Calculating direction derivative to a line in 2D

[member 428835]

Homework Statement

Compute $\partial_n f$ where $n$ is normal to $f$, and $f$ lies in the $x-z$ plane and is parameterized by $$x(s) = \frac{1}{c} \sin (c s);\\
z(s) = \frac{1}{c} (1-\cos (c s))
$$

Homework Equations

$\partial_n f = \nabla f \cdot \hat n$

The Attempt at a Solution

I was thinking of defining vector $\hat f \equiv x(s) \hat i + z(s) \hat k$. Then tangent would be $\hat f '(s)$. Now rotate this by $90^\circ$. We know $$
\begin{bmatrix}
\cos\theta & -\sin \theta\\
\sin\theta & \cos \theta\\
\end{bmatrix}$$
is the rotational matrix where $\theta = 90^\circ$. Thus the unit normal $\hat n$ to the surface would be $$\hat n = \frac{z'(s) \hat i - x'(s) \hat k}{\sqrt{z'(s)^2+x'(s)^2}}$$
Once we have $\hat n$, we should be able to construct $\partial_n f$ via $\nabla f \cdot \hat n$, where I assume $\nabla f = x'(s) \hat i + z'(s) \hat k$. Then $$
\partial_n f = \frac{z'(s)x'(s) - x'(s)z'(s) }{\sqrt{z'(s)^2+x'(s)^2}} = 0
$$
I think I went wrong in computing $\nabla f$. Can someone help?

 

[haruspex]

Then tangent would be $\hat f '(s)$.

Why? s is only a parameter. It has no physical meaning in the shape of the curve.

By the way, it is not clear to me whether the answer is expected in terms of x and z or in terms of s.

 

[member 428835]

Why? s is only a parameter. It has no physical meaning in the shape of the curve.

I'm not sure, but I know this is correct. I should specify that $s$ is said to be the arclength.

By the way, it is not clear to me whether the answer is expected in terms of x and z or in terms of s.

I'd like the answer as a function of $s$, as I'll have to integrate w.r.t $s$.

 

[haruspex]

I think I may have misunderstood the question...
Is f a curve in the XZ plane or a function of x and z?

 

[Orodruin]

What is $f$?

 

[member 428835]

Oh shoot, thank you both for responding! Okay, by asking the question you've given me the answer! I need to compute $\int_s \partial_n\phi(x,z)$ along a curve in the $x-z$ plane, where $x = x(s)$ and $z=z(s)$. Since I have an explicit formula for $\phi(x,z)$, all I need is $\nabla \phi \cdot \hat n$, where $\hat n$ is above in post 1.