Vector calculus, normals to plane curves

miglo
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Homework Statement


show that n(t)=-g'(t)i+f'(t)j and -n(t)=g'(t)i-f'(t)j are both normal to the curve r(t)=f(t)i+g(t)j at the point (f(t),g(t)).


Homework Equations





The Attempt at a Solution


i tried finding the unit normal of r(t) in hopes that it would be exactly what n(t) and its negative are, but after going through a lot of algebra i didnt get what i was hoping would be the solution, unless i did some algebra mistake. Anyways i doubt I am approaching this problem the right way and would just like a hint at what to do.
 
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hi miglo! :smile:

hint: what is the tangent to f(t)i+g(t)j ? :wink:
 
f'(t)i+g'(t)j?
 
Yes. Are the supposed normals perpendicular to the tangent?
 
yeah i think i just figured it out like a minute ago
i take the dot product of r'(t) with n(t) which equals 0, therefore n(t) is normal to the curve at (f(t),g(t)) and then i do the same with -n(t)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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