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anonymous

  • 5 years ago

A normal line is a line perpendicular to a tangent line at the point where the tangent meets the function. Give the equation of the normal line to the graph of y=3xsqrt{x^2+6}+4 at the point (0,4)

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  1. TuringTest
    • 5 years ago
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    so we're gonna need the calculus version of point-slope form y-y1=f'(x1)(x-x1) which means we need to find dy/dx (at x=x1) then we need the fact that the normal line to this line has the slope of the negative reciprocal -1/(f'(x1) between those facts you're good

  2. TuringTest
    • 5 years ago
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    so step one (as usual in calc) find dy/dx

  3. anonymous
    • 5 years ago
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    so dy/dx = (3)(1/2(x^2+6)^-1/2(2x)

  4. anonymous
    • 5 years ago
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    or 6x/2(x^2+6)

  5. anonymous
    • 5 years ago
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    or 3x/x^2+6

  6. TuringTest
    • 5 years ago
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    I hadn't tried yet, but that's not what I'm getting...

  7. anonymous
    • 5 years ago
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    oh

  8. anonymous
    • 5 years ago
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    i did the chain rule for sqrt{x^2+6}

  9. anonymous
    • 5 years ago
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    oh so would it be 3x/sqrt{x^2+6}

  10. TuringTest
    • 5 years ago
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    Yeah, but you need to use it in conjunction with the product rul so be very careful... I even have messed up a little according to wolfram, so I'm looking where I lost a 2

  11. TuringTest
    • 5 years ago
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    that's closer^

  12. anonymous
    • 5 years ago
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    oh would the 3x be part of sqrt{x^2+6} in the beginning

  13. anonymous
    • 5 years ago
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    would i have to use the chain rule for (3xsqrt{x^2+6}) or just sqrt{x^2+6}

  14. TuringTest
    • 5 years ago
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    product then chain rule (3x)'(sqrt(x^2+6)+3x(sqrt(x^2+6))' ^^^^^^^^ chain rule here

  15. anonymous
    • 5 years ago
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    hmm i kinda got this for the product rule i used the chain rule in the product rule... 3(sqrt{x^2+6})+1/2(x^2+6)^-1/2(2x)(3x)

  16. TuringTest
    • 5 years ago
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    yeah, that looks good now the simplification is the annoying bit, I only just found my earlier mistake :/

  17. TuringTest
    • 5 years ago
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    simplify the obvious stuff then factor out (x^2+6)^(-1/2) with finess

  18. anonymous
    • 5 years ago
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    dy/dx= 3sqrt{x^2+6}+6x^2/2sqrt{x^2+6}

  19. TuringTest
    • 5 years ago
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    well, you could just use a calculator I suppose... right what you have so far though, just a pain to evaluate

  20. TuringTest
    • 5 years ago
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    so you should keep going and simplify if you can

  21. anonymous
    • 5 years ago
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    dy/dx= 3sqrt{x^2+6}+3x^2/sqrt{x^2+6}

  22. TuringTest
    • 5 years ago
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    you can factor out 3(x^2+6)^(-1/2) if you want to do it to look nice

  23. anonymous
    • 5 years ago
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    hmm how?

  24. TuringTest
    • 5 years ago
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    dy/dx=3(x^2+6)^(1/2)+3x^2(x^2+6)^(-1/2) =3(x^2+6)^(-1/2)(x^2+6+x^2)=3(2x^2+6)/sqrt(x^2+6) =6(x^2+3)/sqrt{x^2+6} watch the exponents carefully if that lost you

  25. TuringTest
    • 5 years ago
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    I have to go eat.... Evaluate that at x=0 to get the derivative at that point, which is your slope of the tangent f'(0) the slope perpendicular to that will be the normal slope, so that is the one you want. We know it will be -1/f'(0) from algebra. so use -1/f'(0) in the the slope m in the point-slope formula for a line y-y1=m(x-x1) See you later perhaps

  26. anonymous
    • 5 years ago
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    okay thank you so much! have a great dinner :) thank you.

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