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calyne

Find the points on the lemniscate 2(x^2 + y^2)^2 = 25(x^2 - y^2) where the tangent is horizontal. The derivative using implicit differentiation is (-4x^3 - 4xy^2 + 25x)/(4x^3 + 4x^2 y + 25y)

  • 2 years ago
  • 2 years ago

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  1. calyne
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    leminscate is just the name of that particular type of curve but makes no difference

    • 2 years ago
  2. Mani_Jha
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    Where the tangent is horizontal, dy/dx=0 Just differentiate it and set the result equal to zero.

    • 2 years ago
  3. Rohangrr
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    follow this !! cims.nyu.edu/~kiryl/Calculus/.../Smith-implicit-differentiation.pdf

    • 2 years ago
  4. calyne
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    damn it i'd love to because my textbook doesn't actually explain it just provides examples but i've done lright so far anyway

    • 2 years ago
  5. calyne
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    but what do i do after i set it equal to zero i solve for x or y??

    • 2 years ago
  6. calyne
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    PLEASE HELP ME

    • 2 years ago
  7. calyne
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    so ok what 4(x^2 + y^2)(2x - 2yy') = 25(2x - 2yy') and then y'=0 so 4(x^2 + y^2)(2x) = 25(2x) NOW WHAT

    • 2 years ago
  8. Rohangrr
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    it will give u a better explanation cims.nyu.edu/~kiryl/Calculus/.../Smith-implicit-differentiation.pdf

    • 2 years ago
  9. calyne
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    yeah no i can't access it i'm now allowed oh well

    • 2 years ago
  10. calyne
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    i don't like go to nyu maybe if you wanna download it and upload it here go for it but i can't check that url

    • 2 years ago
  11. Mani_Jha
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    Find y' from that result. Set that equal to zero Then solve for x and y.

    • 2 years ago
  12. calyne
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    right well that's what i did at first and either way i was stuck there too that's why i tried this

    • 2 years ago
  13. phi
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    You got this 4(x^2 + y^2)(2x - 2yy') = 25(2x - 2yy') I think the 2nd factor on the left-hand side should be (2x+2y y'). So it should be 4(x^2 + y^2)(2x + 2yy') = 25(2x - 2yy') Now solve for y'

    • 2 years ago
  14. phi
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    you get \[ y' = \frac{-4x^3-4xy^2+25x}{25y+4y(x^2+y^2)}=0 \] It looks like we can solve the numerator for the x values that make it zero.

    • 2 years ago
  15. calyne
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    show me pleasee

    • 2 years ago
  16. phi
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    Are you asking how to solve for x in the expression \[ -4x^3-4xy^2+25x =0 \]

    • 2 years ago
  17. phi
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    x=0 is one solution now solve \[ 4x^2+4y^2-25=0 \] moving the \( 4y^2-25 \) to the right-hand side, and dividing by 4: \[x^2= \frac{24-4y^2}{4} \]

    • 2 years ago
  18. phi
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    Now try each of these x values in the original equation. Notice that x=0 results in an imaginary value for y, so we can discard that point. with x^2 substituted into the original equation, solve for y, and then find x. you should get y=±5/4

    • 2 years ago
  19. calyne
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    oh no i'd gotten 2x+2yy' for the second term where you said i got - btw

    • 2 years ago
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