Here's the question you clicked on:
math_proof
is this vector field F tangent to or normal to the curve C
F=<y,-x> where C={(x,y):x^2+y^2=1} and n=<x,y>
i think is tangent but i'm not sure
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.
so its normal at all points to C?
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
yea C is only circle and the vector F goes in circles so that means its tangent at all points on C?
F is tangent because it is orthogonal to the normal vector n.
what if F=<Y,x> and n=<x,y>
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.
The F and n would not be orthogonal since their dot product=2xy. |dw:1353384419533:dw|They would look something like this
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.
When x=y then F=n; the vectors are parallel when x=y.
thanks a lot for that explanation
no prob...been awhile since I've done these