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anonymous

  • one year ago

I need a step by step tutorial on this basically i have no background on how to solve this. Would really appreciate the help. Find the equation of the tangent line and normal line to the curve 2x^2+y^2=4 at (1, -sqrt(2)).

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  1. wolf1728
    • one year ago
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    2x^2+y^2=4 Wouldn't you first have to put this into the form y = ???

  2. anonymous
    • one year ago
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    I have to isolate y first?

  3. wolf1728
    • one year ago
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    Okay isolating y y^2=-2x^2 +4 y = Square root (-2x^2 +4 )

  4. anonymous
    • one year ago
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    then what?

  5. wolf1728
    • one year ago
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    valtajaros When someone asks a question then just leaves, I am not sticking around

  6. wolf1728
    • one year ago
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    It helps if the asker sticks around so that I can ask some things

  7. anonymous
    • one year ago
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    the equation for a line passing through a point can be written as \[y-y _{1}=(x-x _{1})m\] where m is the slope of the line and \[y _{1}\] and \[x _{1}\] are the coordinates of the point that are known to us , in this case (1,-sqrt(2)) since the point lies both on the curve and on the line, the slope at this point will be same for curve and the line. we know slope=dy/dx derive the given equation and put the values of x and y as, 1 and sqrt(2) respectively, you get the value of m put the values of m, x, and y in the general equation and you get the equation for tangent

  8. wolf1728
    • one year ago
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    Okay, nitishdua31 seems aware of this so I will leave.

  9. anonymous
    • one year ago
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    really sorry im having internet problems thank you tho

  10. anonymous
    • one year ago
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    so first i have to get the derivative of y=sqrt(4-2x^2)

  11. anonymous
    • one year ago
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    http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-tannorm-2009-1.pdf see if this can help :)

  12. anonymous
    • one year ago
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    thanks man ill have a look in to this! :)

  13. anonymous
    • one year ago
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    yes find the derivative and that gives you the value of slope

  14. anonymous
    • one year ago
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    happy to help :D

  15. anonymous
    • one year ago
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    so like sqrt(u) = du/2sqrt(u) from y = sqrt(4-2x^2) i get y' =(-4x/2sqrt(4-2x^2) = -2/sqrt(4-2x^2)

  16. anonymous
    • one year ago
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    *-2x/sqrt(4-2x^2)

  17. anonymous
    • one year ago
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    -2(1)/sqrt(4-2(1)^2) = -2/sqrt(2) ???

  18. anonymous
    • one year ago
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    yes ! so its basically -sqrt(2)

  19. anonymous
    • one year ago
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    Attachment from Mathematica 9 may be a bit short on comments.

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  20. wolf1728
    • one year ago
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    valtajaros "really sorry im having internet problems thank you tho" ************************************************************** sorry :-(

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