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anonymous

  • 5 years ago

a line is parametrized by x=2+3t and y=4+7t a) what part of the line is obtained by restricting two non-negative numbers? b) what part of the line is obtained if is restricted to -1<t<0? c) how should t be restricted to give the part of the line to the left of the y-axis?

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  1. anonymous
    • 5 years ago
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    The wording for part a is kind of weird. I'd imagine if a nonnegative value of t is restricted, then there would be a hole in the graph at the corresponding (x, y) point. If t is restricted to an interval of two nonnegative numbers such as 2<t<3, then 8<x<11 and 18<y<25. I'm not sure what part a is getting at. Part b is more concrete, just put t = -1 into both the x and y equations to find the lower limits of x and y; and put t = 0 into both the x and y equations to find the upper limits of x and y. In this case, -1<x<2 and -3<y<4. To restrict the line to the left of the y-axis in part c, make sure that x < 0 and choose no t's that will make x > or = 0. Put 0 in for x and make sure it is greater than 2 + 3t. 0 > 2 + 3t -2/3 > t t < -2/3 (Answer)

  2. anonymous
    • 5 years ago
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    i understood part b buy im confused about part c

  3. anonymous
    • 5 years ago
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    Okay, to restrict the line to the left side of the y-axis, we have to restrict it to wherever x is negative. x is negative in quadrants 2 and 3, right? We start with the equation x < 0 because all numbers less than 0 are negative. SUBSTITUTE 2 + 3t in for x in the x > 0 equation and solve for t. Sorry, I didn't make that step clear.

  4. anonymous
    • 5 years ago
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    by choosing values greater than zero for x will allow there to be only y values on the left

  5. anonymous
    • 5 years ago
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    Oh, shoot. That was a typo. The first time I wrote it I put x < 0. That's what it's supposed to be. x < 0.

  6. anonymous
    • 5 years ago
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    sorry but why does it have to be less than zero

  7. anonymous
    • 5 years ago
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    x has to be less than 0 because the 2nd and 3rd quadrants are on the left side of the y axis. In both of these quadrants x is negative. Pick any point on the left side of the y-axis. (-2, 1), for example, is in the 2nd quadrant on the left side of the y-axis. The x value is negative. Because all x-values on the left side of the y-axis are negative, all the x values are less than zero.

  8. anonymous
    • 5 years ago
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    ohh okay thank you so much . for part a if it helps the answer is right of (2,4)

  9. anonymous
    • 5 years ago
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    Now part a makes sense. The wording of the sentence made me think something else. The same idea for part c applies here, kind of. Part a, however, wants t to be nonnegative: zero or positive. So t is greater than or equal to 0. Solve for t in both the x and y equations. You should get t = (x - 2)/3 and t = (y - 4)/7. Substitute both of these into t is greater than or equal to 0. \[t \ge0\] \[(x-2)/3\ge0\] \[(y-4)/7\ge0\] Solve for x and y in each. You should get x is greater than or equal to 0 and y is greater than or equal to 4. This actually translates to the right of (2, 4) and above (2, 4). By the way, to calculate this point, plug t = 0 into both the equations to get it. A question to be raised is: Why did the answer say we only had to be to the right of (2, 4)? Well, if you graph the equation of the line, all points to the right of (2, 4) are also above (2, 4), so "to the right" is a sufficient answer.

  10. anonymous
    • 5 years ago
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    x should be greater than or equal to 2... haha, just noticed a typo above.

  11. anonymous
    • 5 years ago
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    i think the reason the answer said right of 2,4- is because the points have to be greater that 2,4

  12. anonymous
    • 5 years ago
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    The math says that \[x \ge2\] AND \[y \ge4\], so we'd be tempted to conclude that all points we choose MUST be greater than x = 2 in the x-direction (so to the right) AND be greater than y = 4 in the y-direction (so above). If I could draw it it might make more sense. However, if you graph the line, it will intersect (2, 4) and go up and to the right after (2, 4). To say the points would have to be above (2, 4) in the y-direction would be redundant. Redundance is frowned upon usually.

  13. anonymous
    • 5 years ago
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    that makes a lot of sense. i just had a question in part a we used both equations but in part c we on use the x= equation why is that

  14. anonymous
    • 5 years ago
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    See, I learned when I took Calc 2 that the majority of problems become "case-by-case" problems. Every problem is different. The reason why we use different equations in each part amounts to what we're told to work with in each part. In part a, we were told that we had to restrict t. We restricted it to t is greater than or equal to 0. In part c, we were told that we had to restrict x. We restricted it to x < 0. Whatever restriction we have, we need to substitute that variable for the other ones. When our restriction is related to t, we solve for t in both the x and y equations and substitute them separately to ultimately find x and y. When our restriction is related to x (or y), we just substitute that to ultimately find t.

  15. anonymous
    • 5 years ago
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    thank you so much for spending your time and explaining this problem so well . i am a fan:)

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