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IsTim

  • 3 years ago

Determine the equation of the tangent to the curve at a given point. x^2+9y^2=37 (1,2)

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  1. IsTim
    • 3 years ago
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    I've gotten the slope of -1/18, but I don't know if that's correct. And if so, how do I continue?

  2. IsTim
    • 3 years ago
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    So, first, dy/dx=-x/dy. I then subbed in (1,2) and got -1/18.

  3. CHAD159753
    • 3 years ago
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    Lol uhh so I cant give answers so ill just leave now but what youv done so far is correct :D

  4. IsTim
    • 3 years ago
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    I don't need the answer. I need to know what my process afterwards should be.

  5. inkyvoyd
    • 3 years ago
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    What's the derivative?

  6. inkyvoyd
    • 3 years ago
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    And Chad is in 11th grade, he won't be able to help you

  7. IsTim
    • 3 years ago
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    1. Derivative is 2x+18y(dy/dx)=0. Right? 2. I figured that out already...

  8. CHAD159753
    • 3 years ago
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    lol ink I can give him an answer ;D

  9. inkyvoyd
    • 3 years ago
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    x^2+9y^2=37 dy/dx 37=dy/dx x^2+dy/dx 9y^2 0=2x+18y dy/dx

  10. inkyvoyd
    • 3 years ago
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    No, you won't give him the answer he's looking for. THe answer he's looking for is an explanation, because, as everyone knows, it's the explanantion that matters in calc

  11. IsTim
    • 3 years ago
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    I already have the answer in my textbook and potentially Wolfram Alpha. I need to know how to GET there though.

  12. inkyvoyd
    • 3 years ago
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    Normally at this point I try to isolate dy/dx for convenience

  13. shruti
    • 3 years ago
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    from conic section?

  14. inkyvoyd
    • 3 years ago
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    0=2x+18y dy/dx (-2x)/(18y)=dy/dx x/(-9y)=dy/dx

  15. IsTim
    • 3 years ago
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    Yeah, I think I did that.It's -x/dy right? @ Shruti: I don't think so...

  16. inkyvoyd
    • 3 years ago
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    And, my answer agrees with your answer here.

  17. inkyvoyd
    • 3 years ago
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    Although my work here is moot, because i'm in 9th grade.

  18. IsTim
    • 3 years ago
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    Ok. Thanks for the help so far.

  19. IsTim
    • 3 years ago
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    Ok. How would I get to the answer of x+18y-37=0 from knowing that the slope is -1/18?

  20. inkyvoyd
    • 3 years ago
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    eh?

  21. inkyvoyd
    • 3 years ago
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    If you had the derivative and wanted to get the original equation?

  22. IsTim
    • 3 years ago
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    I guess?

  23. IsTim
    • 3 years ago
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    @shruti By conics did you mean http://www.purplemath.com/modules/circle3.htm

  24. amistre64
    • 3 years ago
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    since this is an equation of an ellipse, y can be considered an implicit function of x right?

  25. IsTim
    • 3 years ago
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    I guess. My notes only reveal the steps. This work is not from a textbook.

  26. IsTim
    • 3 years ago
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    In fact, it's not even addressed in my textbook. That's strange.

  27. amistre64
    • 3 years ago
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    it can be .... to derive implicitly, ignore the form of the variable and treat everything as tho it acted like you normally think of deriving x^2+9y^2=37 ; just derive it all, but dont toss out the derived bits 2x x' + 18y y' = 0 now lets consider this when we derive it with respect to x x' = dx/dx = 1 y' = dy/dx like normal so, lets solve it for y'

  28. inkyvoyd
    • 3 years ago
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    If you have the derivative, and want to get the original equation, you take the anti-derivative of it

  29. inkyvoyd
    • 3 years ago
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    If you're asking about taking anti-derivatives (indefinite integrals), and have not yet learned about rieman sums, don't worry about the question

  30. inkyvoyd
    • 3 years ago
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    Otherwise, I might have to look into it.

  31. IsTim
    • 3 years ago
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    Y' is -x/dy, which, when we sub in (1,2) is -1/18 right?

  32. dpaInc
    • 3 years ago
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    yes tim, -1/18 is right

  33. IsTim
    • 3 years ago
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    What would I do next then?

  34. inkyvoyd
    • 3 years ago
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    Well, you have a point, and a slope.

  35. inkyvoyd
    • 3 years ago
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    (1,2) -> point -1/18 -> slope

  36. inkyvoyd
    • 3 years ago
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    remember, y=mx+b

  37. IsTim
    • 3 years ago
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    Would I sub those into y=mx+b?

  38. dpaInc
    • 3 years ago
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    you have slope, you have a point, put it in point slope form...

  39. inkyvoyd
    • 3 years ago
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    put in the point and the slope, and then you have b

  40. inkyvoyd
    • 3 years ago
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    then, put in y=mx+b, cept don't put in that point

  41. inkyvoyd
    • 3 years ago
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    This isn't a calculus problem, this is an algebra problem, :D

  42. IsTim
    • 3 years ago
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    Sorry, what is point slope form? It's algebra? I must really have been "dumbing" down these years...

  43. inkyvoyd
    • 3 years ago
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    Ok.

  44. inkyvoyd
    • 3 years ago
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    There's multiple ways of writing the equation for a line.

  45. inkyvoyd
    • 3 years ago
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    This is point slope: y-y1=m(x-x1)

  46. inkyvoyd
    • 3 years ago
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    this is slope intecept: y=mx+b

  47. inkyvoyd
    • 3 years ago
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    This is standard form: ax+by=c

  48. inkyvoyd
    • 3 years ago
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    I normally work with slope intercept, because i'm more used to it.

  49. inkyvoyd
    • 3 years ago
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    However, point slope is the fastest way to go here.

  50. inkyvoyd
    • 3 years ago
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    y-y1=m(x-x1) (x1,y1)=(1,2) m=-1/18

  51. IsTim
    • 3 years ago
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    The answer is in the form of ax+by=c I guess, but I am also more familiar with slope intercept.

  52. IsTim
    • 3 years ago
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    Wait, I already have the slope... I know it is -1/18. But how do I get to x+18y-37?

  53. inkyvoyd
    • 3 years ago
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    y-1=(-1/18)(x-2) now, just isolate y, and distribute x

  54. inkyvoyd
    • 3 years ago
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    They have given you the point that the line must pass through, and you have found the slope,

  55. IsTim
    • 3 years ago
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    Oh. I finally understand. Thank you very much. I will call out again if I need help.

  56. inkyvoyd
    • 3 years ago
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    Ok. :D

  57. IsTim
    • 3 years ago
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    Wait, 1 and 2 in y-1=(-1/18)(x-2)/ Shouldn't it be y-2=(-1/18)(x-1)? Because (1,2)?

  58. inkyvoyd
    • 3 years ago
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    Yes

  59. inkyvoyd
    • 3 years ago
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    I got it confused myself. xD

  60. IsTim
    • 3 years ago
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    Why doesn't my answer look like that provided on the sheet, which is x+18y-37?

  61. IsTim
    • 3 years ago
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    Oh. I figured that out. Sorry.

  62. IsTim
    • 3 years ago
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    Thank you all for putting forth the effort to assist me.

  63. IsTim
    • 3 years ago
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    Or at the very least, looking over the question.

  64. inkyvoyd
    • 3 years ago
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    No problem :)

  65. inkyvoyd
    • 3 years ago
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    This is good premptive practice for me ;)

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