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y 6 4 t=1 t=0 2 x 6 2 t = 1 4 t = 2 2 4 6 t=2
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Figure 16-1
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Cartesian-coordinate graph of the parametric equations x = 2t and y = 3t
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Parametric Equations in Two-Space
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which is identical to the slope-intercept equation y = (3/2)x
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The polar-coordinate counterpart Let s see what happens if we change our pair of parametric equations to polar form We ll put q in place of x, and put r in place of y Now we have
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q = 2t and r = 3t To create the polar graph, we can input various values of t, just as we did before To keep things from getting messy, let s restrict ourselves to values of t such that we see only the part of the graph corresponding to the first full counterclockwise rotation of the direction angle, so 0 q 2p Consider the following cases: When t = 0, we have q = 2 0 = 0 and r = 3 0 = 0 When t = p /4, we have q = 2 p /4 = p /2 and r = 3p /4 When t = p /2, we have q = 2 p /2 = p and r = 3p /2 When t = 3p /4, we have q = 2 3p /4 = 3p /2 and r = 3 3p /4 = 9p /4 When t = p, we have q = 2p and r = 3p
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Our graph is a spiral, as shown in Fig 16-2 Its equation can be derived with algebra exactly as we did in the Cartesian plane, substituting q for x and r for y to get r = (3/2)q
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Figure 16-2
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Polar-coordinate graph of the parametric equations q = 2t and r = 3t Each radial division represents p units For simplicity, we restrict q to the interval [0,2p ]
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What s a Parameter
Are you confused
If you re having trouble understanding the concept of a parameter, imagine the passage of time In science and engineering, elapsed time t is the parameter on which many things depend In the Cartesian situation described above, as time flows from the past (t < 0) through the present moment (t = 0) and into the future (t > 0), a point moves along the line in Fig 16-1, going from the third quadrant (lower left, in the past) through the origin (right now) and into the first quadrant (upper right, in the future) In the polar case, as time flows from the present (t = 0) into the future (t > 0), a point travels along the spiral in Fig 16-2, starting at the center (right now) and going counterclockwise, arriving at the outer end when t = p (a little while from now)
Here s a challenge!
Find a pair of parametric equations that represent the line shown in Fig 16-3
Solution
We re given two points on the line One of them, (0,3), tells us that the y-intercept is 3 We can deduce the slope from the coordinates of the other point When we move 4 units to the right from (0,3), we must go downward by 3 units (or upward by 3 units) to reach (4,0) The rise over run ratio is 3 to 4, so the slope is 3/4 The slope-intercept form of the equation for our line is y = ( 3/4)x + 3
y 6 (0, 3) 4 2 (4, 0) x 6 4 2 2 4 6 2 4 6
Figure 16-3
How can we represent this line as a pair of parametric equations
Parametric Equations in Two-Space
We can let x vary directly with the parameter t We describe that relation simply as x=t That s one of our two parametric equations We can substitute t for x into the point-slope equation to get y = ( 3/4)t + 3 That s the other parametric equation
Here s an experiment!
Do you suspect that the pair of equations x=t and y = ( 3/4)t + 3 isn t the only parametric way we can represent the line in Fig 16-3 If so, maybe you re right Let x = 2t, or x = t + 1, or x = 2t + 1, and see what happens when you generate the equation for y in terms of t on that basis When you put the two parametric equations together, do you get the same line as the one shown in Fig 16-3