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anonymous

  • 5 years ago

find parametric equations for the tangent line at t=2 for x=t^2, y=t^3

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  1. anonymous
    • 5 years ago
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    Okay, sounds like you need to find the tangent line of this curve at t=2 and state that tangent line in terms of some parameter. Give me a minute to write something out and scan, okay?

  2. anonymous
    • 5 years ago
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    k

  3. anonymous
    • 5 years ago
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    scanning

  4. anonymous
    • 5 years ago
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  5. anonymous
    • 5 years ago
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    k ty

  6. anonymous
    • 5 years ago
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    brb

  7. anonymous
    • 5 years ago
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    k

  8. anonymous
    • 5 years ago
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    I know it might look a bit full-on. Does that stuff look familiar?

  9. anonymous
    • 5 years ago
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    yah i did the problem but there is a discrepancy bw my answer an the answer found at the end of the book

  10. anonymous
    • 5 years ago
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    the answer i got was x=4+4t and y=8+12t but at the end of the book the answer was x=4+4(t-2) and y=8+12(t-2_ do u know why?

  11. anonymous
    • 5 years ago
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    Well, just looking at that, I don't see any difference between what you have and what the book does, because if you identify t-2 as a parameter, what's being said between you guys is\[x=4+4(paramater)\]\[y=8+12(parameter)\]

  12. anonymous
    • 5 years ago
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    If you were to take ALL the real numbers and plug them into t and t-2, you'd end up with the same set of numbers, and because the form of both your equations is the same, you'd end up with the same lines.

  13. anonymous
    • 5 years ago
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    so there is no difference

  14. anonymous
    • 5 years ago
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    no real difference

  15. anonymous
    • 5 years ago
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    but do u know how it becomes t-2

  16. anonymous
    • 5 years ago
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    Well, I had a look at my solution and their solution and they're identical, just parametrized differently. It's the case with parametrization that you can set x equal to ANY kind of concoction of some parameter you want, and because of the relationship between y and x, derive the associated parametrization for y. Here, I chose s, but they chose 4+4(t-2). My choosing s forced my y to be 3s-4, and their choosing x=4+4(t-2) forced their y to be 8 + 12(t-2). Why they chose that parametrization, I cannot say. They might be setting the problem up for something to come.

  17. anonymous
    • 5 years ago
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    okay thank u for ur help

  18. anonymous
    • 5 years ago
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    np

  19. anonymous
    • 5 years ago
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    Incidentally, how did you get your own parametrizations?

  20. anonymous
    • 5 years ago
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    u mean how did i find the answer

  21. anonymous
    • 5 years ago
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    Well, how did you go about parametrizing? What did you say 'let x equal' to?

  22. anonymous
    • 5 years ago
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    I guess I mean, what was your motivation for 4+4t?

  23. anonymous
    • 5 years ago
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    well i first took the derivative , plugged in 2 into each equation to find my slope . then plugged in two in two into the original equations to find a point and plugged everything into the x and y formula

  24. anonymous
    • 5 years ago
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    By x and y formula, do you mean the point-slope formula for a straight line?

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