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.

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.
