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bmelyk Group Title

Looking for someone to check my answer: Find an equation of the tangent line through the given point. x 2 y 3 + 15y = 34x, (3, 2)

  • 2 years ago
  • 2 years ago

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  1. bmelyk Group Title
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    \[\frac{ 34-2x }{ 2y+15 }\]

    • 2 years ago
  2. bmelyk Group Title
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    thats the answer i got.

    • 2 years ago
  3. bmelyk Group Title
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    well that's what i had for y'

    • 2 years ago
  4. asnaseer Group Title
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    You first need to use implicit differentiation to find an expression for \(\displaystyle\frac{dy}{dx}\). Then substitute x=3 and y=2 into that expression to get the slope of the tangent line at that point. So then you will know the slope of the tangent line and you also know it passes through the point (3,2) - use this information to calculate the equation of the tangent line.

    • 2 years ago
  5. bmelyk Group Title
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    was my y' equation correct?

    • 2 years ago
  6. asnaseer Group Title
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    it doesn't look correct to me - can you please list your steps so that I can help spot where you may have made a mistake?

    • 2 years ago
  7. bmelyk Group Title
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    \[x ^{2}y ^{2}+15y=34x\] \[2x*2yy'+15y'=34\] \[y'(2y+15)=34-2x\] \[y'=\frac{ 34-2x }{ 2y+15 }\]

    • 2 years ago
  8. asnaseer Group Title
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    I thought you had \(y^3\) in the equation listed in your question?

    • 2 years ago
  9. bmelyk Group Title
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    lol so it is.

    • 2 years ago
  10. bmelyk Group Title
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    \[y'=\frac{ 34-2x }{ 3y ^{2}+15 }\]

    • 2 years ago
  11. bmelyk Group Title
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    did i have it right the second time (right above what you just wrote)

    • 2 years ago
  12. asnaseer Group Title
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    sorry I meant:\[\frac{d}{dx}(x^2y^3)=(x^2)\frac{d}{dx}(y^3)+y^3\frac{d}{dx}(x^2)\]

    • 2 years ago
  13. asnaseer Group Title
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    you haven't used the chain rule correctly

    • 2 years ago
  14. asnaseer Group Title
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    I mean "product rule"

    • 2 years ago
  15. asnaseer Group Title
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    have a look here: http://www.1728.org/chainrul.htm

    • 2 years ago
  16. bmelyk Group Title
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    \[x ^{2}*3y ^{2}y'+y ^{3}2x+15y'=34\]

    • 2 years ago
  17. asnaseer Group Title
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    that is correct :)

    • 2 years ago
  18. bmelyk Group Title
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    okay so now i isolate y'

    • 2 years ago
  19. asnaseer Group Title
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    exactly

    • 2 years ago
  20. bmelyk Group Title
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    \[y'=\frac{ 34-2xy ^{3} }{ x ^{2}3y ^{2}+15 }\]

    • 2 years ago
  21. asnaseer Group Title
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    yup - now follow the other steps that I had listed above.

    • 2 years ago
  22. asnaseer Group Title
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    NOTE: we don't usually write an expression as \(x^23y^2\) - it is better to write it as \(3x^2y^2\)

    • 2 years ago
  23. bmelyk Group Title
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    ok.

    • 2 years ago
  24. asnaseer Group Title
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    the general rule of thumb is to write in this order: 1. Constants first 2. Then letters in alphabetical order

    • 2 years ago
  25. bmelyk Group Title
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    okay so now i sub in my points right?

    • 2 years ago
  26. asnaseer Group Title
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    correct

    • 2 years ago
  27. bmelyk Group Title
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    so i had: -14/123

    • 2 years ago
  28. asnaseer Group Title
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    perfect! just a couple of more steps to go now :)

    • 2 years ago
  29. bmelyk Group Title
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    is the equation: \[y-2=-14/123x+42/123\]

    • 2 years ago
  30. asnaseer Group Title
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    yes - that looks correct. I wouldn't have separated the two constants here (the -2 and the 42/123)

    • 2 years ago
  31. asnaseer Group Title
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    you may also want to multiply both sides by 123 to remove the fractions from the final equation.

    • 2 years ago
  32. bmelyk Group Title
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    so how would you make it look?

    • 2 years ago
  33. asnaseer Group Title
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    ok, you got to this equation:\[y-2=-14/123x+42/123\]first add 2 to both sides to get:\[y=-14x/123 + 288/123\]then multiply both sides by 123 - what will you get then?

    • 2 years ago
  34. bmelyk Group Title
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    y123=-14x+288

    • 2 years ago
  35. asnaseer Group Title
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    correct - but again, remember to write constants first - so 123y instead of y123

    • 2 years ago
  36. bmelyk Group Title
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    that's not really how you write the equation of a line though.

    • 2 years ago
  37. asnaseer Group Title
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    so I would write the final equation as:\[123y=288-14x\]

    • 2 years ago
  38. asnaseer Group Title
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    it is still an equation of a line. you can write it in "standard form" as follows:\[14x+123y=288\]

    • 2 years ago
  39. asnaseer Group Title
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    maybe you are only used to seeing it in the form: \(y=mx +c\)

    • 2 years ago
  40. asnaseer Group Title
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    One last advice before I leave - you wrote one of the terms in your original equation as: -14/123x this can sometimes be confused for: \[-\frac{14}{123x}\]so it is usually better to write it as: -14x/123

    • 2 years ago
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