## math_proof 3 years ago is this vector field F tangent to or normal to the curve C

1. math_proof

F=<y,-x> where C={(x,y):x^2+y^2=1} and n=<x,y>

2. math_proof

n is a normal to C

3. math_proof

i think is tangent but i'm not sure

4. eseidl

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

I mean tangent...lol

6. math_proof

so its normal at all points to C?

7. eseidl

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

yea C is only circle and the vector F goes in circles so that means its tangent at all points on C?

9. eseidl

F is tangent because it is orthogonal to the normal vector n.

10. eseidl

|dw:1353384188503:dw|

11. math_proof

what if F=<Y,x> and n=<x,y>

12. eseidl

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

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

14. eseidl

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

When x=y then F=n; the vectors are parallel when x=y.

16. eseidl

|dw:1353385035892:dw|

17. math_proof

thanks a lot for that explanation

18. eseidl

no prob...been awhile since I've done these