us the parametric equations ` ` `
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The argument on the last slide shows in a similar fashion that the cycloid generated by a circle of radius R has parametric equations
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THEOREM 2 Slope of the Tangent Line Let c (t) = (x (t), y (t)), where x (t) and y (t) are differentiable. Assume that CAUTION Do not confuse dy/dx with the derivatives dx/dt and dy/dt, which are derivatives with respect to the parameter t. Only dy/dx is the slope of the tangent line.
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Let c (t) = (t2 + 1, t3 − 4t). Find:
(a) An equation of the tangent line at t = 3 (b) The points where the tangent is horizontal. ` ` `
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