Question

Find the unit normal vector to the surface $4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z=40$ at the point $\left( 2,-1,2 \right)$.

Answer 1

Hint: We use the fact that gradient of the surface at a given point is the normal vector for that surface. So, we first find the normal vector for the given surface. Once we find the normal vector, we convert the normal vector to a unit vector (vector of magnitude ‘1’) to get the required result.

Complete step by step answer:
We have the equation of the surface given as $4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z=40$. We need to find the unit normal vector at the point $\left( 2,-1,2 \right)$ for the given surface.
Let us assume the equation of the surface be $f(x,y,z)=0\Leftrightarrow 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40=0$.
We first find the normal vector for the given surface f. We know that the gradient of the surface ‘f’ at a given point $\left( x,y,z \right)$ is vector normal to the surface.
The gradient of the surface is defined as ${{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)}}={{\left. \left( \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z} \right) \right|}_{\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)}}$.
Now we find the gradient of the surface $f(x,y,z)=4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40$.
Normal vector of the surface is ${{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z} \right) \right|}_{\left( 2,-1,2 \right)}}$.
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \dfrac{\partial \left( 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40 \right)}{\partial x},\dfrac{\partial \left( 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40 \right)}{\partial y},\dfrac{\partial \left( 4x{{z}^{3}}-3{{x}^{2}}{{y}^{2}}z-40 \right)}{\partial z} \right) \right|}_{\left( 2,-1,2 \right)}}\].
We know that while doing partial derivative w.r.t x $\left( \dfrac{\partial }{\partial x} \right)$, we take all other terms as constant and apply the derivative only for ‘x’.
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \left( 4{{z}^{3}}-6x{{y}^{2}}z-0 \right),\left( 0-6{{x}^{2}}yz-0 \right),\left( 12x{{z}^{2}}-3{{x}^{2}}{{y}^{2}}-0 \right) \right) \right|}_{\left( 2,-1,2 \right)}}\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}={{\left. \left( \left( 4{{z}^{3}}-6x{{y}^{2}}z \right),\left( -6{{x}^{2}}yz \right),\left( 12x{{z}^{2}}-3{{x}^{2}}{{y}^{2}} \right) \right) \right|}_{\left( 2,-1,2 \right)}}\].
Normal vector of the surface $f(x,y,z)$ is\[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( \left( 4{{\left( 2 \right)}^{3}}-6\left( 2 \right){{\left( -1 \right)}^{2}}\left( 2 \right) \right),\left( -6{{\left( 2 \right)}^{2}}\left( -1 \right)\left( 2 \right) \right),\left( 12\left( 2 \right){{\left( 2 \right)}^{2}}-3{{\left( 2 \right)}^{2}}{{\left( -1 \right)}^{2}} \right) \right)\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( \left( 4.\left( 8 \right)-6.\left( 2 \right).\left( 1 \right).\left( 2 \right) \right),\left( -6.\left( 4 \right).\left( -1 \right).\left( 2 \right) \right),\left( 12.\left( 2 \right).\left( 4 \right)-3.\left( 4 \right).\left( 1 \right) \right) \right)\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( \left( 32-24 \right),48,\left( 96-12 \right) \right)\].
Normal vector of the surface $f(x,y,z)$ is \[{{\left. grad\left( f\left( x,y,z \right) \right) \right|}_{\left( 2,-1,2 \right)}}=\left( 8,48,84 \right)\] ---(1).
We know that the unit vector of any given vector $\left( x,y,z \right)$ is $\pm \dfrac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}\times \left( x,y,z \right)$. Using this we find the unit normal vector for the surface $f(x,y,z)$.

So, unit normal vector of the surface $f(x,y,z)$ is $\pm \dfrac{1}{\sqrt{{{8}^{2}}+{{48}^{2}}+{{84}^{2}}}}\times \left( 8,48,84 \right)$.

Note: We can verify that the magnitude of the obtained unit normal vector is ‘1’. We can see that the normal vector and unit normal vectors are parallel to each other having different magnitudes. We should not write a normal vector as a unit normal vector as its magnitude is not ‘1’.