solution to # 38
The reason there are two equations is because they are parametric (both x and y are given as a function of t). I'm still a little hazy about when parametric graphing is helpful, but I think an example we used in class was the height and width of a function changing at different rates during time t. (? Maybe?) Don't trust me on that one...
But in terms of graphing it, it isn't too hard. You press the MODE button on your calculator and switch from "function" to "parametric" in the row labeled "graph." Then when you go to the y= screen, you will see xt1, yt1, xt2, yt2, etc. instead of y1, y2, etc. For #38, enter 16t-42 in the xt1 row and t^3-12t^2 in the yt1 row.
When I graph these, I'm always worried that I'm missing part of the graph, because sometimes a parametric curve is a tangled mess. My viewing window was -50 < x < 100 and -50 < y < 10, with 0 < t < 5 and tstep=.1. With these dimensions, I found at least one x-intercept (x=-24) and at least one y-intercept (y=-23.625). I'm wary that I'm missing something though, since Mr. Karafiol asked us to find "all x- and y- intercepts."
But in terms of graphing it, it isn't too hard. You press the MODE button on your calculator and switch from "function" to "parametric" in the row labeled "graph." Then when you go to the y= screen, you will see xt1, yt1, xt2, yt2, etc. instead of y1, y2, etc. For #38, enter 16t-42 in the xt1 row and t^3-12t^2 in the yt1 row.
When I graph these, I'm always worried that I'm missing part of the graph, because sometimes a parametric curve is a tangled mess. My viewing window was -50 < x < 100 and -50 < y < 10, with 0 < t < 5 and tstep=.1. With these dimensions, I found at least one x-intercept (x=-24) and at least one y-intercept (y=-23.625). I'm wary that I'm missing something though, since Mr. Karafiol asked us to find "all x- and y- intercepts."
posted by Casey Blue at 3:13 PM
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