# Thread: Normals to plane curves

1. Hello!!!

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 $(f(t), g(t))$.

$n(t) \cdot r(t)=-g'(t) f(t)+ f'(t) g(t)$

But is this equal to $0$ ? If so why?

2. Originally Posted by evinda
Hello!!!

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 $(f(t), g(t))$.

$n(t) \cdot r(t)=-g'(t) f(t)+ f'(t) g(t)$

But is this equal to $0$ ? If so why?
Hey evinda!!

That will not generally be 0.
But then, shouldn't n(t) be normal to the direction or velocity of the curve, which is r'(t)?

Originally Posted by I like Serena
Hey evinda!!

That will not generally be 0.
But then, shouldn't n(t) be normal to the direction or velocity of the curve, which is r'(t)?
It holds that $r'(t)=f'(t) i+ g'(t) j$.

And $n(t) \cdot r'(t)=-g'(t) f'(t)+f'(t) g'(t)=0$.

So $n$ is normal to $r'$. But doesn't this imply that $n$ is parallel to $r$ ?

Or am I wrong?

4. Originally Posted by evinda
It holds that $r'(t)=f'(t) i+ g'(t) j$.

And $n(t) \cdot r'(t)=-g'(t) f'(t)+f'(t) g'(t)=0$.

So $n$ is normal to $r'$. But doesn't this imply that $n$ is parallel to $r$ ?

Or am I wrong?
No.

$r$ is merely a vector from the origin to a point on the curve, and $r$ is generally not parallel or perpendicular to the curve.
See:

The fact that $n$ is normal to $r'$ means that $n$ is normal to the curve given by $r(t)$.

Originally Posted by I like Serena
No.

$r$ is merely a vector from the origin to a point on the curve, and $r$ is generally not parallel or perpendicular to the curve.
See:

The fact that $n$ is normal to $r'$ means that $n$ is normal to the curve given by $r(t)$.
Could you explain it further to me? I haven't really understood it...

6. Originally Posted by evinda
Could you explain it further to me? I haven't really understood it...
What is it that you do not understand?

If we draw a vector $n$ in the drawing that is normal to the curve at point $r(t)$, that vector $n$ will be neither parallel nor perpendicular to the vector $r(t)$.

Originally Posted by I like Serena
What is it that you do not understand?

If we draw a vector $n$ in the drawing that is normal to the curve at point $r(t)$, that vector $n$ will be neither parallel nor perpendicular to the vector $r(t)$.
Ah I see now what you mean... Thank you!!!

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