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Equation of normal line and tangent plane

  1. Nov 5, 2015 #1
    1. The problem statement, all variables and given/known data
    A cup is represented by the surface -(z-1)2 + x2 + y2 = 1
    and it is on a table represented by the plane z=0

    a) find the angle at which the cup intersects the table

    b) find the equation of the normal line to the cup at the point (0, √2 , 2)

    c) find the equation the tangent plane at the point (0, √2 , 2)

    2. Relevant equations


    3. The attempt at a solution
    for a) I use cos θ = a⋅b / |a||b|

    a and b are normals of the cup and table at the point of intersect, which are found by finding the gradients.

    ∇ƒcup = <Fx, Fy, Fz> = < 2x , 2y, -2(z-1) >

    ∇ƒtable = <Fx, Fy, Fz> = < 0 , 0 , 1 >

    so at the point of contact between the cup and table, z is 0.

    plugging that into the equation of the surface of the cup :

    -(0-1)2 + x2 + y2 = 1

    x2 + y2 = 2

    so the intersection between the cup and table is a circle with radius √2

    then I find a point on the circle x2 + y2 = 2 , setting x = 0 then y = √2
    then a point is (0,√2, 0)

    plugging this point into < 2x , 2y, -2(z-1) > and < 0 , 0 , 1 >:

    < 0, 2√2, 2 > and < 0 , 0 , 1 >

    ∴ cos θ = a⋅b / |a||b|


    b) plugging the point (0,√2, 2) into < 2x , 2y, -2(z-1) >:

    < 0, 2√2, -2 >

    equation of the normal line to the cup at point (0,√2, 2) =

    (0,√2, 2) + t < 0, 2√2, -2 > = ( 0 , √2 + 2√2 t , 2-2t )

    c) equation of tangent plane at point (0,√2, 2) :

    2√2(y- √2) -2(z-2) = 0


    can someone check my work?
     

    Attached Files:

    Last edited: Nov 6, 2015
  2. jcsd
  3. Nov 6, 2015 #2

    ehild

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    You forgot to take the square root when calculating the angle.
     
  4. Nov 6, 2015 #3
     

    Attached Files:

    Last edited: Nov 6, 2015
  5. Nov 6, 2015 #4

    ehild

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  6. Nov 6, 2015 #5
    upload_2015-11-5_23-19-30.png
    sorry, it is late over here :(
     
  7. Nov 6, 2015 #6

    ehild

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    It is right now!
     
  8. Nov 6, 2015 #7
    b and c are correct as well?
     
  9. Nov 6, 2015 #8

    ehild

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    They are correct but you can simplify the equation of the tangent plane.
     
  10. Nov 6, 2015 #9
    √2(y- √2) -(z-2) = 0
     
  11. Nov 6, 2015 #10

    ehild

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    Expand the parentheses. What will be the resultant constant?
     
  12. Nov 6, 2015 #11
    √2y- 2 - z + 2 = 0

    √2y - z = 0
     
  13. Nov 6, 2015 #12

    ehild

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    It is much nicer, isn't it?
     
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