# qr code vb.net open source Rectangular-coordinate graph of the parametric equations x = t 1 and y = ln t in .NET framework Paint Code 39 Extended in .NET framework Rectangular-coordinate graph of the parametric equations x = t 1 and y = ln t

Rectangular-coordinate graph of the parametric equations x = t 1 and y = ln t
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We can construct a rectangular-coordinate graph of the relation between x and y by tabulating the values for several points, based on various values of t Let s break things down into the following cases: When t 0, ln t is undefined, so there are no points to plot When t = e 2 014, we have x = (e 2) 1 = e2 739 and y = ln (e 2) = 2 When t = e 1 037, we have x = (e 1) 1 = e 272 and y = ln (e 1) = 1 When t = 1, we have x = 1 1 = 1 and y = ln 1 = 0 When t = 2, we have x = 2 1 = 1/2 and y = ln 2 069 When t = e 272, we have x = e 1 037 and y = ln e = 1 When t = e2 739, we have x = (e2) 1 = e 2 014 and y = ln (e2) = 2
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Figure 16-6 shows the curve we obtain when we plot these points and connect the dots To keep the picture clean, the points aren t labeled As in Fig 16-4, we use distorted rectangular coordinates to help us fit more of the curve on the page than we could with a true Cartesian grid
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Polar-coordinate graph: example 2 We can directly substitute q for x and r for y in the above example, tabulate some values, graph the results, and get the curve shown in Fig 16-7 Let s restrict t to keep q within the closed interval [0,2p] so we see only the first full positive revolution The situation breaks down as follows:
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When t = e5 148, we have q = (e5) 1 = e 5 00067 and r = ln (e5) = 5 When t = e2 739, we have q = (e2) 1 = e 2 014 and r = ln (e2) = 2
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Figure 16-7
Polar-coordinate graph of the parametric equations q = t 1 and r = ln t Each radial division represents 1 unit
When t = 2, we have q = 2 1 = 1/2 and r = ln 2 069 When t = 1, we have q = 1 1 = 1 and r = ln 1 = 0 When t = 1/2, we have q = (1/2) 1 = 2 and r = ln (1/2) 069 When t = p 1 032, we have q = (p 1) 1 = p and r = ln (p 1) 114 When t = (2p ) 1 016, we have q = [(2p ) 1] 1 = 2p and r = ln [(2p ) 1] 184
As we plot the points to obtain this graph, we must remember that when we have a negative radius in polar coordinates, we go outward from the origin by a distance equal to |r|, but in the opposite direction from that indicated by q
Are you confused
The polar graph in Fig 16-7 can be baffling Imagine that we start out facing east, in the direction q = 0 Our graph is infinitely far away in this direction As we turn counterclockwise, the curve approaches us as r becomes finite and decreases When we have turned counterclockwise through an angle of 1 rad (approximately 57 ), the graph has come all the way in and reached the origin As we continue to turn counterclockwise, the radius becomes negative, so the graph is behind us As we rotate farther counterclockwise, r increases negatively When we ve rotated all the way around through a complete circle, the graph is approximately 184 units to our rear