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s3a

  • 4 years ago

(a) What is the equation of the tangent plane to the surface z = 2xy + 3y^2 + 2x + y + 4 at the point P(-4,-3,60)? (b) What are the parametric equations of the normal line at this point?

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  1. s3a
    • 4 years ago
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    For (a), I get z = -12x -25y - 63 For (b), I get that the constant are -4, -3, and 60 for x, y, and z respectively but, I don't know how to find the coefficients of t for x, y, and z and would appreciate it if someone could help me figure this out.

  2. s3a
    • 4 years ago
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    Basically, I get it all except the coefficients of t for part (b).

  3. Zarkon
    • 4 years ago
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    you might want to try part (a) again

  4. s3a
    • 4 years ago
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    My answer is correct. I checked.

  5. Zarkon
    • 4 years ago
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    did you type the question correctly....(-4,-3,60) is not a solution to the equation you gave above

  6. s3a
    • 4 years ago
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    Sorry, z != 2xy + 3y^2 + 2x + y + 4 z = x^2 + 2xy + 3y^2 + 2x + y + 4

  7. s3a
    • 4 years ago
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    (I'm sleepy.)

  8. Zarkon
    • 4 years ago
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    do you know how to find the normal vector to the plane?

  9. s3a
    • 4 years ago
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    Is it -12 i - 25 j - 63k ?

  10. s3a
    • 4 years ago
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    and do you mean *a* normal vector?

  11. Zarkon
    • 4 years ago
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    no that is not a normal vector to the surface

  12. s3a
    • 4 years ago
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    So, I don't know. :(

  13. Zarkon
    • 4 years ago
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    \[a(x-x_0)+b(y-y_0)+c(z-z_0)=0\] is a plane with normal vector \(\vec{n}=<a,b,c>\)

  14. Zarkon
    • 4 years ago
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    find the coefficients infront of the x,y,z variables

  15. s3a
    • 4 years ago
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    Oh wait! Is the normal vector: 12 i + 25 j + k ?

  16. s3a
    • 4 years ago
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    (I did that based on what you just said.)

  17. Zarkon
    • 4 years ago
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    yes

  18. s3a
    • 4 years ago
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    Oh so the coefficients of t are the coefficients of i, j, k.

  19. s3a
    • 4 years ago
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    I get it then :) Thanks!

  20. Zarkon
    • 4 years ago
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    don't forget to include your original point (-4,-3,60)

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