A smooth function f satisfies $\left|\operatorname{ grad}\ f \right|=1$ ,then the integral curves of $\operatorname{grad}\ f$ are geodesics $M$ is riemannian manifold, if a smooth function $f$ satisfies $\left| \operatorname{grad}\ f \right|=1,$ then prove the integral curves of $\operatorname{grad}\ f$ are geodesics.
 A: Well $\text{grad}(f)$ is a vector such that $g(\text{grad}(f),-)=df$, therefore  integral curves satisfy
$$
\gamma'=\text{grad}(f)\Rightarrow
g(\gamma',X)=df(X)=X(f)
$$
Now let $X,Y$ be a vector fields
$$
XYf=Xg(\text{grad}(f),Y)=
g(\nabla_X\text{grad}(f),Y)+g(\text{grad}(f),\nabla_XY)=
g(\nabla_X\text{grad}(f),Y)+\nabla_XY(f)
$$
and
$$
YXf=Yg(\text{grad}(f),X)=
g(\nabla_Y\text{grad}(f),X)+g(\text{grad}(f),\nabla_YX)=
g(\nabla_Y\text{grad}(f),X)+\nabla_YX(f)
$$
which gives after subtraction and vanishing torsion
$$
[X,Y]f-\nabla_XY(f)+\nabla_YX(f)=0=g(\nabla_X\text{grad}(f),Y)-g(\nabla_Y\text{grad}(f),X)
$$
It follows that
$$
g(\nabla_X\text{grad}(f),Y)=g(\nabla_Y\text{grad}(f),X)
$$
Now the easy part, substitute $X=\text{grad}(f)$ and conclude that for every $Y$
$$
g(\nabla_{\text{grad}(f)}\text{grad}(f),Y)=g(\nabla_Y\text{grad}(f),\text{grad}(f))=0
$$
The last one because $g(\text{grad}(f),\text{grad}(f))=1$ is constant, so
$$
0=Yg(\text{grad}(f),\text{grad}(f))=2g(\nabla_Y\text{grad}(f),\text{grad}(f))
$$
A: I'll use $\nabla$ for the gradient. 
If $|\nabla f| = 1$, we have that $g(\nabla f,\nabla f) = 1$ where $g$ is the metric. Taking the covariant derivatve of the expression you have
$$ 0 = \nabla (1) = \nabla\left( g(\nabla f,\nabla f)\right) = 2 g(\nabla f, \nabla^2 f) = 2 \nabla_{\nabla f} (\nabla f) $$
The third equality used that $\nabla g = 0$ for the Levi-Civita connection of a Riemannian metric, and the fourth inequality uses that the Hessian of a scalar function is symmetric. 
Since $\nabla_{\nabla f} \nabla f = 0$, we have that the vector field $\nabla f$ is geodesic, and hence the integral curves are geodesic curves. 
A: Let $Y$ be any vector field.
$$\Bbb{grad}f=(df)^\sharp, g(\Bbb{grad}f,Y)=df(Y)$$
Since $df$ is closed, $$df[\Bbb{grad}f,Y]=\Bbb{grad}f(df(Y))-Y(df(\Bbb{grad}f))$$
$$=\Bbb{grad}f(df(Y))-Yg(\Bbb{grad}f,\Bbb{grad}f)$$
$$=\Bbb{grad}f(df(Y))-Y(1)=\Bbb{grad}f(df(Y))$$
Torsion tensor vanishes identically, hence $$[\Bbb{grad}f,Y]=\nabla _{\Bbb{grad}f}Y-\nabla_Y\Bbb{grad}f$$
Then $$g(\nabla _{\Bbb{grad}f}\Bbb{grad}f,Y)=(\Bbb{grad}f)g(\Bbb{grad}f,Y)-g(\Bbb{grad}f,\nabla_{\Bbb{grad}f}Y)$$
$$=(\Bbb{grad}f)[(df)Y]-(df)(\nabla_{\Bbb{grad}f}Y)$$
$$=(\Bbb{grad}f)[(df)Y]-(df)([\Bbb{grad}f,Y]+\nabla_Y\Bbb{grad}f)$$
$$=(\Bbb{grad}f)[(df)Y]-(\Bbb{grad}f)[(df)Y]+(df)\nabla_Y\Bbb{grad}f$$
$$=g(\Bbb{grad}f,\nabla_Y\Bbb{grad}f)=\frac{1}{2}\nabla_Yg(\Bbb{grad}f,\Bbb{grad}f)=0$$
Therefore $$\nabla _{\Bbb{grad}f}\Bbb{grad}f=0$$
Result follows.
