Given the one-form $a= xy^2 dx − \sqrt{ yz}dy + (x + z)^{2/3} dz$, and $p=(2,-1,3)$ find $α_p(v_p)$ where $(i)v_p=(2,-2,1)$, $(ii)v_p=2\frac{\partial}{\partial x} − 3\frac{\partial}{\partial y} + 4\frac{\partial}{\partial z}$
$(i), a_p(v_p)=( 2(-1)^2 dx − (-1)1dy + (2 + 3)^{2/3} dz)(2,-2,1)=4dx-2dy+5^{2/3}dz$
$(ii), a_p(v_p)=( 2(-1)^2 dx − (-1)1dy + (2 + 3)^{2/3} dz)(2\frac{\partial}{\partial x} − 3\frac{\partial}{\partial y} + 4\frac{\partial}{\partial z})=4-3+4\cdot 5^{2/3}$
Am I doing this correct ?
We have the one form $$\Omega^1(\mathbb R^3)\ni\alpha(x,y,z)=xy^2dx-\sqrt{yz} dy+(x+z)^{2/3}dz$$ and the tangent vector of $T_p\mathbb R^3$ is $$v_p=2\partial_x\vert_p-2\partial_y\vert_p+\partial_z\vert_p$$ Remember that if we call $x^1,x^2,x^3$ the coordinates of the real space, we have $dx^j(\partial_{x^i})=\begin{cases}1&i=j\\0&i\ne j \end{cases},\text{ for }i,j=1,2,3$. $$\alpha(v_p)=(xy^2dx-\sqrt{yz}dy+(x+z)^{2/3}dz)(v_p)=\\xy^2dx(v_p)-\sqrt{yz}dy(v_p)+(x+z)^{2/3}dz(v_p)=\\xy^2dx(2\partial_x\vert_p-2\partial_y\vert_p+\partial_z\vert_p)-\sqrt{yz}dy(2\partial_x\vert_p-2\partial_y\vert_p+\partial_z\vert_p)+(x+z)^{2/3}dz(2\partial_x\vert_p-2\partial_y\vert_p+\partial_z\vert_p)=\\2xy^2-2 \sqrt{yz}+(x+z)^{2/3}$$ and you can now evaluate this in $p$. Are you sure the point is $(2,-1,3)?$