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math_proof

  • 3 years ago

is this vector field F tangent to or normal to the curve C

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  1. math_proof
    • 3 years ago
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    F=<y,-x> where C={(x,y):x^2+y^2=1} and n=<x,y>

  2. math_proof
    • 3 years ago
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    n is a normal to C

  3. math_proof
    • 3 years ago
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    i think is tangent but i'm not sure

  4. eseidl
    • 3 years ago
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    Take the dot product of the normal vector and F: \[<y, -x>*<x,y>=xy-xy=0\]Since the vectors are orthogonal, the vector field cannot be normal to the curve (if it was, F and n would be parallel). Since your only other choice is parallel, they are parallel.

  5. eseidl
    • 3 years ago
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    I mean tangent...lol

  6. math_proof
    • 3 years ago
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    so its normal at all points to C?

  7. eseidl
    • 3 years ago
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    Another way to see this is to note that C is just a unit circle:|dw:1353383917635:dw|No, F is tangent to all point of C

  8. math_proof
    • 3 years ago
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    yea C is only circle and the vector F goes in circles so that means its tangent at all points on C?

  9. eseidl
    • 3 years ago
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    F is tangent because it is orthogonal to the normal vector n.

  10. eseidl
    • 3 years ago
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    |dw:1353384188503:dw|

  11. math_proof
    • 3 years ago
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    what if F=<Y,x> and n=<x,y>

  12. eseidl
    • 3 years ago
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    The only way F can be perpendicular to n is if F is tangent to the circle. We showed they are perpendicular because their dot product is zero. Thus F is tangent to C.

  13. eseidl
    • 3 years ago
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    The F and n would not be orthogonal since their dot product=2xy. |dw:1353384419533:dw|They would look something like this

  14. eseidl
    • 3 years ago
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    However, when either x=0 or y=0 the dot product is zero and so n and F would be perpendicular at those points (F would be tangent to C at these points)...but in general 2xy doesn't equal zero so F wouldn't be tangent or normal to C at those points.

  15. eseidl
    • 3 years ago
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    When x=y then F=n; the vectors are parallel when x=y.

  16. eseidl
    • 3 years ago
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    |dw:1353385035892:dw|

  17. math_proof
    • 3 years ago
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    thanks a lot for that explanation

  18. eseidl
    • 3 years ago
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    no prob...been awhile since I've done these

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