Question

Consider the surface $xyz=30$. How do you find the unit normal vector to the surface at (2, 5, 3)?

Answer 1

Hint: First of all find the vector perpendicular to the surface $xyz=30$ at (2, 5, 3) by finding its gradient at this point. Gradient is given as $\nabla f=\dfrac{\partial f}{\partial x}\hat{i}+\dfrac{\partial f}{\partial y}\hat{j}+\dfrac{\partial f}{\partial z}\hat{k}$ where $f=xyz-30$. Now, after finding the normal vector in the form $a\hat{i}+b\hat{j}+c\hat{k}$ by substituting the coordinates divide it by its magnitude to get the required unit normal vector. Magnitude of the vector is given as $\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$.

Complete step by step answer:
Here we have been provided with the surface $xyz=30$ and we are asked to find the unit normal vector to this surface at the point (2, 5, 3).
Now, we can write $xyz=30$ as $xyz-30=0$, so assuming $xyz-30=f$ we have the surface of the form $f\left( x,y,z \right)$. To find the normal vector to this surface first we have to find the gradient of this function. Gradient of a function $f$ is given as $\nabla f=\dfrac{\partial f}{\partial x}\hat{i}+\dfrac{\partial f}{\partial y}\hat{j}+\dfrac{\partial f}{\partial z}\hat{k}$. So we get,
$\Rightarrow \nabla f=\dfrac{\partial \left( xyz-30 \right)}{\partial x}\hat{i}+\dfrac{\partial \left( xyz-30 \right)}{\partial y}\hat{j}+\dfrac{\partial \left( xyz-30 \right)}{\partial z}\hat{k}$
In the first term we have to partially differentiate the function with respect to $x$ while keeping $yz$ as the constant. Similarly, in the second term we will have $xz$ as the constant and in the third term $xy$ as the constant while partially differentiating with respect to $y$ and $z$ respectively. Also, the partial derivative of a constant term is 0, so keeping these points in mind we get,
$\Rightarrow \nabla f=yz\hat{i}+xz\hat{j}+xy\hat{k}$
Now, we have to substitute the given coordinates to find the vector at that particular point.
$\begin{align}
  & \Rightarrow \nabla f\text{ at }\left( 2,3,5 \right)=\left( 5\times 3 \right)\hat{i}+\left( 2\times 3 \right)\hat{j}+\left( 2\times 5 \right)\hat{k} \\
 & \Rightarrow \nabla f\text{ at }\left( 2,3,5 \right)=15\hat{i}+6\hat{j}+10\hat{k} \\
\end{align}$
Let us assume this normal vector as $\vec{n}$so we have,
\[\Rightarrow \vec{n}=15\hat{i}+6\hat{j}+10\hat{k}\]
We know that unit vector is given as \[\hat{n}=\dfrac{{\vec{n}}}{n}\] where $n$ is the magnitude of $\vec{n}$. The magnitude of a vector $a\hat{i}+b\hat{j}+c\hat{k}$ is given as $\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$, so we have,
\[\begin{align}
  & \Rightarrow \hat{n}=\dfrac{15\hat{i}+6\hat{j}+10\hat{k}}{\sqrt{{{15}^{2}}+{{6}^{2}}+{{10}^{2}}}} \\
 & \Rightarrow \hat{n}=\dfrac{15\hat{i}+6\hat{j}+10\hat{k}}{\sqrt{361}} \\
 & \therefore \hat{n}=\dfrac{1}{19}\left( 15\hat{i}+6\hat{j}+10\hat{k} \right) \\
\end{align}\]

Note: You must know about the symbol of partial derivative and the process to calculate it. Just like the derivative of a constant is 0 the partial derivative of a constant is also 0. This question is of the topic ‘vector calculus’ and generally we study them in engineering mathematics. Remember the process to find the unit vector of a given vector.