anonymous
  • anonymous
is this vector field F tangent to or normal to the curve C
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
F=where C={(x,y):x^2+y^2=1} and n=
anonymous
  • anonymous
n is a normal to C
anonymous
  • anonymous
i think is tangent but i'm not sure

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anonymous
  • anonymous
Take the dot product of the normal vector and F: \[*=xy-xy=0\]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.
anonymous
  • anonymous
I mean tangent...lol
anonymous
  • anonymous
so its normal at all points to C?
anonymous
  • anonymous
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
anonymous
  • anonymous
yea C is only circle and the vector F goes in circles so that means its tangent at all points on C?
anonymous
  • anonymous
F is tangent because it is orthogonal to the normal vector n.
anonymous
  • anonymous
|dw:1353384188503:dw|
anonymous
  • anonymous
what if F=and n=
anonymous
  • anonymous
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.
anonymous
  • anonymous
The F and n would not be orthogonal since their dot product=2xy. |dw:1353384419533:dw|They would look something like this
anonymous
  • anonymous
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.
anonymous
  • anonymous
When x=y then F=n; the vectors are parallel when x=y.
anonymous
  • anonymous
|dw:1353385035892:dw|
anonymous
  • anonymous
thanks a lot for that explanation
anonymous
  • anonymous
no prob...been awhile since I've done these

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